text:
\begin{document}

\title[]{A note on restriction of Characters of Alternating groups to Sylow
Subgroups}

\author[Eugenio Giannelli]{Eugenio Giannelli}
\address[E. Giannelli]{Trinity Hall, University of Cambridge, Trinity Lane, CB21TJ, UK}
\email{eg513@cam.ac.uk}

\thanks{The first author's research was funded by Trinity Hall, University of Cambridge.}

\begin{abstract}
We restrict irreducible characters of alternating groups
of degree
divisible by $p$  to
their Sylow $p$-subgroups and study the number
of linear constituents.
\end{abstract}

\maketitle

\section{Introduction}   \label{sec:intro}

Let $p$ be a prime number and let $G$ be a finite group. We denote by $\mathrm{Irr}(G)$ the set of ordinary irreducible characters of $G$ and we let $P$ be a Sylow $p$-subgroup of $G$.
It seems a very natural problem to study the restriction to $P$ of irreducible characters of $G$. Our interest in this line of research was initially motivated by the long-standing McKay Conjecture (see \cite{McKay} for the original statement and/or \cite[Section 2]{Malle} for a complete survey of the state of the art on this problem).
For some special classes of groups it has been possible to show that the restrictions to a Sylow $p$-subgroup of irreducible characters of degree coprime to $p$ admit a unique linear (i.e. coprime to $p$) constituent. This allowed to show stronger forms of the McKay conjecture. This happens for example in \cite{G} and \cite{N}.

In \cite{GN} we studied the number of linear constituents of the restricted character $\chi_P$ for any $\chi\in\mathrm{Irr}(G)$ such that $p$ divides $\chi(1)$.
As a consequence of the evidences collected,
the following conjecture was proposed.

 \begin{conjD}
 Suppose that $\chi \in \irr G$ has degree divisible by $p$, and let $P \in \syl pG$.
 If $\chi_P$ has a linear constituent, then $\chi_P$ has at least $p$ different linear constituents.
 \end{conjD}

The above statement has been verified for various classes of groups in \cite{GN}. In particular, when $G$ is the symmetric group $\fS_n$ it is shown that the restriction to a Sylow $p$-subgroup of any irreducible character of degree divisible by $p$ has at least $p$ distinct linear constituents.

The aim of this note is to prove that the above statement holds for alternating groups.
In particular for any natural number $n$ we let $\fA_n$ be the alternating group of degree $n$ and we denote by $Q_n$ a Sylow $p$-subgroup of $\fA_n$.
The main result of this note is the following.

\begin{thmA}
Let $n$ be a natural number and let $p$ be a prime.
If $\chi \in {\rm Irr}(\fA_n)$  has
degree divisible by $p$, then
the restriction $\chi_{Q_n}$ has at least $p$ different
linear constituents.
\end{thmA}

When $p$ is an odd prime, Theorem A follows quite directly from the corresponding statement for symmetric groups (see Section \ref{sec: 3.1} below).
On the other hand in order to prove Theorem A for $p=2$, we will need to use some new algebraic and combinatorial ideas (this is done in Section \ref{sec: 3.2}).

 \section{Background}

We start by recalling some very basic combinatorial definitions and notation in the framework of the representation theory of symmetric groups. We refer the reader to \cite{James} or \cite{OlssonBook} for a more detailed account.

A composition $\lambda=(\lambda_1,\lambda_2,\dots ,\lambda_s)$ is a finite  sequence of positive integers. We say that $\lambda_i$ is a part of $\lambda$ and that $\lambda$ is a composition of $|\lambda|=\sum\lambda_i$. We say that $\lambda$ is a partition if its parts are non-increasingly ordered and we denote by $\lambda'$ the conjugate partition of $\lambda$. The Young diagram associated to $\lambda$ is the set $[\lambda]:=\{(i,j)\in{\mathbb N}\times{\mathbb N}\mid 1\leq i\leq s,1\leq j\leq\lambda_i\}$.
For any natural number $n$ we denote by $\mathcal{P}(n)$ (respectively $\mathcal{C}(n)$) the set of partitions (respectively compositions) of $n$.
We will sometimes write $\lambda\vdash n$ if $\lambda\in\mathcal{P}(n)$.

Following the notation used in \cite{OlssonBook}, given $(i,j)\in [\lambda]$ we denote by $H_{i,j}(\lambda)$ the corresponding $(i,j)$-hook. Moreover we let $h_{i,j}(\lambda)=|H_{i,j}(\lambda)|$. Finally, for any natural number $e$ we denote by $\mathcal{H}_e(\lambda)$ the subset of $[\lambda]$ defined by
$$\mathcal{H}_e(\lambda)=\{(i,j)\in [\lambda]\ :\ e\mid h_{i,j}(\lambda)\}.$$
Given an $e$-hook $h:=H_{i,j}(\lambda)$ we denote by $\lambda - h$ the partition of $n-e$ obtained from $\lambda$ by removing its $(i,j)$-rim hook (we refer the reader to \cite{OlssonBook} for the formal definition).
We denote by $C_e(\lambda)$ and $Q_e(\lambda)$ the $e$-core and the $e$-quotient of $\lambda$, respectively. Finally we let $w_e(\lambda)$ be the $e$-weight of $\lambda$.

Let $\mu$ and $\lambda$ be partitions. We say that $\mu$ is a \textit{subpartition} of $\lambda$, written $\mu\subseteq\lambda$, if $\mu_i\leq\lambda_i$, for all $i\geq1$. When this occurs, we let

$[\lambda\smallsetminus\mu]=\{(i,j)\in{\mathbb N}\times{\mathbb N}\mid 1\leq i\leq s,\mu_i<j\leq\lambda_i\}.$
This is called a skew Young diagram.

\subsection{Characters of $\fS_n$ and $\fA_n$}
Irreducible characters of $\fS_n$ are naturally labelled by partitions of $n$. Given $\lambda\in\mathcal{P}(n)$ we denote by $\chi^\lambda$ the corresponding element of $\mathrm{Irr}(\fS_n)$.
If $\lambda\neq \lambda'$, then we denote by $\phi^\lambda$ the irreducible character of $\mathfrak{A}_n$ defined by $\phi^\lambda=(\chi^\lambda)_{\mathfrak{A}_n}$. On the other hand, if $\lambda=\lambda'$ then we let $\phi^\lambda_+$ and $\phi^\lambda_-$ be the two irreducible characters of $\mathfrak{A}_n$ such that $(\chi^\lambda)_{\mathfrak{A}_n}=\phi^\lambda_++\phi^\lambda_-$. Clearly we have that $(\phi_+^\lambda)^{\sigma}=\phi^\lambda_-$, for any $\sigma\in\fS_n\smallsetminus\fA_n$.

\medskip

A key ingredient in our proof will be a rather sophisticated use of the Littlewood-Richardson rule (see \cite[Chapter 16]{James}). For the convenience of the reader we recall this here.

\begin{defn}
Let ${\mathcal A}=a_1,\dots,a_k$ be a sequence of positive integers. The type of $\mathcal{A}$ is the sequence of non-negative integers $m_1,m_2,\dots$ where $m_i$ is the number of occurrences of $i$ in $a_1,\dots,a_k$. We say that $\mathcal{A}$ is a \textit{reverse lattice sequence} if the type of its prefix $a_1,\dots,a_j$ is a partition, for all $j\geq1$.

Let $\mu\vdash n$ and $\nu\vdash m$. The outer tensor product $\chi^\mu\times\chi^\nu$ is an irreducible character of ${\mathfrak S}_n\times{\mathfrak S}_m$. Inducing this character to ${\mathfrak S}_{n+m}$ we may write
$$
(\chi^\mu\times\chi^\nu)^{{\mathfrak S}_{n+m}}=\sum_{\lambda\vdash(n+m)}C_{\mu,\nu}^\lambda\chi^\lambda.
$$
The \textit{Littlewood-Richardson rule} asserts that $C_{\mu,\nu}^\lambda$ is zero if $\mu\not\subseteq\lambda$ and otherwise equals the number of ways to replace the nodes of the skew diagram $[\lambda\smallsetminus\mu]$ by natural numbers such that:
the numbers are weakly increasing along rows;
the numbers are strictly increasing down the columns;
the sequence obtained by reading the numbers from right to left and top to bottom is a reverse lattice sequence of type $\nu$.
We call any such configuration a \textit{Littlewood-Richardson configuration of type} $\nu$ for $[\lambda\smallsetminus\mu]$.
\end{defn}

The Murnaghan-Nakayama rule (see \cite[2.4.7]{JK}) allows to recursively compute the character table of symmetric groups.
The following fact (see \cite[2.7.33]{JK}) is a very useful consequence of this.

\begin{lem}\label{lem:farahat}
Let $\la\vdash n$, $Q_e(\la)=(\la_0,\ldots,\la_{e-1})$.
Assume that $w\ge w_e(\la)$, and let $\rho \in \fS_n$
be  the product of $w$ $e$-cycles;
let $\gamma \in \fS_n$ act on the fixed points of $\rho$.
Then

$$\chi^{\la}(\rho\gamma)=
\begin{cases}
\pm \binom{w}{|\la_0|,|\la_1|,\ldots,|\la_{e-1}|}\chi^{C_e(\la)}(\gamma)\prod_{i=0}^{e-1} \chi^{\la_i}(1) &  \text{if } w=w_e(\la) \\
 \rm{0} &  \text{if } w>w_e(\la)
\end{cases} \:.$$
\end{lem}

\section{The proof of Theorem A}

Let $n$ be a natural number and let $P_n$ be a Sylow $p$-subgroup of $\fS_n$.
Denote by $Q_n=P_n\cap\fA_n$ the unique Sylow $p$-subgroup of $\fA_n$ contained in $P_n$. Clearly $Q_n\trianglelefteq P_n$. We recall that if $n=a_1p^{n_1}+a_2p^{n_2}+\cdots+a_kp^{n_k}$ is the $p$-adic expansion of $n$ then $P_n=(P_{p^{n_1}})^{\times a_1}\times (P_{p^{n_2}})^{\times a_2}\times\cdots\times (P_{p^{n_k}})^{\times a_k}.$

\subsection{Odd primes}\label{sec: 3.1} Let us start by assuming that $p$ is an odd prime.
In this case we clearly have that $Q_n=P_n$.
If $n=\sum_{j=1}^ka_jp^{n_j}$ is the $p$-adic expansion of $n$ then we denote by $\omega_n$ a chosen element of $P_n$ obtained as the product of $a_j$ cycles of length $p^{n_j}$, for all $j\in\{1,\ldots, k\}$.
The following Lemma is a consequence of \cite[Lemma 3.11]{GN}.

\begin{lem}\label{lem:omega_n}
Let $n, p$ and $\omega_n$ be as above. Then
\begin{itemize}
\item[(i)] $\theta(\omega_n)$ is a $p$-th root of unity for every linear character $\theta$ of $P_{n}$.
\item[(ii)] $\delta(\omega_n)=0$ for all  $\delta\in\mathrm{Irr}(P_{n})$ such that $p\ |\ \delta(1)$.
\end{itemize}
\end{lem}

The following result is a corollary of \cite[Theorem 3.1]{GN}.

\begin{lem}\label{lem:uniquelinear}
Let $\phi\in\mathrm{Irr}(\fA_n)$.
Then the restriction of $\phi$ to $\fA_n$ has a linear constituent.
\end{lem}
\begin{proof}
Let $\lambda$ be a partition of $n$ such that $\phi$ is an irreducible constituent of $(\chi^\lambda)_{\fA_n}$. Let $\theta$ be a linear constituent of $(\chi^\lambda)_{P_n}$ (this exists by  \cite[Theorem 3.1]{GN}).
If $\lambda\neq \lambda'$ then $\theta$ is a constituent of $\phi_{P_n}$.
Suppose that $\lambda=\lambda'$ and (without loss of generality) that $\phi=\phi^\lambda_+$. Let $\sigma\in N_{\fS_n}(P_n)\smallsetminus \fA_n$. Such a $\sigma$ exists (choose for example an element in $N_{\fS_n}(P_n)$ of cycle type given by the product of $p^{k-1}$ cycles of length $p-1$, where $p^k$ is a $p$-adic digit of $n$).
Then if we assume that $\theta$ is not a constituent of $\phi_{P_n}$ we deduce that $\theta$ is a constituent of $(\phi^\lambda_-)_{P_n}$. Hence $\theta^\sigma$ is a linear constituent of $((\phi^\lambda_-)_{P_n})^{\sigma}=((\phi^\lambda_-)^{\sigma})_{P_n}=\phi_{P_n}$.
\end{proof}

We are ready to prove Theorem A of the introduction for odd primes.

\begin{prop}
Let $p$ be an odd prime, let $n$ be a natural number and let $\phi\in\mathrm{Irr}(\fA_n)$ be such that $p$ divides $\phi(1)$. Then the restriction of $\phi$ to $P_n$ has at least $p$
distinct linear constituents.
\end{prop}
\begin{proof}
Let $\lambda$ be a partition of $n$ such that  $\phi$ is an irreducible constituent of $(\chi^\lambda)_{\fA_n}$. If $\lambda\neq\lambda'$ then the statement follows from \cite[Theorem A]{GN}.
Assume now that  $\lambda=\lambda'$ and (without loss of generality) $\phi=\phi^\lambda_+$. Let $n=\sum_{j=0}^ka_jp^j$ be the $p$-adic expansion of $n$ and let
$\omega_n\in P_{n}$ be the product of $a_j$ $p^j$-cycles for $j\in\{0,1,\ldots, k\}$.
Since $p$ divides $\phi(1)$ we have that $p$ divides $\chi^\lambda(1)$ and therefore that $\chi^\lambda(\omega_n)=0$. This follows from the description of irreducible characters of symmetric groups whose degree is not divisible by $p$ (see for instance \cite{Mac} or \cite[Proposition 6.4]{OlssonBook}). Moreover, by \cite[Theorem 2.5.13]{JK} we obtain that $\phi(\omega_n)=\frac{\chi^\lambda(g)}{2}=0$. Suppose for a contradiction that $\phi_{P_n}$ has $\ell$ distinct linear constituents $\theta_1, \ldots, \theta_\ell$, for some $1\leq \ell<p$. From Lemma \ref{lem:omega_n} we deduce that
$$0=\phi(\omega_n)=c_1\theta_1(\omega_n)+\cdots c_\ell\theta_\ell(\omega_n)=c_1\zeta_1+\cdots c_\ell\zeta_\ell,$$
for some $p$-th roots of unity $\zeta_1, \ldots, \zeta_\ell$ and some positive natural numbers $c_1, \ldots, c_\ell$.
This is a contradiction, since no $\mathbb{N}$-linear combination of $\zeta_1, \ldots, \zeta_\ell$ can equal $0$.

\end{proof}

\subsection{The prime $2$}\label{sec: 3.2}
To deal with the remaining $p=2$ case, we need to fix some more precise notation.
Let $n\in\mathbb{N}$, for $k\in\{0,1,\ldots, n-1\}$ we let $g_k$ be the element of $\fS_{2^n}$ defined by $$g_k=\prod_{i=1}^{2^k}(i, i+2^k).$$
It is not too difficult to see that $P_{2^n}:=\left\langle g_0,\ldots, g_{n-1}\right\rangle$ is a Sylow $2$-subgroup of $\fS_{2^n}$. Moreover the element $\gamma_n\in P_{2^n}$ defined by $\gamma_n=g_1g_2\cdots g_{n-1}$ has cycle type $(2^{n-1}, 2^{n-1})$. In particular $\gamma_n\in\fA_{2^n}\cap P_{2^n}=Q_{2^n}$.
Similarly, the element $\omega_n\in P_{2^n}$ defined by $\omega_n=g_0\gamma_n$ is a $2^n$-cycle.

\begin{lem}\label{lem:2-root}
Let $\gamma_n$ and $\omega_n$ be the elements defined above and let $g\in\{\gamma_n,\omega_n\}$. Then:
\begin{itemize}
\item[(i)] $\theta(g)\in\{-1,+1\}$ for every linear character $\theta$ of $P_{2^n}$.
\item[(ii)] $\delta(g)=0$ for all $\delta\in\mathrm{Irr}(P_{2^n})$ such that $\delta(1)$ is even.
\end{itemize}
\end{lem}
\begin{proof}
Let $g=\gamma_n$. It is well known that $P_{2^n}=P_{2^{n-1}}\wr C_2=(P_{2^{n-1}}\times P_{2^{n-1}})\rtimes C_2$. Let $\sigma$ be a generator of the top group $C_2$, acting on the base group by swapping the two direct factors. From this point of view we observe that $\gamma_n=(\gamma_{n-1}, 1; \sigma)$. To prove (i) we now proceed by induction on $n$. If $n=1$, the result is obviously true. Hence assume that $n\geq 2$ and let $\theta$ be a linear character of $P_{2^n}$. By Gallagher's Corollary (see \cite[6.17]{I}) we know that $\theta$ is the extension by an irreducible character $\psi$ of $C_2$ of an irreducible character $\phi\times\phi\in\mathrm{Irr}(P_{2^{n-1}}\times P_{2^{n-1}})$, for some linear character $\phi$ of $P_{2^{n-1}}$. Using \cite[Lemma 4.3.9]{JK}, it follows that $\theta(\gamma_n)=\phi(\gamma_{n-1})\psi(\sigma)$. Using the inductive hypothesis we conclude that $\phi(\gamma_{n-1})\in\{-1,+1\}$ and therefore that $\theta(\gamma_n)=\pm 1$.

The proof of case (ii) is again done by induction on $n$. We observe that if $\delta$ is an irreducible character of $P_{2^n}$ of even degree then either $\delta=(\phi_1\times\phi_2)^{P_{2^n}}$, for some $\phi_1,\phi_2\in\mathrm{Irr}(P_{2^{n-1}}\times P_{2^{n-1}})$ such that $\phi_1\neq\phi_2$;
or otherwise $\delta$  is the extension by an irreducible character $\psi$ of $C_2$ of an irreducible character $\phi\times\phi\in\mathrm{Irr}(P_{2^{n-1}}\times P_{2^{n-1}})$, for some even-degree character $\phi$ of $P_{2^{n-1}}$.
In the first case we have that $\delta(\gamma_n)=0$ because $\gamma_n\notin P_{2^{n-1}}\times P_{2^{n-1}}$. In the second case we can write again $\delta(\gamma_n)=\phi(\gamma_{n-1})\psi(\sigma)=0$, by induction.
If $g=\omega_n$ then the result follows from \cite[Lemma 3.11]{GN}.
\end{proof}

It is useful to recall here the characterization of those partitions of a natural number n labelling irreducible characters of odd degree of symmetric groups. This was first observed in \cite{Mac}. As usual we adopt the symbol $\mathrm{Irr}_{2'}(G)$ to denote the set of irreducible characters of the finite group $G$ whose degree is odd. Moreover, we let $\mathrm{Lin}(G)$ be the notation for the set of linear characters of $G$.

\begin{thm}\label{thm:oddMac}
Let $n$ be a natural number with binary expansion $n=2^{n_1}+2^{n_2}+\cdots+2^{n_t}$, for some $n_1>\cdots >n_t\geq 0$ and let $\lambda\in\mathcal{P}(n)$. Then, $\chi^{\lambda}\in\mathrm{Irr}_{2'}(\fS_{n})$ if and only if there exists a unique removable $2^{n_1}$-hook $h_1$ in $\lambda$ and  $\chi^{(\lambda - h_1)}\in\mathrm{Irr}_{2'}(\fS_{n-2^{n_1}})$.
\end{thm}

For any $n\in\mathbb{N}$ we denote by $\mathcal{L}(n)$ the subset of $\mathcal{P}(n)$ consisting of \textit{hook partitions} (i.e. partitions of the form $(n-x, 1^x)$ for $0\leq x\leq n-1$). An useful consequence of Theorem \ref{thm:oddMac} is that $\chi^\lambda\in\mathrm{Irr}_{2'}(\fS_{2^n})$ if and only if $\lambda\in\mathcal{L}(n)$.

In this article we will need a little bit of control on the maximal power of $2$ dividing the degrees of irreducible characters. We say that $\nu_2(n)=a$ if $n=2^am$, for some odd $m\in\mathbb{N}$.
The theory of $2$-core and $2$-quotient towers, introduced in \cite{Mac} and fully developed and beautifully explained in \cite{OlssonBook}, implies the following fact.

\begin{lem}\label{lem: towers8}
Let $n$ be natural number and let $\lambda\in\mathcal{P}(2^n)$. Then:
\begin{itemize}
\item[(i)] $\nu_2(\chi^\lambda(1))=1$ if and only if $\mathcal{H}_{2^n}(\lambda)=\emptyset$ and $|\mathcal{H}_{2^{n-1}}(\lambda)|=2$.
\item[(ii)] $\nu_2(\chi^\lambda(1))\geq 2$ if and only if $|\mathcal{H}_{2^{n-1}}(\lambda)|\leq 1$.
\end{itemize}
\end{lem}

The Murnaghan-Nakayama formula together with Theorem \ref{thm:oddMac} imply the following fact.

\begin{cor}\label{cor:g lambda}
Let $n$ be a natural number with binary expansion $n=2^{n_1}+\cdots+2^{n_t}+a_02^0,$ for some $t\geq 2$,  $n_1>\cdots >n_t> 0$ and $a_0\in\{0,1\}$.
Let $\lambda\in\mathcal{P}(n)$ be such that $\chi^\lambda(1)$ is even.
Then there exists $g_\lambda\in Q_n$ such that the following hold.

\medskip

\begin{itemize}
\item[(i)] $\theta(g_\lambda)\in\{-1,+1\}$ for every linear character $\theta$ of $P_n$.
\item[(ii)] $\delta(g_\lambda)=0$ for all  $\delta\in\mathrm{Irr}(P_n)$ such that $\delta(1)$ is even.
\item[(iii)] $\chi^\lambda(g_\lambda)=0$.
\end{itemize}
\end{cor}

\begin{proof}
Since $\chi^\lambda(1)$ is even, Theorem \ref{thm:oddMac} implies that either $\mathrm{core}_{2^{n_1}}(\lambda)=\lambda$ (i.e. $\lambda$ does not have $2^{n_1}$-hooks) or
that there exists $r\in\{2,\ldots, t\}$ such that
$$|\mathrm{core}_{2^{n_{j}}}(\lambda)|=|\mathrm{core}_{2^{n_{j-1}}}(\lambda)|-2^{n_j},\ \ \text{for all}\ \ j\in\{2, \ldots, r-1\},$$ and such that
$\mathrm{core}_{2^{n_{r}}}(\lambda)=\mathrm{core}_{2^{n_{r-1}}}(\lambda)$ (i.e. $\mathrm{core}_{2^{n_{r-1}}}(\lambda)$ does not have $2^{n_r}$-hooks).

If $t$ is even, then we let $g_\lambda\in Q_n$ be any element of cycle type $(2^{n_1},2^{n_2},\ldots, 2^{n_t})$.
Statements (i) and (ii) now follow from Lemma \ref{lem:2-root}. Moreover, statement (iii) is a consequence of the Murnaghan-Nakayama rule used together with Theorem \ref{thm:oddMac}.

If $t$ is odd and $\mathrm{core}_{2^{n_1}}(\lambda)=\lambda$ then we let $g_\lambda\in Q_n$ be an element of cycle type $(2^{n_1},2^{n_2-1}, 2^{n_2-1},2^{n_3},\ldots, 2^{n_t})$.
Notice that we can always find an element of this form in $Q_n$ because $t\geq 3$.
On the other hand, if $t$ is odd but $\mathrm{core}_{2^{n_1}}(\lambda)\neq\lambda$,
then we choose $g_\lambda\in Q_n$ to be an element of cycle type $(2^{n_1-1},2^{n_1-1}, 2^{n_2},\ldots, 2^{n_r},\ldots,  2^{n_t})$.
Statements (i) and (ii) now follow from Lemma \ref{lem:2-root}. Statement (iii) is again a consequence of the Murnaghan-Nakayama rule used together with Theorem \ref{thm:oddMac}.
\end{proof}

\begin{lem}\label{lem:lin1}
Let $n$ be a natural number, let $\chi\in\mathrm{Irr}(\fS_n)$ and let $\phi\in\mathrm{Irr}(\fA_n)$ be an irreducible constituent of $\chi_{\fA_n}$. Let $\theta\in\mathrm{Lin}(P_n)$ be a constituent of $\chi_{P_n}$. Then $\theta_{Q_n}$ is a constituent of $\phi_{Q_n}$.
\end{lem}

\begin{proof}
Let $\lambda\in\mathcal{P}(n)$ be such that $\chi=\chi^\lambda$.
If $\lambda\neq \lambda'$ then $\phi=\chi_{\fA_n}$ and the statement clearly holds.
Suppose that $\lambda=\lambda'$ and assume (without loss of generality) that $\phi=\phi^\lambda_+$. Let $\sigma\in P_n\smallsetminus Q_n$. Clearly $\sigma$ acts by conjugation on $\mathrm{Irr}(Q_n)$ since $Q_n\trianglelefteq P_n$. Moreover we have that $\phi^\sigma=\phi^\lambda_-$, since $\sigma\in\fS_n\smallsetminus\fA_n$.
Therefore we have that $(\phi_{Q_n})^\sigma=(\phi^\sigma)_{Q_n}=(\phi^\lambda_-)_{Q_n}$.
Since $(\theta_{Q_n})^\sigma=(\theta^\sigma)_{Q_n}=\theta_{Q_n}$, we deduce that $\theta_{Q_n}$ is a constituent of both $\phi_{Q_n}$ and $(\phi^\sigma)_{Q_n}$.
\end{proof}

We are now in the position to prove Theorem A of the introduction for the prime $2$ and for all natural numbers having at least $2$ even binary digits.

\begin{prop}\label{prop:n not power of 2}
Let $n\in\mathbb{N}$ be such that $n\notin\{2^k, 2^k+1\ |\ k\in\mathbb{N}_0\}$ and let $\phi\in\mathrm{Irr}(\fA_n)$ be such that $\phi(1)$ is even. Then the restriction of $\phi$ to $P_n$ has at least $2$
distinct linear constituents.
\end{prop}
\begin{proof}
Let $\lambda$ be a partition of $n$ such that $\phi$ is an irreducible constituent of $(\chi^\lambda)_{\fA_n}$. Observe that $\chi^\lambda(1)$ is even and let $\theta_1,\ldots \theta_\ell$ be the distinct linear constituents of $(\chi^\lambda)_{P_n}$. By \cite[Theorem A]{GN} we know that $\ell\geq 2$. Suppose for a contradiction that $(\theta_j)_{Q_n}=(\theta_1)_{Q_n}$ for all $j\in\{2,\ldots, \ell\}$. Then using Corollary \ref{cor:g lambda} we could pick $g\in Q_n$ such that $$0=\chi^\lambda(g)=m\theta_1(g)=\pm m,$$ where $m$ is the positive number equal to the sum of the multiplicities of the linear characters in the decomposition of $(\chi^\lambda)_{P_n}$ into irreducible constituents. This is clearly a contradiction.
Hence we always have two linear constituents that do not coincide on $Q_n$.

The statement now follows from Lemma \ref{lem:lin1}.
\end{proof}

We are left to deal with the case where $n\in\{2^k, 2^k+1\}$ for some $k\in\mathbb{N}_0$.
In order to prove Theorem A in this final specific case we must use a more combinatorial approach.
In particular we will show that the statement holds for $n=2^k$ and derive the case $n=2^k+1$ as a corollary.

\begin{prop}\label{prop:4divides}
Let $k\in\mathbb{N}_0$, let $n=2^k+\varepsilon$ for some $\varepsilon\in\{0,1\}$, let $\lambda\in\mathcal{P}(n)$ be such that $4$ divides $\chi^\lambda(1)$ and let $\phi$ be an irreducible constituent of $(\chi^\lambda)_{\fA_n}$. Then $\phi_{Q_n}$ has at least two distinct linear constituents.
\end{prop}
\begin{proof}
Let $\gamma_k\in Q_n$ be the element of cycle type $(2^{k-1},2^{k-1}, \varepsilon)$ defined before Lemma \ref{lem:2-root}.
Lemma \ref{lem: towers8} shows that $\mathcal{H}_{2^k}(\lambda)=\emptyset$ and that $|\mathcal{H}_{2^{k-1}}(\lambda)|\leq 1$. This implies that $\chi^\lambda(\gamma_k)=0$. Suppose now for a contradiction that all linear constituents of $(\chi^\lambda)_{P_n}$ coincide on $Q_n$. Evaluation of $(\chi^\lambda)_{P_n}$ at $\gamma_k$ leads to a contradiction. The statement now follows from Lemma \ref{lem:lin1}.
\end{proof}

Proposition \ref{prop:4divides} shows that we just need to study the case where $\phi\in\mathrm{Irr}(\fA_n)$ is an even-degree constituent of $(\chi^\lambda)_{\fA_n}$ for some partition $\lambda$ such that $\nu_2(\chi^\lambda(1))=1$.
In particular, since $\phi(1)$ is even we deduce that $\lambda\neq\lambda'$.

\begin{lem}\label{lem: 2linears}
Let $\lambda\in\mathcal{P}(2^k)$ be such that $\nu_2(\chi^\lambda(1))=1$ and suppose that $(\chi^\lambda)_{Q_{2^k}}$ has a unique linear constituent $\theta$.
Then $(\chi^\lambda)_{Q_{2^k}}=2\theta+\Delta$, where $\Delta$ is a sum (possibly empty) of even degree irreducible characters of $Q_{2^k}$. In particular $(\chi^\lambda)_{P_{2^k}}$ has exactly two linear constituents both appearing with multiplicity $1$.
\end{lem}
\begin{proof}
Let $(\chi^\lambda)_{P_{2^k}}=c_1\theta_1+\cdots+c_\ell\theta_\ell+\Omega$, where $\theta_j\in\mathrm{Lin}(P_{2^k})$, $c_j\in\mathbb{N}$ for all $j\in\{1,\ldots, \ell\}$ and $\Omega$ is a sum (possibly empty) of even degree irreducible characters of $P_{2^k}$. From \cite[Theorem A]{GN} we know that $\ell\geq 2$. From the hypothesis we get that $(\theta_j)_{Q_{2^k}}=\theta$ for all  $j\in\{1,\ldots, \ell\}$. Since $\theta^{P_{2^k}}(1)=2$, it follows that $\ell=2$ and that  $(\chi^\lambda)_{P_{2^k}}=c_1\theta_1+c_2\theta_2+\Omega$.
Since $\nu_2(\chi^\lambda(1))=1$ we deduce from Lemma \ref{lem: towers8} that $\mathcal{H}_{2^k}(\lambda)=\emptyset$ and that $|\mathcal{H}_{2^{k-1}}(\lambda)|=2$.
In particular we have that $C_{2^{k-1}}(\lambda)=\emptyset$, $w_{2^{k-1}}(\lambda)=2$ and $Q_{2^{k-1}}(\lambda)$ is a sequence of $2^{k-1}$ partitions such that all except for two are empty. The two non empty partitions are both equal to $(1)$, the unique partition of $1$.
Using Lemma \ref{lem:farahat} we observe that $\chi^\lambda(\gamma_{k})=\pm 2$.
On the other hand we have that $\chi^\lambda(\gamma_{k})=\pm (c_1+c_2)$, by Lemma \ref{lem:2-root}. We obtain that $c_1=c_2=1$.
This concludes the proof.
\end{proof}

\begin{prop}\label{cor: 2k}
Let $\lambda\in\mathcal{P}(2^k)$ be such that $\nu_2(\chi^\lambda(1))=1$ and that $\lambda\neq\lambda'$. Then $(\chi^\lambda)_{Q_{2^k}}$ has at least two distinct linear constituents.
\end{prop}

The proof of Proposition \ref{cor: 2k} is quite technical. For this reason we decided to postpone it. Section \ref{sec:4} below is entirely devoted to prove this statement.

\begin{cor}\label{cor: 2k+1}
Let $n\in\{2^k, 2^{k}+1\}$, for some $k\in\mathbb{N}$ and let $\phi\in\mathrm{Irr}(\fA_n)$ be such that $\phi(1)$ is even. Then $\phi_{Q_{n}}$ has two distinct linear constituents.
\end{cor}
\begin{proof}
Let $\lambda\in\mathcal{P}(n)$ be such that $\phi$ is an irreducible constituent of $(\chi^\lambda)_{\fA_n}$.
Proposition \ref{prop:4divides} and Proposition \ref{cor: 2k} show that it is enough to consider the case where $n=2^k+1$ and where $\nu_2(\chi^\lambda(1))=1$.
Since $\phi(1)$ is even we immediately get that $\lambda\neq\lambda'$ and hence that $\phi=(\chi^\lambda)_{\fA_n}$.
Moreover, in this situation we have that $\mathcal{H}_{2^k}(\lambda)=\emptyset$ and that $|\mathcal{H}_{2^{k-1}}(\lambda)|=2$. It follows that $\lambda$ is not a hook partition of $2^{k}+1$. Therefore there exists a partition $\mu\in\mathcal{P}(2^k)\smallsetminus\mathcal{L}(2^k)$ such that $\chi^\mu$ is an irreducible constituent of $(\chi^\lambda)_{\fS_{n-1}}$. Since $\mu$ is not a hook partition we have that $\chi^\mu(1)$ is even. Hence $(\chi^\mu)_{Q_{n-1}}$ has two distinct linear constituents by Propositions \ref{prop:4divides} and \ref{cor: 2k}.
Since $Q_n=Q_{n-1}$ we deduce that $(\chi^\mu)_{Q_{n-1}}$ is a constituent of $\phi_{Q_n}$ and the statement follows.
\end{proof}

We conclude by observing that Propositions \ref{prop:n not power of 2} and Corollary \ref{cor: 2k+1} show that Theorem A of the introduction holds.

\section{The proof of Proposition \ref{cor: 2k}}\label{sec:4}

In this section we focus on the proof of Proposition \ref{cor: 2k}. This is certainly the most technical part of the article.
We start by recalling a couple of preliminary results.
The first one was proved in \cite[Theorem 1.1]{G}.

\begin{thm}\label{thm: JLMS}
Let $h\in\mathcal{L}(2^k)$ and let $\chi^h\in\mathrm{Irr}_{2'}(\fS_{2^k})$. Then $(\chi^h)_{P_{2^k}}$ has a unique linear constituent $f(h)$. Moreover, $\chi^h\mapsto f(h)$ is a bijection between $\mathrm{Irr}_{2'}(\fS_{2^k})$ and $\mathrm{Lin}(P_{2^k})$.
\end{thm}

In \cite{G3} it was introduced the concept of $2$-sections of a partition. For the proof of Proposition \ref{cor: 2k} we will only need the following specific instance of that more general definition. The extended concept of $q$-sections of partitions was the key idea to prove Theorem A of \cite{GN} for symmetric groups.

\begin{notation}\label{not: Delta}
Let $n$ be a natural number. Given a composition $\mu$ of $n$ we denote by $\mu^\star$ the partition obtained from $\mu$ by reordering its parts. Let $\lambda$ be a partition of $2n$.
We can uniquely write $\lambda$ as $\lambda=(\omega\circ \varepsilon)^\star$ where $\omega$ is the partition consisting of all the odd parts of $\lambda$ and $\varepsilon$ is the partition consisting of all the even parts of $\lambda$.
(Here the symbol $\circ$ denotes the concatenation of partitions.)
In particular we have that $$\lambda=\big((2k_1+1,\ldots, 2k_t+1)\circ (2r_1,\ldots ,2r_s)\big)^\star,\ \ \ \ \ $$
where $k_1\geq \cdots\geq k_t\geq 0$ and $r_1\geq \cdots\geq r_s> 0$.
Since $\lambda\vdash 2n$ we clearly have that $t=2\zeta$ for some $\zeta\in\mathbb{N}_{0}$.
We let $\Delta^2(\lambda)$ be the partition of $n$ defined by
$$\Delta^2(\lambda)=\big((k_1+1, \ldots, k_\zeta+1,  k_{\zeta+1}, \ldots, k_t)\circ (r_1,\ldots ,r_s)\big)^\star.$$
\end{notation}

In  \cite[Theorem 3.9]{GN} the following was proved.

\begin{thm} \label{thm: Delta}
Let $n$ be a natural number and let $\lambda$ be a partition of $2n$.
The irreducible character $\chi^{\Delta^2(\lambda)}\times\chi^{\Delta^2(\lambda)}$ is an irreducible constituent of $(\chi^\lambda)_{\fS_{n}\times\fS_n}$.
\end{thm}

We are ready to prove Proposition \ref{cor: 2k}. For convenience, we recall its statement here.

\begin{prop}
Let $\lambda\in\mathcal{P}(2^k)$ be such that $\nu_2(\chi^\lambda(1))=1$ and that $\lambda\neq\lambda'$. Then $(\chi^\lambda)_{Q_{2^k}}$ has at least two distinct linear constituents.
\end{prop}
 \begin{proof}
To ease the notation we let $q:=2^{k-1}$.
From Lemma \ref{lem: towers8} we know that $\nu_2(\chi^\lambda(1))=1$ if and only if $\mathcal{H}_{2^k}(\lambda)=\emptyset$ and $|\mathcal{H}_{q}(\lambda)|=2$. Hence $\lambda$ is a partition of $2^k$ having exactly two hooks of length $q$. We immediately see that the only possibility for $\lambda$ is to have $h_{1,2}(\lambda)=h_{2,1}(\lambda)=q$. It follows that there exist $x\in\{1,2,\ldots, q-1\}$ and $y\in\{x-1,\ldots, q-2\}$ such that $$\lambda=(q-x+1, q-y, 2^{x-1}, 1^{y-(x-1)}).$$
The best way to proceed is to distinguish two cases, depending on the parity of the number $s:=y-(x-1)$.

\textbf{(i)} Suppose that $s$ is even and let
$$\mu=(q-y+\frac{s}{2}, 1^{y-\frac{s}{2}})\ \ \text{and}\ \ \  \nu=(q-y+(\frac{s}{2}-1), 1^{y-\frac{s}{2}+1}).$$
An easy calculation shows that $\mu\notin\{\nu, \nu'\}$ since $\lambda\neq \lambda'$.
We will now show that $\chi^\rho\times\chi^\rho$ is an irreducible constituent of $(\chi^\lambda)_{\fS_q\times\fS_q}$, for all $\rho\in\{\mu, \nu\}$.

Notice that $y-s/2=x+s/2-1$.
It is easy to see that the skew Young diagram $[\lambda\smallsetminus\mu]$ is the disjoint union of the three disconnected diagrams $Z_1:=H_{1, q-y+\frac{s}{2}+1}(\lambda)$,
$Z_2:=H_{2,2}(\lambda)$ and $Z_3:=H_{x+1+\frac{s}{2},1}(\lambda)$.
Observe that $Z_1$ is a row of shape $(s/2)$, $Z_2$ is a hook of shape $(q-y-1, 1^{x-1})$ and $Z_3$ is a column of shape $(1^{s/2+1})$.
With this in mind, we proceed as follows.

\medskip

\textbf{-} Replace each of the $s/2$ nodes of $Z_1$ by $1$.

\textbf{-} Replace each of the $q-y-1$ nodes of the first row of the hook $Z_2$ by $1$ and

replace the nodes in the leg of $Z_2$ increasingly from top to bottom with the numbers

$2,3,\ldots, x$.

\textbf{-} Replace the top node of the column $Z_3$ by $1$ and replace the remaining $s/2$

nodes
increasingly from top to bottom
with the
numbers
$x+1,x+2,\ldots, x+s/2$.

\medskip

\noindent What we obtain is a Littlewood-Richardson configuration of type $\mu$ for $[\lambda\smallsetminus\mu]$ because the numbers are weakly increasing along rows, stricly increasing along columns and the sequence obtained by reading the numbers right to left and top to bottom is a reverse lattice sequence of type $\mu$. This follows by observing that the number of $1$s in the sequence is exactly $q-y+s/2=\mu_1$.  Moreover, for all $j\in\{2,3,\ldots, x+s/2\}$ we have exactly one $j$. The claim now follows since $x+s/2-1=y-s/2$.
We remark that the process described above works in the case where $s=0$ as well ($Z_1$ is empty in this case).

We conclude that $\chi^\mu\times\chi^\mu$ is an irreducible constituent of $(\chi^\lambda)_{\fS_q\times\fS_q}$.

\smallskip

We proceed in a very similar way to show that $\chi^\nu\times\chi^\nu$ is an irreducible constituent of $(\chi^\lambda)_{\fS_q\times\fS_q}$.
We observe that the skew Young diagram $[\lambda\smallsetminus\nu]$ is the disjoint union of the three disconnected diagrams $W_1:=H_{1, q-y+\frac{s}{2}}(\lambda)$,
$W_2:=H_{2,2}(\lambda)$ and $W_3:=H_{x+2+\frac{s}{2},1}(\lambda)$.
Observe that $W_1$ is a row of shape $(s/2+1)$, $W_2$ is a hook of shape $(q-y-1, 1^{x-1})$ and $W_3$ is a column of shape $(1^{s/2})$.
With this in mind, we proceed as follows.

\medskip

\textbf{-} Replace each of the $s/2+1$ nodes of $W_1$ by $1$.

\textbf{-} Replace node $(2,q-y)$ (this is the most right node of the first row of the hook

$W_2$)
with a $2$ and replace each of the remaining $q-y-2$ nodes in the first row of

$W_2$ by $1$. Moreover,
replace the nodes in the leg of $W_2$ increasingly from top to

bottom with the numbers
$3,4, \ldots, x+1$.

\textbf{-} Replace the $s/2$
nodes of the column $W_3$
increasingly from top to bottom
with the

numbers
$x+2,\ldots, x+s/2+1$.

\medskip

Arguing exactly as above we verify that this process leaves us with a Littlewood Richardson configuration of type $\nu$ for $[\lambda\smallsetminus\nu]$. Again we remark that the steps described above work in the case where $s=0$ (in this case we have that $Z_3$ is empty).
We conclude that $\chi^\nu\times\chi^\nu$ is an irreducible constituent of $(\chi^\lambda)_{\fS_q\times\fS_q}$.

\medskip

Using Theorem \ref{thm: JLMS} we observe that both $f(\mu)\times f(\mu)$ and $f(\nu)\times f(\nu)$ are irreducible constituents of $(\chi^\lambda)_{P_q\times P_q}$.
Hence for all $\rho\in\{\mu, \nu\}$ there exists an extension $\theta_\rho\in\mathrm{Lin}(P_{2^k})$ of $f(\rho)\times f(\rho)$ such that $\theta_\rho$ is a linear constituent of $(\chi^\lambda)_{P_{2^k}}$. Since $\mu\notin\{\nu, \nu'\}$ we immediately deduce that $(\theta_\mu)_{Q_{2^k}}\neq (\theta_\nu)_{Q_{2^k}}$. Hence the proof is concluded in this case.

\bigskip

\textbf{(ii)} Suppose that $s$ is odd. In this case it is not possible to find two distinct hook partitions $\mu, \nu$ of $q$ such that $\chi^\rho\times \chi^\rho$  is an irreducible constituent of $(\chi^\lambda)_{\fS_q\times\fS_q}$, for all $\rho\in\{\mu, \nu\}$ (take for instance $\lambda=(4,3,1)$ to see this in a small example).
Hence we need to proceed with a sligthly different argument. Let
$$\mu=(q-y+\frac{s-1}{2}, 1^{y-\frac{s-1}{2}})\ \ \text{and}\ \ \  \nu=\Delta^2(\lambda),$$
where $\Delta^2$ is the operator defined in Notation \ref{not: Delta}.
We will now show that $\chi^\rho\times\chi^\rho$ is an irreducible constituent of $(\chi^\lambda)_{\fS_q\times\fS_q}$, for all $\rho\in\{\mu, \nu\}$.
We know from Theorem \ref{thm: Delta} that $\chi^\nu\times\chi^\nu$ is an irreducible constituent of $(\chi^\lambda)_{\fS_q\times\fS_q}$.

In order to show that the same holds for $\mu$ we proceed with a very similar strategy to the one used in the previous case.
Notice that $y-(s-1)/2=x+(s-1)/2$.
It is easy to see that the skew Young diagram $[\lambda\smallsetminus\mu]$ is the disjoint union of the three disconnected diagrams $V_1:=H_{1, q-y+\frac{s+1}{2}}(\lambda)$,
$V_2:=H_{2,2}(\lambda)$ and $V_3:=H_{x+1+\frac{s+1}{2},1}(\lambda)$.
Observe that $V_1$ is a row of shape $((s+1)/2)$, $V_2$ is a hook of shape $(q-y-1, 1^{x-1})$ and $V_3$ is a column of shape $(1^{(s+1)/2})$.
With this in mind, we proceed as follows.

\medskip

\textbf{-} Replace each of the $(s+1)/2$ nodes of $V_1$ by $1$.

\textbf{-} Replace each of the $q-y-1$ nodes of the first row of the hook $V_2$ by $1$ and

replace the nodes in the leg of $V_2$ increasingly from top to bottom with the numbers

$2,3,\ldots, x$.

\textbf{-} Replace the $(s+1)/2$
nodes of the column $W_3$
increasingly from top to bottom

with the
numbers
$x+1,\ldots, x+(s+1)/2$.

\smallskip

Arguing exactly as before we deduce that the process described above gives a Littlewood -Richardson configuration of type $\mu$ for $[\lambda\smallsetminus \mu]$.
Hence we obtain that $\chi^\mu\times \chi^\mu$  is an irreducible constituent of $(\chi^\lambda)_{\fS_q\times\fS_q}$.

\medskip

Whenever $\nu=\Delta^2(\lambda)\notin\mathcal{L}(q)$ then we have that $\chi^\nu(1)$ is even (see the discussion after Theorem \ref{thm:oddMac}) and therefore we know from \cite[Theorem A]{GN} that $(\chi^\nu)_{P_q}$ admits two distinct linear constituents $\eta_1, \eta_2$. Moreover, using again \cite[Theorem 1.1]{G} we know that $f(\mu)\in\mathrm{Lin}(P_q)$ is the unique linear constituent of $(\chi^\mu)_{P_q}$. This shows that $(\chi^\lambda)_{P_q\times P_q}$ has at least $3$ (not necessarily distinct) linear constituents of the form $\phi\times\phi$, for some $\phi\in\mathrm{Lin}(P_q)$. It follows that $(\chi^\lambda)_{P_{2^k}}$ has at least $3$ (not necessarily distinct) linear constituents. Since $\nu_2(\chi^\lambda(1))=1$ we can now use Lemma \ref{lem: 2linears} to conclude that $(\chi^\lambda)_{Q_{2^k}}$ can not have a unique linear constituent.

\medskip

If $\Delta^2(\lambda)\in\mathcal{L}(q)$ we have that $\lambda_2\in\{2,3\}$ by the construction described in Notation \ref{not: Delta}. Suppose for a contradiction that $\lambda_2=3$. Then we have that $y$ and $x$ are odd. Hence $\lambda_1$ is even and therefore $(\Delta^2(\lambda))_2=2$. Hence $\Delta^2(\lambda)\notin\mathcal{L}(q)$. It follows that $\lambda_2=2$ and hence that $y=q-2$. In particular $s=q-x-1$ and we can write $\lambda$ as
$\lambda=(s+2, 2^x, 1^s).$
Consider the conjugate partition $\lambda'$. We certainly have that $\nu_2(\chi^{\lambda'}(1))=1$. Moreover it is easy to see that $\lambda'=(q, x+1, 1^s)$. Since $(\lambda')_2=x+1\geq 3$ and $s>0$ we have that $\Delta^2(\lambda')\notin\mathcal{H}(q)$. Therefore we deduce that $\lambda'$ is one of the partitions we already considered above in case (ii). It follows that $(\chi^{\lambda'})_{P_{2^k}}$ has at least $3$ (not necessarily distinct) linear constituents. Since $(\chi^{\lambda})_{Q_{2^k}}=(\chi^{\lambda'})_{Q_{2^k}}$ the proof is now concluded, by Lemma \ref{lem: 2linears}.

\end{proof}

\medskip


text:
\begin{document}

\begin{center}
{\Large \textbf{A Gamma Distribution Hypothesis \\ for Prime $k$-tuples}} \vskip 4em

{ J. LaChapelle}\\
\vskip 2em
\end{center}

\begin{abstract}
We conjecture average counting functions for prime $k$-tuples based on a gamma distribution hypothesis for prime powers. The conjecture is closely related to the Hardy-Littlewood conjecture for $k$-tuples. Possessing average counting functions along with the exact  functions introduced in \cite{LA1} allows to implicitly define pertinent $k$-tuple zeta functions. The $k$-tuple zeta functions in turn allow construction of $k$-tuple analogs of explicit formulae.
\end{abstract}

\section{Introduction}

The motivation and desire to better understand prime $k$-tuples hardly needs introduction. For a small sample of the literature see \cite{HL}--\cite{Z} and references therein. Of course there would be no suspense if there existed Euler products for $k$-tuples like that for single primes. Their absence seems to indicate that there do not exist generating Dirichlet series whose summands are completely multiplicative in these cases. It is fair to say that this is at the heart of the difficulty in generalizing the single prime case.

This paper hopefully takes a step toward better understanding the distribution of prime $k$-tuples. It is based on two ideas.

The first idea \cite{LA1} uses the arithmetic function $\mu(n)\Lambda(n)/\log(n)$ to represent the exact prime counting function up to some cut-off $\mathsf{x}\in\R_+$,
\begin{equation*}
\pi(\mathsf{x})=-\sum_{n\leq \mathsf{x}}\mu(n)\frac{\Lambda(n)}{\log(n)}\;.
\end{equation*}
This simple representation can be readily extended to prime $k$-tuples determined by an admissible set $\mathcal{H}_k=\{0,h_2,\ldots,h_k\}$:
\begin{equation*}
\pi_{(k)}(\mathsf{x})
=(-1)^k\sum_{n\leq \mathsf{x}}\mu(n)\cdots\mu(n+h_k)\frac{\Lambda(n)}{\log(n)}
\cdots\frac{\Lambda(n+h_k)}{\log(n+h_k)}
\end{equation*}

The second idea is the gamma distribution hypothesis:  \emph{prime powers} are random variables on the \emph{positive reals}, and counting them is a random process following a non-homogenous gamma distribution \cite{LA3}. The resulting probability model --- a non-homogenous Poisson process --- yields quite accurate average counting functions associated with the primes.

Therefore it is natural to generalize to prime $k$-tuples to test the hypothesis. The obvious tack is to consider a joint gamma distribution on $\R_+^k$. But, in light of the exact $k$-tuple counting function, the counting is modeled by a probability distribution along a ray $\mathbf{r}_k\in\R_+^k$ determined by an admissible set $\mathcal{H}_k=\{0,h_2,\ldots,h_k\}$. Taking this into account leads to an ansatz for the density of \emph{$k$-tuples of prime powers} along $\mathbf{r}_k$ up to some cut-off $\mathsf{x}$
\begin{equation*}
    {P_{(k)}(n;\mathsf{x})}:=\frac{(-1)^{n-1}}{n!}
    \,\int_{0}^{\mathsf{x}}(\log(r))^{n-1}\cdots(\log(r+h_k))^{n-1}\,dr
\end{equation*}
where the integral is defined by the Cauchy principal value.

There is no reason to expect the probability measure on $\mathbf{r}_k$ to coincide with the probability measure on $\R_+$ for the single prime case, and we will argue that
\begin{equation*}
dr=C_{(k)}(\mathsf{x})\,dt
\end{equation*}
where $C_{(k)}(\mathsf{x})$ is a suitable normalization defined in (\ref{prime constant}).  For admissible $k$-tuples, the ansatz leads to accurate counting functions because asymptotically $C_{(k)}(\mathsf{x})\sim C_{(k)}$ where $C_{(k)}$ is the appropriate singular series, i.e. the prime $k$-tuple constant.

But the enumeration is secondary. The primary goal is to extract information about prime $k$-tuple distributions, which means we need to discover pertinent $k$-tuple zeta functions\footnote{The zeta functions implicitly  defined by the log-zeta functions will not be studied here, but they are of obvious interest.} implicitly defined by
\begin{equation*}
\log\left(\zeta_{(k)}(s)\right):=\sum_{n=1}^\infty\frac{\Lambda_{(k)}(n)}{\log_{(k)}(n)\,n_{(k)}^s}
\end{equation*}
where $\Lambda_{(k)}(n):=\Lambda(n)\cdots\Lambda(n+h_k)$, similarly $\log_{(k)}(n):=\log(n)\cdots\log(n+h_k)$, and $n_{(k)}$ denotes the geometric mean of $n_k:=(n, n+h_2,\ldots, n+h_k)$. These log-zeta functions are just what one would guess from the structure of the $k$-tuple analogs of the first Chebyshev function \cite{LA1}.

Unfortunately we haven't found an explicit representation of $\zeta_{(k)}(s)$ that would allow the prime $k$-tuple issue to be settled. But in the final section we motivate the fairly obvious conjecture that
\begin{equation*}
\zeta_{(k)}(s)=\sum_{n_{(k)}}\frac{1}{n_{(k)}^s}=\sum_{n=1}^\infty\frac{1}{n_{(k)}^s}\;.
\end{equation*}
 The first sum is over the geometric means of points along a ray in the pair-wise coprime $k$-lattice denoted $\mathfrak{N}_+^k\subset\mathbb{N}_+^k$.

This point bears repeating: Possessing exact counting functions (in terms of standard arithmetic functions) and a model probability distribution facilitates constructing $k$-tuple zeta functions and, subsequently, explicit integral representations of certain counting functions.\footnote{We emphasize that our explicit formulae are left as integral representations. Since we do not determine the complex analytic properties of the $k$-tuple zeta functions, we cannot express the integrals in terms of residues.} This opens the possibility to attack the problem of  prime $k$-tuple distributions using more-or-less elementary methods borrowed from the single prime case.

\section{Prime $k$-tuple conjecture}
According to \cite{LA3}, events along a directed graph can be modeled by a suitable gamma distribution. For prime $k$-tuples the events occur along a ray $\mathbf{r}_k\in\R_+^k$ determined by an admissible set $\mathcal{H}_k=\{0,h_2,\ldots,h_k\}$.  We are counting prime-power events up to some cut-off $\mathsf{x}$, and according to the gamma hypothesis this is a scaled Poisson process.

To learn how to apply the gamma hypothesis for $k$-tuples, let's briefly review the single prime case.  Consider a Poisson process with trivial mean. In this case, counting corresponds to the observation of integers because the events are evenly distributed with unit density. Accordingly, the trivial gamma distribution on $\R_+$ and its associated Poisson process yield a model of the positive integers $\mathbb{Z}_+$ as the cut-off $\mathsf{x}\rightarrow\infty$, because they are in one-to-one correspondence with the positive natural numbers $\mathbb{N}_+$.

Less trivially, the gamma hypothesis posits the expected number of prime-power events along $\R_+$ is given by
\begin{equation}
\overline{N(\mathsf{x})}=\sum_{n=1}^\infty\frac{(-1)^{1}}{n!}
    \,\gamma(n,-\log(\mathsf{x}))\;.
\end{equation}
 This has the expected form of a scaled Poisson expectation \cite{LA3}; suggesting we write the incomplete gamma function as an integral in order to infer the associated prime-power probability distribution on $\R_+$:
 \begin{eqnarray}\label{integral}
 \overline{N(\mathsf{x})}&=&\sum_{n=1}^\infty\frac{(-1)^{1}}{n!}\int d\gamma(n,-\log(\mathsf{x}))\notag\\
 &=&\sum_{n=1}^\infty\frac{1}{n!}\int(-\log(\mathsf{x}))^{n-1}\,dx\notag\\
 &=&-\sum_{n=1}^\infty\frac{(-1)^{n}}{n!}\int_0^\mathsf{x}(\log(t))^{n-1}\,dt\;.
 \end{eqnarray}
 where the integral in the third line is taken as the Cauchy principal value. We infer the probability distribution of prime powers goes like $\log(t)^{-1}$. Interchanging the sum and integral yields\footnote{We use $\mathrm{Ei}(\log(\mathsf{x}))$ instead of $\mathrm{li}(\mathsf{x})$ to remind that the gamma hypothesis applies to the more general case of complex cut-off $\mathsf{x}\in\C_+$.} $\overline{N(\mathsf{x})}=\mathrm{Ei}(\log(\mathsf{x}))-\log(\log(\mathsf{x}))$.

 Return now to the general case. The integral in (\ref{integral}) becomes a multiple integral on $\R_+^k$, and the probability distribution will be a $k$-fold product of logarithms restricted to the appropriate ray determined by $\mathcal{H}_k$. Consequently,  for the general $k$-tuple case  the mean number of prime-power $k$-tuple events is expected to be
\begin{eqnarray}\label{k-density}
\overline{N_{(k)}(\mathsf{x})}&=&-\sum_{n=1}^\infty\frac{(-1)^{n}}{n!}\int_{0}^{\mathsf{x}}(\log(r))^{n-1}\cdots (\log(r+h_k))^{n-1}\,dr\notag\\
&\propto&-\sum_{n=1}^\infty\frac{(-1)^{n}}{n!}\int_{0}^{\mathsf{x}}(\log(t))^{n-1}\cdots (\log(t+h_k))^{n-1}\,dt
\end{eqnarray}
where $r\in\mathbf{r}_k$ and $t\in\R_+$. In Appendix A we argue the ratio of the two integrators is given by
 \begin{equation}
C_{(k)}(\mathsf{x}):=\prod_{p\leq \mathsf{x}_{(k)}^k} \left(1-\frac{\nu_p(\mathcal{H}_k)}{p}\right)\,\prod_p\left(1-\frac{1}{p}\right)^{-k}\;.
\end{equation}

This leads to
\begin{ansatz}
Let $\mathcal{H}_k=\{0,h_2,\ldots,h_k\}$ be admissible. The expected number of events associated with counting admissible prime-power $k$-tuples up to some cutoff $\mathsf{x}$ is
\begin{equation}
    \overline{N_{(k)}(\mathsf{x})}=\sum_{n=1}^\infty{P_{(k)}(n;\mathsf{x})}
:=-\sum_{n=1}^\infty\frac{(-1)^{n}}{n!}
    \,I_{(k)}(n;\mathsf{x})
\end{equation}
where
\begin{eqnarray}
I_{(k)}(n;\mathsf{x})&:=&\int_{0}^{\mathsf{x}}(\log(r))^{n-1}\cdots(\log(r+h_k))^{n-1}\,dr\notag\\
&=&C_{(k)}(\mathsf{x})\int_{0}^{\mathsf{x}}(\log(t))^{n-1}\cdots(\log(t+h_k))^{n-1}\,dt\notag\\
&=:&C_{(k)}(\mathsf{x})\int_{0}^{\mathsf{x}}\log^{n-1}_{(k)}(t)\,dt\notag\\
\end{eqnarray}
with the integral defined by the principal value and $C_{(k)}(\mathsf{x})\sim C_{(k)}$ the singular series.

\end{ansatz}

The analysis in \cite{LA3} suggests that
\begin{equation}
   \overline{N_{(k)}(\mathsf{x})}\approx\sum_{n\leq{\mathsf{x}}}\lambda_{(k)}(n)
   -\sum_{n\,\mid\,\mathsf{x}}\lambda_{(k)}(n)\;.
 \end{equation}
 where
\begin{eqnarray}
\lambda_{(k)}(n)&:=&\frac{\Lambda(n)\cdots\Lambda(n+h_k)}{\log(n)\cdots\log(n+h_k)}\notag\\
&=:&\frac{\Lambda_{(k)}(n)}{\log_{(k)}(n)}\;.
\end{eqnarray}
Moreover, since the sum and integral can be interchanged,
\begin{equation}
    \overline{N_{(k)}(\mathsf{x})}= C_{(k)}(\mathsf{x})\int_{0}^{\mathsf{x}}\frac{1}{\log_{(k)}(t)}\,dt-\mathrm{\;small\; remainder}
\end{equation}
for all $k\in\mathbb{N}_+$. Again, \cite{LA3} suggests the small remainder term represents $\sum_{n\,\mid\,\mathsf{x}}\lambda_{(k)}(n)$ while
\begin{eqnarray}\label{average J}
  \overline{J_{(k)}(\mathsf{x})}:= C_{(k)}(\mathsf{x})\int_{0}^{\mathsf{x}}\frac{1}{\log_{(k)}(t)}\,dt&=:&C_{(k)}(\mathsf{x})\mathrm{Ei}_{(k)}(\log(\mathsf{x}))\notag\\
  &\approx& \sum_{n\leq{\mathsf{x}}}\lambda_{(k)}(n)
\end{eqnarray}
is the average $k$-tuple analog of Riemann's counting function.

In particular, for prime doubles
\begin{eqnarray}
    \overline{N_{(2)}(\mathsf{x})}&=&C_{(2)}(\mathsf{x})\sum_{n=1}^\infty\frac{(-1)^{n-1}}{n!}
    \,\int_{0}^{\mathsf{x}}\left(\log(t)\log(t+h_{2i})\right)^{n-1}\,dt\notag\\
&=:&\overline{J_{(2)}(\mathsf{x})}-\overline{\omega_{(2)}(\mathsf{x})}\;.
\end{eqnarray}
where $C_{(2)}=C_2\prod_{p\mid i}\frac{p-1}{p-2}$ with $C_2$ the twin prime constant.

Given this heuristic motivation, we conjecture:
\begin{conjecture}\label{average primes}
Given an admissible $\mathcal{H}_k$, the average number of admissible prime $k$-tuples up to some cut-off integer $ \mathsf{x}$ is
 \begin{equation}\label{average k-tuples}
 \overline{\pi_{(k)}(\mathsf{x})}=\sum_{m=1}^\infty\frac{\mu(m)}{m}\overline{J_{(k)}(\mathsf{x}^{1/m})}\;.
 \end{equation}
\end{conjecture}
The reader is encouraged to numerically explore (\ref{average k-tuples}). Note that, whereas the Hardy-Littlewood conjecture is asymptotic, (\ref{average k-tuples}) holds for all $\mathsf{x}>2$. The difference between predicted counts for Conjecture \ref{average primes} and the Hardy-Littlewood conjecture is especially stark when $\mathsf{x}\ll h_k$.

Analogous reasoning helps to define the average prime double Chebyshev function:
\begin{definition}
\begin{eqnarray}\label{average Chebyshev}
    \overline{\psi_{(2)}(\mathsf{x})}&:=&C_{(2)}(\mathsf{x})\int_{0}^{\mathsf{x}}\frac{\log(t_{(2)})}{\log(t)\log(t+h_{2i})}\,dt\notag\\
    &\approx&\sum_{n\leq \mathsf{x}}\frac{\Lambda_{(2)}(n)}{\log_{(2)}(n)}\log(n_{(2)})
\end{eqnarray}
where $t_{(2)}:=\left(t(t+h_2)\right)^{1/2}$ is the geometric mean of $(t, t+h_2)$.
\end{definition}
This has obvious extensions to higher $k$-tuples:
\begin{definition}
\begin{eqnarray}
    \overline{\psi_{(k)}(\mathsf{x})}&:=&C_{(k)}(\mathsf{x})\int_{0}^{\mathsf{x}}\frac{\log(t_{(k)})}{\log_{(k)}(t)}\,dt\notag\\
    &\approx&\sum_{n\leq \mathsf{x}}\frac{\Lambda_{(k)}(n)}{\log_{(k)}(n)}\log(n_{(k)})\;.
\end{eqnarray}
\end{definition}

\begin{conjecture}\label{average theta}
\begin{equation}
\overline{\theta_{(2)}(\mathsf{x})}=\sum_{m=1}^\infty\mu(m)\overline{\psi_{(2)}(\mathsf{x}^{1/m})}\;.
\end{equation}
\end{conjecture}
Note that $\overline{\psi_{(2)}(\mathsf{x})}
=C_{(2)}(\mathsf{x})(\mathrm{Ei}\left(\log({ \mathsf{x}})\right)+\mathrm{Ei}\left(\log({\mathsf{x}+h_2})\right)-\mathrm{Ei}\left(\log({h_2})\right))/2$ follows from (\ref{average Chebyshev}).  Hence $\overline{\theta_{(2)}(\mathsf{x})}\sim C_{(2)}\left( \mathsf{x}/\log( \mathsf{x})\right)$ which is consistent with the Hardy-Littlewood twin prime conjecture.

\section{Explicit formulae}
Having  both exact and average summatory functions allows  to deduce associated $k$-tuple zeta functions and subsequent explicit formulae. Here we will confine attention to prime doubles but indicate the generalization to higher $k$-tuples.

Define the prime double zeta function \emph{implicitly} by
\begin{definition}
\begin{equation}
\log\left(\zeta_{(2)}(s)\right):=\sum_{n=1}^\infty\frac{\lambda_{(2)}(n)}{n^{s/2}(n+2i)^{s/2}}
=\sum_{n=1}^\infty\frac{\Lambda_{(2)}(n)}{\log_{(2)}(n)\,n_{(2)}^{s}}\;,
\;\;\;\;\Re (s)>1\;.
\end{equation}
\end{definition}
It follows that
\begin{eqnarray}
\log'(\zeta_{(2)}(s))=\frac{\zeta'_{(2)}(s)}{\zeta_{(2)}(s)}
&=&-\sum_{n=1}^\infty\frac{\Lambda_{(2)}(n)}{\log_{(2)}(n)\,n_{(2)}^{s}}\log(n_{(2)})\;.
\end{eqnarray}

Using this log-zeta function, along with the gamma hierarchy from \cite{LA3} as a guide, we construct an explicit formula  for
  \begin{equation}
\psi_{(2)}(\mathsf{x})=\sum_{n\leq \mathsf{x}}\frac{\Lambda_{(2)}(n)}{\log_{(2)}(n)}\log(n_{(2)})\;.
\end{equation}

\begin{proposition}\label{explicit formula}
Put $\widetilde{ \mathsf{x}}= \mathsf{x}+\epsilon$ with $ \mathsf{x}\in\N_+$ and $0<\epsilon<1$. Let $\sigma_a$ be the abscissa of absolute convergence of $\sum_{n=1}^\infty\frac{\lambda_{(2)}(n)\log(n_{(2)})}{n^{s/2}(n+2i)^{s/2}}$. Then, for $c>\sigma_a$,
\begin{eqnarray}\label{explicit eq}
 \psi_{(2)}(\mathsf{x})&=&-\lim_{\epsilon\rightarrow0}\lim_{T\rightarrow\infty}\frac{1}{2\pi i}\int_{c-iT}^{c+iT}\Gamma\left(0,-\log(\widetilde{{ \mathsf{x}}_{(2)}}^s)\right)
 \;d\log'(\zeta_{(2)}(s))\,,
 \;\,\;\;\;\;\;c>\sigma_a\notag\\
 &=&\lim_{\epsilon\rightarrow0}\lim_{T\rightarrow\infty}\frac{1}{2\pi i}\int_{c-iT}^{c+iT}\mathrm{Ei}\left(\log(\widetilde{{ \mathsf{x}}_{(2)}}^s)\right)
 \;d\log'(\zeta_{(2)}(s))\,,
 \;\,\;\;\;\;\;c>\sigma_a\notag\\
 &=&\sum_{n\leq \mathsf{x}}\frac{\Lambda_{(2)}(n)}{\log_{(2)}(n)}\log(n_{(2)})\;.
 \end{eqnarray}
\end{proposition}

\emph{Proof}: First integrate (\ref{explicit eq}) by parts. The boundary term does not contribute because i) a comparison test yields a finite $\sigma_a$ (in fact $\sigma_a=1$) so $\lim_{t\rightarrow\infty}|\log'(\zeta_{(2)}(c+it))|<\infty$ for $c>\sigma_a$;  and ii) $\lim_{t\rightarrow\infty}|\mathrm{Ei}\left(\log(\widetilde{{ \mathsf{x}}_{(2)}}^s)\right)|=0$ since
\begin{eqnarray}
\lim_{t\rightarrow\infty}\left|\mathrm{Ei}\left(\log({ \mathsf{x}}^{(c+it)})\right)\right|
&=&\lim_{t\rightarrow\infty}\left|\frac{{ \mathsf{x}}^{(c+it)}}{(c+it)\log({ \mathsf{x}})}
\left(1+O\left(\frac{1}{(c+it)\log({ \mathsf{x}})}\right)\right) \right|\notag\\
&\leq&\frac{{ \mathsf{x}}^c}{\log({ \mathsf{x}}))}\lim_{t\rightarrow\infty}
\left|\frac{1}{(c+it)}
\left(1+O\left(\frac{1}{(c+it)}\right)\right) \right|=0\;.\notag\\
\end{eqnarray}

Next, following standard arguments, use the truncating integral
\begin{lemma}
\begin{equation}
\frac{1}{2\pi i}\int_{c-i T}^{c+i T}\frac{\mathsf{x}^s}{s}ds=\left\{\begin{array}{l}
                                                                1+O\left(\frac{\mathsf{x}^c}{T\log(\mathsf{x})}\right)\;\;\;\;\mathsf{x}>1 \\
                                                                O\left(\frac{\mathsf{x}^c}{T\log(\mathsf{x})}\right)\;\;\;\;0<\mathsf{x}<1
                                                              \end{array}\right.\;.
\end{equation}
\end{lemma}
\emph{proof}: We include the proof for completeness.
For $\mathsf{x}>1$ integrate over a rectangle with left edge $(L-i T,L+i T)$ such that $L<c$. We have
\begin{equation}
\lim_{L\rightarrow-\infty}\left|\int_{L-i T}^{L+i T}\frac{\mathsf{x}^s}{s}ds\right|\leq\lim_{L\rightarrow-\infty}\int_{-T}^{T}\frac{\mathsf{x}^L}{|L+ i t|}dt<\lim_{L\rightarrow-\infty}\frac{T \mathsf{x}^L}{L}=0\;.
\end{equation}
The top and bottom contribute
\begin{eqnarray}
\left|\int_{-\infty\pm i T}^{c\pm i T}\frac{\mathsf{x}^s}{s}ds\right|&\leq&\int_{-\infty}^{0}\frac{-\mathsf{x}^{c-r}}{|(c-r)\pm i T|}dr\notag\\
&=&\mathsf{x}^c\int_{-\infty}^{0}\frac{-\mathsf{x}^{-r}}{|(c-r)\pm i T|}dr\notag\\
&<&\mathsf{x}^c\int_{-\infty}^{0}\frac{-\mathsf{x}^{-r}}{T}dr\notag\\
&=&\frac{\mathsf{x}^{c}}{T\log(\mathsf{x})}\;.
\end{eqnarray}
Finally, the pole at $s=0$ contributes $\mathrm{Res}=1$.

Now, for $\mathsf{x}<1$ integrate over the right edge $(R-i T,R+i T)$ with $c< R$. Then
\begin{equation}
\lim_{R\rightarrow\infty}\left|\int_{R-i T}^{R+i T}\frac{\mathsf{x}^s}{s}ds\right|\leq\lim_{R\rightarrow\infty}\int_{-T}^{T}\frac{e^{-R|\log(\mathsf{x})|}}{|R+ i t|}dt<\lim_{R\rightarrow\infty}\frac{T e^{-R|\log(\mathsf{x})|}}{R}=0\;.
\end{equation}
The top and bottom contribute the same order as for $\mathsf{x}>1$, so the well-known lemma is established. $\qed$

Hence, for $c>\sigma_a$,
\begin{eqnarray}
- \lim_{\epsilon\rightarrow0}\lim_{T\rightarrow\infty}\frac{1}{2\pi i}\int_{c-iT}^{c+iT}\log'(\zeta_{(2)}(s))\frac{\widetilde{{ \mathsf{x}}_{(2)}}^{s}} {s}\,ds\notag\\
&&\hspace{-1.25in}=\lim_{\epsilon\rightarrow0}\lim_{T\rightarrow\infty}\frac{1}{2\pi i}\int_{c-iT}^{c+iT}\sum_{n=1}^\infty\frac{\lambda_{(2)}(n)\log(n_{(2)})}{n^{s/2}(n+2i)^{s/2}}
 \frac{\widetilde{{ \mathsf{x}}_{(2)}}^{s}}
 {s}\,ds\notag\\
 &&\hspace{-1.25in}=\lim_{\epsilon\rightarrow0}\lim_{T\rightarrow\infty}
 \sum_{n=1}^\infty\frac{\Lambda_{(2)}(n)}{\log_{(2)}(n)}\log(n_{(2)})
 \frac{1}{2\pi i}\int_{c-iT}^{c+iT}
 \frac{\frac{\left(\widetilde{ \mathsf{x}}^{1/2}(\widetilde{ \mathsf{x}}+2i)^{1/2}\right)^{s}}{n^{s/2}(n+2i)^{s/2}}}
 {s}\,ds\notag\\
 &&\hspace{-1.25in}=\lim_{\epsilon\rightarrow0}\sum_{n\leq\lfloor\widetilde{\mathsf{x}}\rfloor}
 \frac{\Lambda_{(2)}(n)}{\log_{(2)}(n)}\log(n_{(2)})\notag\\
 &&\hspace{-1in}=\sum_{n\leq \mathsf{x}}\frac{\Lambda_{(2)}(n)}{\log_{(2)}(n)}\log(n_{(2)})
 \end{eqnarray}
where the third equality follows from the lemma. (Justifying the interchange of the sum and integral is straightforward, and interchange of the $T$-limit and sum is allowed because the summand contains $O(n^{-c})$ with $c>1$.)
$\QED$

Clearly this result only has teeth if one possesses an explicit representation of $\zeta_{(2)}(s)$. \emph{But if} a suitable representation of $\zeta_{(2)}(s)$ can be found and it enjoys analytic properties similar to $\zeta(s)$, then we might expect
\begin{eqnarray}
 \psi_{(2)}(\mathsf{x})&=&\lim_{T\rightarrow\infty}\frac{1}{2\pi i}\int_{c-iT}^{c+iT}
 \mathrm{Ei}(\log({ \mathsf{x}}_{(2)}^{s}))\;d\log'(\zeta_{(2)}(s))\notag\\
 &=&\lim_{T\rightarrow\infty}\frac{1}{2\pi i}\int_{c-iT}^{c+iT}\mathrm{Ei}(\log({ \mathsf{x}}_{(2)}^{s}))\log''(\zeta_{(2)}(s))\;ds
 \end{eqnarray}
would lead to something like
\begin{equation}
\psi_{(2)}( \mathsf{x})\sim C_{(2)}\mathrm{Ei}(\log(\mathsf{x}))
-C_{(2)}\sum_{\rho_{(2)}}\mathrm{Ei}(\log(\mathsf{x}^{\rho_{(2)}}))
 +\mathrm{small\,terms}
 \end{equation}
 where the sum would include conjugate pairs of nontrivial zeros of $\zeta_{(2)}(s)$. \emph{And if} non-trivial zeros of $\zeta_{(2)}(s)$ are confined within its critical strip, then the same proof strategy used for the PNT would appear to apply to prime doubles. Evidently, if this scenario plays out the prime double constant $C_{(2)}$ will have to come from $\log''(\zeta_{(2)}(s))$.

Remark that for higher $k$-tuples one should define
\begin{definition}\begin{equation}
\overline{ \varphi_{(k)}(\mathsf{x})}:=C_{(k)}(\mathsf{x})\int_{0}^{\mathsf{x}}\frac{\log^{k-1}(t_{(k)})}{\log_{(k)}(t)}\,dt\;.
\end{equation}
\end{definition}
and the $k$-tuple log-zeta function
\begin{definition}
\begin{equation}
\log\left(\zeta_{(k)}(s)\right):=\sum_{n=1}^\infty\frac{\lambda_{(k)}(n)}{n_{(k)}^s}
=\sum_{n=1}^\infty\frac{\Lambda_{(k)}(n)}{\log_{(k)}(n)\,n_{(k)}^s}
\end{equation}
so that
\begin{equation}
\log^{(k-1)'}\left(\zeta_{(k)}(s)\right)
=(-1)^{k-1}\sum_{n=1}^\infty\frac{\Lambda_{(k)}(n)}{\log_{(k)}(n)n_{(k)}^s}\log^{k-1}(n_{(k)})\;.
\end{equation}
\end{definition}

Then to construct an explicit formula at level $k$,  consider
\begin{equation}
\lim_{T\rightarrow\infty}\frac{(-1)^{k-1}}{2\pi i}\int_{c-iT}^{c+iT}\Gamma(0,-\log({ \mathsf{x}}_{(k)}^{s}))
 \;d\log^{(k-1)'}\left(\zeta_{(k)}(s)\right)\;.
\end{equation}
Assuming integration by parts to be valid, Perron's formula would apply yielding
\begin{equation}
\varphi_{(k)}(\mathsf{x})=\sum_{n\leq \mathsf{x}}\frac{\Lambda_{(k)}(n)}{\log_{(k)}(n)}\log^{k-1}(n_{(k)})\;.
\end{equation}
Finally, assuming favorable analytic properties for $\zeta_{(k)}(s)$ similar to $\zeta(s)$, one would expect to find $\varphi_{(k)}(\mathsf{x})\sim C_{(k)}\mathrm{Ei}(\log(\mathsf{x}))$.

\section{Searching for $\zeta_{(k)}(s)$}
The manipulations in the previous section point to a possible representation for $\zeta_{(k)}(s)$ which we formulate as another conjecture. First note that
\begin{eqnarray}
\log\left(\zeta_{(2)}(s)\right)
&>&\sum_{\mathfrak{p}_2\in\mathfrak{P}_2}
          \sum_{\omega,\omega':\omega=\omega'}\frac{1}{\omega p^{\omega s/2}}\frac{1}{\omega' (p^\omega+2i)^{\omega's/2}}
          \;,\;\;\;\;\;\;\Re(s)>1\notag\\\notag\\
&>&\sum_{\mathfrak{p}_2\in\mathfrak{P}_2}
          \sum_{\omega,\omega':\omega=\omega'}\frac{1}{\omega (p^{\omega'})^{\omega s/2}}\frac{1}{\omega' ((p+2i)^\omega)^{\omega's/2}}\notag\\\notag\\
&=&\sum_{\mathfrak{p}_2\in\mathfrak{P}_2}\sum_{\omega^2}\frac{1}{\omega^2 p_{(2)}^{s\,\omega^2}}\notag\\
&=&-\sum_{\mathfrak{p}_2\in\mathfrak{P}_2}\log\left(1-p_{(2)}^{-s}\right)\;.
\end{eqnarray}
This suggests to define $Z_{(k)}(s):=\prod_{\mathfrak{p}_k\in\mathfrak{P}_k}\left(1-p_{(k)}^{-s}\right)^{-1}$ which motivates
\begin{conjecture}
Let $\mathfrak{r}_k$ be a ray in the pair-wise coprime $k$-lattice $\mathfrak{N}_+^k$ and $n_{(k)}$ the geometric mean of a point $n_k=(n, n+h_2, \ldots, n+h_k)\in\mathfrak{r}_k$. Then
\begin{equation}
\zeta_{(k)}(s)=\sum_{n_{(k)}}\frac{1}{n_{(k)}^s}=\sum_{n=1}^\infty\frac{1}{n_{(k)}^s}
\end{equation}
where the first sum is over the geometric means of all points along $\mathfrak{r}_k\in\mathfrak{N}_+^k$\;.
\end{conjecture}

If the conjecture is correct, the $k$-tuple zeta function is evidently a restriction of the multiple zeta function which motivates
\begin{conjecture}
$\zeta_{(k)}(s)$ is meromorphic on $\C$, and the singular part of $(-1)^k\log^{(k)'}(\zeta_{(k)}(s))$ is given by
\begin{equation}
\frac{1}{s-1}\,\frac{(-1)^k}{2\pi i}\oint\log^{(k)'}(\zeta_{(k)}(s))\,ds=\frac{C_{(k)}}{s-1}\;.
\end{equation}
\end{conjecture}

This conjecture is equivalent to the Hardy-Littlewood prime $k$-tuple conjecture in the following sense. If we believe the gamma hypothesis, then being a sum of $1/n_{(k)}^s$ along a ray in the pair-wise coprime $k$-lattice \emph{strongly suggests} $\zeta_{(k)}(s)$ has a first order pole at $s=1$ and there are no other poles, while its zeros are determined by conspiring projections of almost periodic exponentials.

Together with the explicit formula at level $k$ the conjecture implies
\begin{eqnarray}
\overline{\varphi_{(k)}(\mathsf{x})}
 &=&\left.\lim_{T\rightarrow\infty}\frac{(-1)^k}{2\pi i}\int_{c-iT}^{c+iT}
 \mathrm{Ei}(\log({ \mathsf{x}}_{(k)}^{s}))\log^{(k)'}(\zeta_{(k)}(s))\;ds\right|_{s=1}\notag\\\notag\\
 &\sim&C_{(k)}\,\mathrm{Ei}(\log(\mathsf{x}))\;.
 \end{eqnarray}
Equivalently,
\begin{equation}
\lim_{N\rightarrow\infty}\frac{1}{N}\sum_{n\leq N}\frac{\Lambda_{(k)}(n)}{\log_{(k)}(n)}\log^k(n_{(k)})
= C_{(k)}\;.
\end{equation}

Of course, possessing poles and zeros of $\zeta_{(k)}(s)$ would be tantamount to evaluating the exact summatory functions. And their singular part would presumably furnish the prime $k$-tuple constants. The goal would be to express the integral in Proposition \ref{explicit formula} as a sum over $k$-tuple zeta residues in the usual way; which would presumably verify Conjecture \ref{average primes} and validate the gamma hypothesis.

\vspace{.5in}\noindent
\textbf{Acknowledgments:}
I thank A. Granville for offering some helpful advice.

\appendix
\section{The $k$-tuple normalization}

In this appendix we argue that $C_{(k)}(\mathsf{x})$ is asymptotically the prime $k$-tuple constant. Since the argument is based on the gamma distribution hypothesis, it is very close to the standard probability argument. However, the details are a bit different since the counting occurs in the pair-wise coprime lattice.

According to the probability model \cite{LA3} for the case $k=1$, there is a distinction between the positive integer lattice $\mathbb{Z}_+$ and the strictly positive natural numbers $\mathbb{N}_+$ that characterize the counting process associated with the trivial gamma distribution on $\R_+$. However, the trivial gamma distribution yields a Poisson process with unit integer density, and this allows for a meaningful identification.

But in general the probability model is a counting process along a ray $\mathfrak{r}_{k}$ in the $k$-lattice, and there is no guarantee that the trivial gamma distribution along $\mathbf{r}_{k}$ leads to a unit integer density. Accordingly, in virtue of restricting to the pair-wise coprime $k$-lattice, the integer density may depend on $\mathcal{H}_k$ and $\mathsf{x}$; and it is necessary to renormalize the probability distribution along $\mathbf{r}_{k}$ if one wants to compare counting processes and maintain the natural identification of $\mathbb{N}_+$ and $\mathbb{Z}_+$.

In other words, comparing exact counting functions to average counting functions in the single prime case, we can interpret
\begin{equation}
\sum_{n\leq \mathsf{x}}\frac{\Lambda(n)}{\log(n)}\longrightarrow\int_0^\mathsf{x}\frac{dt}{\log(t)}
\end{equation}
as a representation of the averaging process, and then $dt$ is the integrator associated with the probability measure of prime powers --- a kind of smoothed $\Lambda(n)$. For the general case we have
\begin{equation}
\sum_{n\leq \mathsf{x}}\frac{\Lambda_{(k)}(n)}{\log_{(k)}(n)}\longrightarrow\int_{0}^{\mathsf{x}}\frac{dr}{\log_{(k)}(r)}\;.
\end{equation}
But now $dr$ encodes both the `smoothed' $\Lambda_{(k)}(n)$ and the density of integers along $\mathbf{r}_k$.

In order to put the density of integers along $\R_+$ and $\mathbf{r}_k$ on equal footing, we use the gamma hypothesis for prime powers together with the fundamental theorem of arithmetic. Evidently, it suffices to deduce their ratio for the particular case of counting prime powers. So the task is to determine the prime-power density along $\mathfrak{r}_{k}$ relative to the prime-power density along $\R_+$.

We are looking for the probability that the $k$-th power of the geometric mean of a point $n_k=(n,n+h_2,\ldots,n+h_k)\in\mathfrak{r}_{k}$ is coprime to some prime, since this is a necessary condition for $n_{(k)}^k$ to be a prime power. Our main tool is a theorem by T\'{o}th \cite{T}:
Let $k,m,u\geq1$ and
\begin{equation}
P_k^{(u)}(m)=\!\!\!\!\!\!\!\!\sum_{\begin{array}{c}
            \scriptstyle{1\leq a_1, \ldots, a_k\leq m} \\
            \scriptstyle{(a_i,a_j)=1, i\neq j} \\
            \scriptstyle{(a_i,u)=1}
          \end{array}}\!\!\!\!\!\!\!\!1
\end{equation}
be the number of $k$-tuples $(a_1,\ldots, a_k)$ on the pair-wise coprime lattice with $1\leq a_i\leq m$ and $(a_i,u)=1$ for all $i\in\{1,\ldots, k\}$.
\begin{theorem}(\cite{T})
For a fixed $k\geq1$, we have uniformly for $m,u\geq1$,
\begin{equation}
P_k^{(u)}(m)=A_k f_k(u) m^k+O\left(\theta(u)m^{k-1}\log^{k-1}(m)\right)
\end{equation}
where\begin{eqnarray*}
       A_k &=& \prod_{p'}\left(1-\frac{1}{p'}\right)^{k-1}\left(1+\frac{k-1}{p'}\right) \\
       f_k(u) &=&\prod_{p'|u}\left(1-\frac{k}{p'+k-1}\right)
     \end{eqnarray*}
and $\theta(u)$ is the number of squarefree divisors of $u$.
\end{theorem}
Restrict $P_k^{(u)}(m)$ to the ray $\mathfrak{r}_{k}$ and choose $m>n+h_k$. Then the density of points along $\mathfrak{r}_{k}$ that are coprime to a given prime (or prime power) $p$ is
\begin{equation}
D_k^{(p)}(m):=P_k^{(p)}(m)/m^k=A_k f_k(p)+O\left(\log^{k-1}(m)/m\right)\;.
\end{equation}
Of course $m$ is automatically coprime to all primes $p> m$. Consequently, the density of prime powers along $\mathfrak{r}_{k}$ relative to $\mathbb{Z}_+$ is
\begin{equation}
\lim_{m\rightarrow\infty}\prod_{p\leq m} D_k^{(p)}(m)=\prod_p\left(1-\frac{1}{p}\right)^{-k}\;.
\end{equation}

Meanwhile, to determine the contribution to $dr$ from the `smoothed' $\Lambda_{(k)}(n)$, follow the standard argument. There are $p-\nu_p(\mathcal{H}_k)$ residue classes mod $p$ that $n$ can occupy that guarantee $n^k_{(k)}$ is not divisible by $p$. So the probability that $\Lambda_{(k)}(n)$ does not vanish given that $n$ is a prime power is
 \begin{equation}
 \prod_{p< n_{(k)}^k} \left(1-\frac{\nu_p(\mathcal{H}_k)}{p}\right)\;.
\end{equation}

It follows that the integrator $dr$ associated with the probability measure on $\mathbf{r}_k$ is
\begin{equation}\label{prime constant}
dr=C_{(k)}(\mathsf{x})\,dt:=\prod_{p< \mathsf{x}_{(k)}^k} \left(1-\frac{\nu_p(\mathcal{H}_k)}{p}\right)\prod_p\left(1-\frac{1}{p}\right)^{-k}\,dt
\end{equation}
where $dt$ is integrator associated with the normalized
Haar measure on $\R_+$. Finally, define $C_{(k)}$ to be the asymptote
\begin{equation}
C_{(k)}:=\lim_{\mathsf{x}\rightarrow\infty}C_{(k)}(\mathsf{x}) =\prod_p\left(1-\frac{\nu_p(\mathcal{H}_k)}{p}\right)\left(1-\frac{1}{p}\right)^{-k}\;.
\end{equation}

config: arxiv


TITLE: What method can i use to find the first 3 roots of y(t)=tan(t)+t?
QUESTION [3 upvotes]: Just by looking at the function:
$$y(t) = \tan(t)+t$$
I can immediately see that there is a root at $t=0$, though after graphing it I can see many more roots and I can calculate them using computer aided programs; i.e. the next root is at $t=2.0288$.
What I want to know is:  what methods I can use to solve this second, third.... $n$th root with only a scientific calculator.

REPLY [2 votes]: As you know, equations which mix polynomials and trigonometric functions do not have explicit solutions and numerical methods should be used for solving them.
As you noticed from the graph, the solutions are closer and closer to the vertical asymptotes and there are an infinite number say $$t_n=(2n-1)\frac{\pi}{2}+\epsilon_n$$ Because of the vertical asymptotes, the function is quite poorly conditions and, personally, I should solve instead $$f(t)=\sin(t)+t\cos(t)=0$$ which looks much better (have a look to the plot).
For the solution, Newton method is simple. Starting from a "reasonable" guess $t_0$, the method will update it according to $$t_{n+1}=t_n-\frac{f(t_n)}{f'(t_n)}$$ which, in your case would write $$t_{n+1}=t_n+\frac{\sin (t_n)+t_n \cos (t_n)}{t_n \sin (t_n)-2 \cos (t_n)}$$
Added after robjohn's answer
If we take into account the first remark, we can see that a good estimate can be generated according to $$t_k^0=\frac{2}{(2k-1) \pi }+\frac{(2k-1) \pi }{2}$$ so, for the first root, the very first estimate is $2.20741609916248$ and, for the tenth root, the very first estimate is $29.8786365129119$.
For the first root,Newton iterates are then $2.03597131803008$, $2.02877485165832$, $2.02875783820645$, $2.02875783811043$ which is the solution for fifteen significant digits.
For the tenth root, Newton iterates are then $29.8785865061909$, $29.8785865061074$ which is the solution for fifteen significant digits.


TITLE: SLLN for V-statistics - Strong consistency of distance covariance in metric spaces
QUESTION [5 upvotes]: I'm trying to understand the proof for strong consistency of distance covariance in metric spaces (Proposition 2.6 Distance covariance in metric spaces by Russel Lyons published in The Annals of Probability).
Let $(\mathcal{X},d_1)$ and $(\mathcal{Y},d_2)$ be two metric spaces and $((X_k,Y_k))_{k\in\mathbb{N}}$ be an i.i.d sequence of random elements with values in $\mathcal{X}\times \mathcal{Y}$ such that
$$
Ed_1(X_1,x')<\infty \quad \quad and \quad \quad Ed_2(Y_1,y')<\infty,
$$
for any $x'\in \mathcal{X}$ and $y'\in\mathcal{Y} $. The distance covariance $dcov(X,Y)$ is given by
$$
dcov(X,Y) = Eh((X_{1},Y_1),...,(X_6,Y_6)),
$$
where
\begin{align*}
h((x_{1},y_1),...,(x_6,y_6))=&f_1(x_1,x_2,x_3,x_4)f_2(y_1,y_2,y_5,y_6) ,\\
\end{align*}
and
$$
f_i(z_1,z_2,z_3,z_4) =d_i(z_1,z_2)-d_i(z_1,z_3)-d_i(z_2,z_4)+d_i(z_3,z_4).
$$
The emperical distance covariance hence becomes a $V$-statistic with (in general) non-symmetric kernel $h$ of degree $6$, i.e.
$$
dcov_n(X,Y) = \frac{1}{n^6} \sum_{i_1=1}^n \cdots \sum_{i_6=1}^n h((X_{i_1},Y_{i_1}),...,(X_{i_6},Y_{i_6})).
$$
Now it is stated without reference to a SLLN that $dcov_n(X,Y) \to dcov(X,Y)$ almost surely.
Problem:
The weakest conditions for SLLN for $V$-statistics I know require that

The kernel $h$ is symmetric and \begin{align*} E| h((X_{i_1},Y_{i_1}),...,(X_{i_6},Y_{i_6}))|^{\frac{\#\{i_1,...,i_6\}}{6}} < \infty, \end{align*} for any $1 \leq i_1 \leq \cdots \leq i_6 \leq 6$.

So my approach was to let $\tilde{h}$ be the symmetrized version of $h$: $$\tilde{h}((x_1,y_1),...,(x_6,y_6))=\frac{1}{6!} \sum_{\sigma\in \Pi_6} h((x_{\sigma(1)},y_{\sigma(1)}),...,(x_{\sigma(6)},y_{\sigma(6)})),$$
where $\Pi_6$ is the set of all permutations on $\{1,...,6\}$. Then the above conditions of the SLLN reduces to
\begin{align*}
E| h((X_{i_1},Y_{i_1}),...,(X_{i_6},Y_{i_6}))|^{\frac{\#\{i_1,...,i_6\}}{6}} < \infty,
\end{align*}
for all $(i_1,...,i_6)\in \{1,...,6\}^6$.
The only way (that i can see) we can create upper bounds for $h$ is by the triangle inequality, which amounts to the following inequalities (suppressing the $i$ index in the metric)
\begin{align}
\frac{|f_i(z_1,z_2,z_3,z_4)|}{2}\leq \left\{
\begin{array}{lll}
d(z_1,z_4), & d(z_2,z_3), & d(z_1,z_2) \lor d(z_1,z_3), \\
d(z_1,z_2) \lor d(z_1,z_4), & d(z_1,z_2) \lor d(z_2,z_3), & d(z_1,z_2) \lor d(z_2,z_4), \\
d(z_1,z_4)  \lor d(z_1,z_3), & d(z_1,z_4)  \lor d(z_2,z_3), & d(z_1,z_4)  \lor d(z_2,z_4), \\
d(z_2,z_3) \lor d(z_1,z_3), & d(z_2,z_3) \lor d(z_2,z_4),  & d(z_3,z_4) \lor d(z_1,z_3), \\
d(z_3,z_4) \lor d(z_1,z_4), & d(z_3,z_4) \lor d(z_2,z_3), &d(z_3,z_4) \lor d(z_2,z_4). \\  \end{array}
\right.
\end{align}
and we realize that if for example $i_1=i_2$ then every combination of the above inequalities result in
$$
E| h((X_{i_1},Y_{i_1}),...,(X_{i_6},Y_{i_6}))|^{\frac{\#\{i_1,...,i_6\}}{6}} \leq 4E([\gamma(X_{i_1},...)\theta(Y_{i_1},...)]^{\frac{\#\{i_1,...,i_6\}}{6}}),
$$
where $\gamma$ and $\theta$ is any of the above upper bounds.
We know that $E\gamma(X_{i_1},...)<\infty$ and $E\theta(Y_{i_1},...)<\infty$, for example
$$
Ed(X_{i_2},X_{i_3}) \lor d(X_{i_1},X_{i_3}) \leq E d(X_{i_2},x')+d(X_{i_3},x')+d(X_{i_1},x')+d(X_{i_3},x') < \infty.
$$
But we can't use independence to split up the expectation since $X_{i_1}$ may not be independent of $Y_{i_1}$. I realize that if $\#\{i_1,...,i_6\}\leq 3$ then Cauchy-Schwarz inequality can be used to say that
\begin{align*}
4E([\gamma(X_{i_1},...)\theta(Y_{i_1},...)]^{\frac{\#\{i_1,...,i_6\}}{6}})&\leq 4E([\gamma(X_{i_1},...)\theta(Y_{i_1},...)]^{1/2})+4 \\
&\leq 4[E\gamma(X_{i_1},...)]^{1/2}[E\theta(Y_{i_1},...)]^{1/2} +4< \infty,
\end{align*}
but if $3<\#\{i_1,...,i_6\}\leq 6$, then we can't utilize this either.
Thus: Is there another version of SLLN for $V$-statistics which is suitable in this situation,  or is there another way to bound the above expectation?

REPLY [1 votes]: You are right, the place is not properly explained in the article.
The problem is to show the integrability of $f_1(X_{i_1},X_{i_2},X_{i_3},X_{i_4})f_2(Y_{i_1},Y_{i_2},Y_{i_5},Y_{i_6})$ for all $i_1,i_2,\dots,i_6$, not just for distinct ones. First assume that $i_1,i_2,i_3,i_4$ are distinct and $i_1,i_2,i_5,i_6$ are distinct. Here one can employ other ideas of the article in order to establish the integrability. Namely, it is easy to see that
$$
f_1(z_1,z_2,z_3,z_4) =d_\mu(z_1,z_2)-d_\mu(z_1,z_3)-d_\mu(z_2,z_4)+d_\mu(z_3,z_4).\tag{1}
$$
Then we can estimate, using the Cauchy-Schwarz inequality,
$$
\mathbb E[f_1(X_{i_1},X_{i_2},X_{i_3},X_{i_4})f_2(Y_{i_1},Y_{i_2},Y_{i_5},Y_{i_6})] \\\le \big(\mathbb E[f_1(X_{i_1},X_{i_2},X_{i_3},X_{i_4})^2]\mathbb E[f_2(Y_{i_1},Y_{i_2},Y_{i_5},Y_{i_6})^2]\big)^{1/2}
$$
Further, by (1)
$$
\mathbb E[f_1(X_{i_1},X_{i_2},X_{i_3},X_{i_4})^2]\\\le 4\big(\mathbb E[d_\mu(X_{i_1},X_{i_2})^2] + \mathbb E[d_\mu(X_{i_1},X_{i_3})^2] + \mathbb E[d_\mu(X_{i_2},X_{i_4})^2] + \mathbb E[d_\mu(X_{i_3},X_{i_4})^2]\big).
$$
But $\mathbb E[d_\mu(X_{i},X_{j})^2] <\infty$ for $i\neq j$ (Lemma 2.1 of the article). Similarly,
$$
\mathbb E[f_2(Y_{i_1},Y_{i_2},Y_{i_5},Y_{i_6})^2]<\infty,
$$
qed.
When there are two coincidences, one can use Cauchy-Schwarz to proceed.
So let there be a single coincidence in some of $i_1,i_2,i_3,i_4$ be equal. If $i_1\neq i_2$, then we can proceed as in the article. Say, if $i_2=i_3$, we can estimate $f_1(X_{i_1},X_{i_2},X_{i_3},X_{i_4})\le g_1(X_{i_2},X_{i_3},X_{i_4})$, $f_2(\dots)\le g_2(Y_{i_1},Y_{i_5},Y_{i_6})$.
So the only remaining case is $i_1=i_2$. Here
$$
h = \big(d_1(X_{i_3},X_{i_4})-d_1(X_{i_1},X_{i_3})-d_1(X_{i_1},X_{i_4})\big) \big(d_2(Y_{i_5},Y_{i_6})-d_2(Y_{i_1},Y_{i_5})-d_2(Y_{i_1},Y_{i_6})\big).
$$
Upon expanding, there are many well-behaved terms. Ill-behaved come with the same (positive) sign and all of the same kind, so it is necessary and sufficient to establish the convergence of
$$
\frac{1}{n^4}\sum_{i,j,k=1}^n d_1(X_i,X_j)d_2(Y_i,Y_k)
$$
to zero. In turn, for the latter it is sufficient to have
$$
\mathbb{E}[d_1(X,X')^{3/4}d_2(Y,Y'')^{3/4}]<\infty,\tag{1}
$$
where $(X,Y)$ has the given distribution, and $X'$, $Y''$ have the given marginal distribution, and $(X,Y),X',Y''$ are independent. Assumption (1) might be necessary too, as for the classical Marcinkiewicz-Zygmund SLLN a similar condition is necessary. It does not seem though that (1) follows from the assumptions of the article. So I suggest you to inquire the author, how he deals with this.

UPD (from the discussion): here is an errata for original paper and Janson's paper, which shows that the consistency holds in the original formulation.