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Graham's number
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This article is about the large number named after Ronald Graham. For the investing term named after Benjamin Graham, see Graham number.
Graham's number is an enormous number that arises as an upper bound on the answer of a problem in the mathematical field of Ramsey theory. It is named after mathematician Ronald Graham, who used the number as a simplified explanation of the upper bounds of the problem he was working on in conversations with popular science writer Martin Gardner. Gardner later described the number in Scientific American in 1977, introducing it to the general public. At the time of its introduction, it was the largest specific positive integer ever to have been used in a published mathematical proof. The number was published in the 1980 Guinness Book of World Records, adding to its popular interest. Other specific integers (such as TREE(3)) known to be far larger than Graham's number have since appeared in many serious mathematical proofs, for example in connection with Harvey Friedman's various finite forms of Kruskal's theorem. Additionally, smaller upper bounds on the Ramsey theory problem from which Graham's number derived have since been proven to be valid.
Graham's number is much larger than many other large numbers such as Skewes' number and Moser's number, both of which are in turn much larger than a googolplex. As with these, it is so large that the observable universe is far too small to contain an ordinary digital representation of Graham's number, assuming that each digit occupies one Planck volume, possibly the smallest measurable space. But even the number of digits in this digital representation of Graham's number would itself be a number so large that its digital representation cannot be represented in the observable universe. Nor even can the number of digits of that number. And so forth, for a number of times far exceeding the total number of Planck volumes in the observable universe. Thus Graham's number cannot even be expressed in this way by power towers of the form a b c ⋅ ⋅ ⋅ {\displaystyle \scriptstyle a^{b^{c^{\cdot ^{\cdot ^{\cdot }}}}}} .
However, Graham's number can be explicitly given by computable recursive formulas using Knuth's up-arrow notation or equivalent, as was done by Graham. As there is a recursive formula to define it, it is much smaller than typical busy beaver numbers. Though too large to be computed in full, the sequence of digits of Graham's number can be computed explicitly through simple algorithms. The last 12 digits are: 262464195387. With Knuth's up-arrow notation, Graham's number is g 64 {\displaystyle g_{64}} where:
g n = { 3 ↑↑↑↑ 3 , n = 1 3 ↑ g n − 1 3 , n ≥ 2 , n ∈ N {\displaystyle g_{n}=\left\{{\begin{matrix}3\uparrow \uparrow \uparrow \uparrow 3,&n=1\\3\uparrow ^{g_{n-1}}3,&n\geq 2,n\in \mathbb {N} \end{matrix}}\right.}
Contents
[hide]
1 Context
2 Publication
3 Definition
3.1 Magnitude
4 Rightmost decimal digits
5 References
5.1 Bibliography
6 External links
Context[edit]
Example of a 2-colored 3-dimensional cube containing one single-coloured 4-vertex coplanar complete subgraph. The subgraph is shown below the cube. Note that this cube would contain no such subgraph if, for example, the bottom edge in the present subgraph were replaced by a blue edge – thus proving by counterexample that N* > 3.
Graham's number is connected to the following problem in Ramsey theory:
Connect each pair of geometric vertices of an n-dimensional hypercube to obtain a complete graph on 2n vertices. Colour each of the edges of this graph either red or blue. What is the smallest value of n for which every such colouring contains at least one single-coloured complete subgraph on four coplanar vertices?
In 1971, Graham and Rothschild proved that this problem has a solution N*, giving as a bound 6 ≤ N* ≤ N, with N being a large but explicitly defined number F 7 ( 12 ) = F ( F ( F ( F ( F ( F ( F ( 12 ) ) ) ) ) ) ) {\displaystyle \scriptstyle F^{7}(12)\;=\;F(F(F(F(F(F(F(12)))))))} , where F ( n ) = 2 ↑ n 3 {\displaystyle \scriptstyle F(n)\;=\;2\uparrow ^{n}3} in Knuth's up-arrow notation; the number is between 4 → 2 → 8 → 2 and 2 → 3 → 9 → 2 in Conway chained arrow notation.[1] This was reduced in 2014 via upper bounds on the Hales–Jewett number to N ′ = 2 ↑↑↑ 6 {\displaystyle \scriptstyle N'\;=\;2\;\uparrow \uparrow \uparrow \;6} .[2] The lower bound of 6 was later improved to 11 by Geoffrey Exoo in 2003,[3] and to 13 by Jerome Barkley in 2008.[4] Thus, the best known bounds for N* are 13 ≤ N* ≤ N'.
Graham's number, G, is much larger than N: f 64 ( 4 ) {\displaystyle \scriptstyle f^{64}(4)} , where f ( n ) = 3 ↑ n 3 {\displaystyle \scriptstyle f(n)\;=\;3\uparrow ^{n}3} . This weaker upper bound for the problem, attributed to an unpublished work of Graham, was eventually published and named by Martin Gardner in Scientific American in November 1977.[5]
Publication[edit]
The number gained a degree of popular attention when Martin Gardner described it in the "Mathematical Games" section of Scientific American in November 1977, writing that Graham had recently established, in an unpublished proof, "a bound so vast that it holds the record for the largest number ever used in a serious mathematical proof." The 1980 Guinness Book of World Records repeated Gardner's claim, adding to the popular interest in this number. According to physicist John Baez, Graham invented the quantity now known as Graham's number in conversation with Gardner. While Graham was trying to explain a result in Ramsey theory which he had derived with his collaborator Bruce Lee Rothschild, Graham found that the quantity now known as Graham's number was easier to explain than the actual number appearing in the proof. Because the number which Graham described to Gardner is larger than the number in the paper itself, both are valid upper bounds for the solution to the problem studied by Graham and Rothschild.[6]
Definition[edit]
Using Knuth's up-arrow notation, Graham's number G (as defined in Gardner's Scientific American article) is
G = 3 ↑↑ ⋯ ⋯ ⋯ ⋯ ⋯ ↑ ⏟ 3 3 ↑↑ ⋯ ⋯ ⋯ ⋯ ↑ ⏟ 3 ⋮ ⏟ 3 ↑↑ ⋯ ⋅ ⋅ ↑ ⏟ 3 3 ↑↑↑↑ 3 } 64 layers {\displaystyle \left.{\begin{matrix}G&=&3\underbrace {\uparrow \uparrow \cdots \cdots \cdots \cdots \cdots \uparrow } 3\\&&3\underbrace {\uparrow \uparrow \cdots \cdots \cdots \cdots \uparrow } 3\\&&\underbrace {\qquad \;\;\vdots \qquad \;\;} \\&&3\underbrace {\uparrow \uparrow \cdots \cdot \cdot \uparrow } 3\\&&3\uparrow \uparrow \uparrow \uparrow 3\end{matrix}}\right\}{\text{64 layers}}}
where the number of arrows in each subsequent layer is specified by the value of the next layer below it; that is,
G = g 64 , where g 1 = 3 ↑↑↑↑ 3 , g n = 3 ↑ g n − 1 3 , {\displaystyle G=g_{64},{\text{ where }}g_{1}=3\uparrow \uparrow \uparrow \uparrow 3,\ g_{n}=3\uparrow ^{g_{n-1}}3,}
and where a superscript on an up-arrow indicates how many arrows there are. In other words, G is calculated in 64 steps: the first step is to calculate g1 with four up-arrows between 3s; the second step is to calculate g2 with g1 up-arrows between 3s; the third step is to calculate g3 with g2 up-arrows between 3s; and so on, until finally calculating G = g64 with g63 up-arrows between 3s.
Equivalently,
G = f 64 ( 4 ) , where f ( n ) = 3 ↑ n 3 , {\displaystyle G=f^{64}(4),{\text{ where }}f(n)=3\uparrow ^{n}3,}
and the superscript on f indicates an iteration of the function, e.g., f 4 ( n ) = f ( f ( f ( f ( n ) ) ) ) {\displaystyle \scriptstyle f^{4}(n)\;=\;f(f(f(f(n))))} . Expressed in terms of the family of hyperoperations H 0 , H 1 , H 2 , ⋯ {\displaystyle \scriptstyle {\text{H}}_{0},{\text{H}}_{1},{\text{H}}_{2},\cdots } , the function f is the particular sequence f ( n ) = H n + 2 ( 3 , 3 ) {\displaystyle \scriptstyle f(n)\;=\;{\text{H}}_{n+2}(3,3)} , which is a version of the rapidly growing Ackermann function A(n,n). (In fact, f ( n ) > A ( n , n ) {\displaystyle \scriptstyle f(n)\;>\;A(n,\,n)} for all n.) The function f can also be expressed in Conway chained arrow notation as f ( n ) = 3 → 3 → n {\displaystyle \scriptstyle f(n)\;=\;3\rightarrow 3\rightarrow n} , and this notation also provides the following bounds on G:
3 → 3 → 64 → 2 < G < 3 → 3 → 65 → 2. {\displaystyle 3\rightarrow 3\rightarrow 64\rightarrow 2<G<3\rightarrow 3\rightarrow 65\rightarrow 2.\,}
Magnitude[edit]
To convey the difficulty of appreciating the enormous size of Graham's number, it may be helpful to express—in terms of exponentiation alone—just the first term (g1) of the rapidly growing 64-term sequence. First, in terms of tetration ( ↑↑ {\displaystyle \scriptstyle \uparrow \uparrow } ) alone:
g 1 = 3 ↑↑↑↑ 3 = 3 ↑↑↑ ( 3 ↑↑↑ 3 ) = 3 ↑↑ ( 3 ↑↑ ( 3 ↑↑ … ( 3 ↑↑ 3 ) … ) ) {\displaystyle g_{1}=3\uparrow \uparrow \uparrow \uparrow 3=3\uparrow \uparrow \uparrow (3\uparrow \uparrow \uparrow 3)=3\uparrow \uparrow (3\uparrow \uparrow (3\uparrow \uparrow \ \dots \ (3\uparrow \uparrow 3)\dots ))}
where the number of 3s in the expression on the right is
3 ↑↑↑ 3 = 3 ↑↑ ( 3 ↑↑ 3 ) . {\displaystyle 3\uparrow \uparrow \uparrow 3\ =\ 3\uparrow \uparrow (3\uparrow \uparrow 3).}
Now each tetration ( ↑↑ {\displaystyle \scriptstyle \uparrow \uparrow } ) operation reduces to a power tower ( ↑ {\displaystyle \scriptstyle \uparrow } ) according to the definition
3 ↑↑ X = 3 ↑ ( 3 ↑ ( 3 ↑ … ( 3 ↑ 3 ) … ) ) = 3 3 ⋅ ⋅ ⋅ 3 where there are X 3s . {\displaystyle 3\uparrow \uparrow X\ =\ 3\uparrow (3\uparrow (3\uparrow \dots (3\uparrow 3)\dots ))\ =\ 3^{3^{\cdot ^{\cdot ^{\cdot ^{3}}}}}\quad {\text{where there are X 3s}}.}
Thus,
g 1 = 3 ↑↑ ( 3 ↑↑ ( 3 ↑↑ … ( 3 ↑↑ 3 ) … ) ) where the number of 3s is 3 ↑↑ ( 3 ↑↑ 3 ) {\displaystyle g_{1}=3\uparrow \uparrow (3\uparrow \uparrow (3\uparrow \uparrow \ \dots \ (3\uparrow \uparrow 3)\dots ))\quad {\text{where the number of 3s is}}\quad 3\uparrow \uparrow (3\uparrow \uparrow 3)}
becomes, solely in terms of repeated "exponentiation towers",
g 1 = 3 3 ⋅ ⋅ ⋅ ⋅ 3 } 3 3 ⋅ ⋅ ⋅ 3 } … 3 3 3 } 3 where the number of towers is 3 3 ⋅ ⋅ ⋅ 3 } 3 3 3 } 3 {\displaystyle g_{1}=\left.{\begin{matrix}3^{3^{\cdot ^{\cdot ^{\cdot ^{\cdot ^{3}}}}}}\end{matrix}}\right\}\left.{\begin{matrix}3^{3^{\cdot ^{\cdot ^{\cdot ^{3}}}}}\end{matrix}}\right\}\dots \left.{\begin{matrix}3^{3^{3}}\end{matrix}}\right\}3\quad {\text{where the number of towers is}}\quad \left.{\begin{matrix}3^{3^{\cdot ^{\cdot ^{\cdot ^{3}}}}}\end{matrix}}\right\}\left.{\begin{matrix}3^{3^{3}}\end{matrix}}\right\}3}
and where the number of 3s in each tower, starting from the leftmost tower, is specified by the value of the next tower to the right.
In other words, g1 is computed by first calculating the number of towers, n = 3 ↑ 3 ↑ 3 … ↑ 3 {\displaystyle \scriptstyle n\;=\;3\uparrow 3\uparrow 3\ \dots \ \uparrow 3} (where the number of 3s is 3 ↑ 3 ↑ 3 = 7625597484987 {\displaystyle \scriptstyle 3\uparrow 3\uparrow 3\;=\;7625597484987} ), and then computing the nth tower in the following sequence:
      1st tower:  3 2nd tower: 3↑3↑3 (number of 3s is 3) = 7625597484987 3rd tower: 3↑3↑3↑3↑...↑3 (number of 3s is 7625597484987) = … ⋮ g1 = nth tower: 3↑3↑3↑3↑3↑3↑3↑...↑3 (number of 3s is given by the n-1th tower)

where the number of 3s in each successive tower is given by the tower just before it. Note that the result of calculating the third tower is the value of n, the number of towers for g1.
The magnitude of this first term, g1, is so large that it is practically incomprehensible, even though the above display is relatively easy to comprehend. Even n, the mere number of towers in this formula for g1, is far greater than the number of Planck volumes (roughly 10185 of them) into which one can imagine subdividing the observable universe. And after this first term, still another 63 terms remain in the rapidly growing g sequence before Graham's number G = g64 is reached. To illustrate just how fast this sequence grows, while g1 is equal to 3 ↑↑↑↑ 3 {\displaystyle 3\uparrow \uparrow \uparrow \uparrow 3} with only four up arrows, the number of up arrows in g2 is this incomprehensibly large number g1.
Rightmost decimal digits[edit]
Graham's number is a "power tower" of the form 3↑↑n (with a very large value of n), so its rightmost decimal digits must satisfy certain properties common to all such towers. One of these properties is that all such towers of height greater than d (say), have the same sequence of d rightmost decimal digits. This is a special case of a more general property: The d rightmost decimal digits of all such towers of height greater than d+2, are independent of the topmost "3" in the tower; i.e., the topmost "3" can be changed to any other non-negative integer without affecting the d rightmost digits.
The following table illustrates, for a few values of d, how this happens. For a given height of tower and number of digits d, the full range of d-digit numbers (10d of them) does not occur; instead, a certain smaller subset of values repeats itself in a cycle. The length of the cycle and some of the values (in parentheses) are shown in each cell of this table:
Number of different possible values of 3↑3↑…3↑x when all but the rightmost d decimal digits are ignored
Number of digits (d)
3↑x
3↑3↑x
3↑3↑3↑x
3↑3↑3↑3↑x
3↑3↑3↑3↑3↑x
1
4
(1,3,9,7) 2
(3,7) 1
(7) 1
(7) 1
(7)
2
20
(01,03,…,87,…,67) 4
(03,27,83,87) 2
(27,87) 1
(87) 1
(87)
3
100
(001,003,…,387,…,667) 20
(003,027,…387,…,587) 4
(027,987,227,387) 2
(987,387) 1
(387)
The particular rightmost d digits that are ultimately shared by all sufficiently tall towers of 3s are in bold text, and can be seen developing as the tower height increases. For any fixed number of digits d (row in the table), the number of values possible for 3 ↑ {\displaystyle \scriptstyle \uparrow } 3↑…3↑x mod 10d, as x ranges over all nonnegative integers, is seen to decrease steadily as the height increases, until eventually reducing the "possibility set" to a single number (colored cells) when the height exceeds d+2.
A simple algorithm[7] for computing these digits may be described as follows: let x = 3, then iterate, d times, the assignment x = 3x mod 10d. Except for omitting any leading 0s, the final value assigned to x (as a base-ten numeral) is then composed of the d rightmost decimal digits of 3↑↑n, for all n > d. (If the final value of x has fewer than d digits, then the required number of leading 0s must be added.)
Let k be the numerousness of these stable digits, which satisfy the congruence relation G(mod 10k)≡[GG](mod 10k).
k=t-1, where G(t):=3↑↑t.[8] It follows that, g63 ≪ k ≪ g64.
The algorithm above produces the following 500 rightmost decimal digits of Graham's number (or of any tower of more than 500 3s):
  …02425950695064738395657479136519351798334535362521
   43003540126026771622672160419810652263169355188780
   38814483140652526168785095552646051071172000997092
   91249544378887496062882911725063001303622934916080
   25459461494578871427832350829242102091825896753560
   43086993801689249889268099510169055919951195027887
   17830837018340236474548882222161573228010132974509
   27344594504343300901096928025352751833289884461508
   94042482650181938515625357963996189939679054966380
   03222348723967018485186439059104575627262464195387

References[edit]
Jump up ^ "Graham's number records". Iteror.org. Retrieved 2014-04-09.
Jump up ^ Lavrov, Mikhail; Lee, Mitchell; Mackey, John (2014). "Improved upper and lower bounds on a geometric Ramsey problem". European Journal of Combinatorics. 42: 135–144. doi:10.1016/j.ejc.2014.06.003.
Jump up ^ Exoo, Geoffrey (2003). "A Euclidean Ramsey Problem". Discrete & Computational Geometry. 29 (2): 223–227. doi:10.1007/s00454-002-0780-5. Exoo refers to the Graham & Rothschild upper bound N by the term "Graham's number". This is not the "Graham's number" G published by Martin Gardner.
Jump up ^ Barkley, Jerome (2008). "Improved lower bound on an Euclidean Ramsey problem". arXiv:0811.1055  [math.CO].
Jump up ^ Martin Gardner (1977) "In which joining sets of points leads into diverse (and diverting) paths". Scientific American, November 1977
Jump up ^ John Baez (2013). "A while back I told you about Graham's number.."
Jump up ^ "The Math Forum @ Drexel ("Last Eight Digits of Z")". Mathforum.org. Retrieved 2014-04-09.
Jump up ^ Ripà, Marco (2011). La strana coda della serie n^n^...^n (in Italian). UNI Service. ISBN 978-88-6178-789-6.
Bibliography[edit]
Gardner, Martin (November 1977). "Mathematical Games" (PDF). Scientific American. 237: 18–28. doi:10.1038/scientificamerican1177-18. ; reprinted (revised) in Gardner (2001), cited below.
Gardner, Martin (1989). Penrose Tiles to Trapdoor Ciphers. Washington, D.C.: Mathematical Association of America. ISBN 0-88385-521-6.
Gardner, Martin (2001). The Colossal Book of Mathematics: Classic Puzzles, Paradoxes, and Problems. New York, NY: Norton. ISBN 0-393-02023-1.
Graham, R. L.; Rothschild, B. L. (1971). "Ramsey's Theorem for n-Parameter Sets" (PDF). Transactions of the American Mathematical Society. 159: 257–292. doi:10.2307/1996010. JSTOR 1996010. The explicit formula for N appears on p. 290. This is not the "Graham's number" G published by Martin Gardner.
Graham, R. L.; Rothschild, B. L. (1978). "Ramsey Theory". In Rota, G-C. Studies in Combinatorics (MAA Studies in Mathematics). 17. Mathematical Association of America. pp. 80–99. ISBN 0-88385-117-2. On p. 90, in stating "the best available estimate" for the solution, the explicit formula for N is repeated from the 1971 paper.
External links[edit]
Sbiis Saibian's article on Graham's number
"A Ramsey Problem on Hypercubes" by Geoff Exoo
Mathworld article on Graham's number
How to calculate Graham's number
Conceptualizing Graham's number
Some Ramsey results for the n-cube prepublication mentions Graham's number
Padilla, Tony; Parker, Matt. "Graham's Number". Numberphile. Brady Haran.
Ron Graham. "What is Graham's Number? (feat Ron Graham)" (video). Numberphile. Brady Haran.
Ron Graham. "How Big is Graham's Number? (feat Ron Graham)" (video). Numberphile. Brady Haran.
The last 16 million digits of Graham's number by the Darkside communication group
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Where Do You Go To (My Lovely)?
From Wikipedia, the free encyclopedia
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This article needs additional citations for verification. Please help improve this article by adding citations to reliable sources. Unsourced material may be challenged and removed. (July 2009) (Learn how and when to remove this template message)
"Where Do You Go To (My Lovely)?"
Single by Peter Sarstedt
from the album Peter Sarstedt
B-side
"Morning Mountain"
Released
January 1969 (1969-01)
Format
7"
Genre
Folk
pop
waltz
Length
4:42
Label
United Artists
Songwriter(s)
Peter Sarstedt
Producer(s)
Ray Singer
Peter Sarstedt singles chronology
"I Am a Cathedral"
(1968) "Where Do You Go To (My Lovely)?"
(1969) "Frozen Orange Juice"
(1969)
"Where Do You Go To (My Lovely)?" is a song by the British singer-songwriter Peter Sarstedt. Its recording was produced by Ray Singer, engineered by John Mackswith at Lansdowne Recording Studios and released in 1969. It was a number-one 1 hit in the UK Singles Chart for four weeks in 1969,[1] and was awarded the 1969 Ivor Novello Award for Best Song Musically and Lyrically.[2] In the United States, the record reached No. 61 on the Cash Box Top 100 Singles. The single also peaked at No. 70 on the Billboard Hot 100 that May.
The music has been described as "a faux European waltz tune"[3] and the arrangement is a very simple one of strummed acoustic guitar and bass guitar, with brief bursts of French-style accordion at the start and the end. The arranger and conductor was Ian Green.[4]
Contents
1 Lyrics
2 Inspiration
3 Reception
4 Sequel
5 References
Lyrics[edit]
This section does not cite any sources. Please help improve this section by adding citations to reliable sources. Unsourced material may be challenged and removed. (March 2014) (Learn how and when to remove this template message)
The song is about a fictional girl named Marie-Claire who grows up on the poverty-stricken backstreets of Naples, becomes a member of the jet set, and goes on to live in Paris. The lyrics describe her from the perspective of a childhood friend; it is left unclear whether they have remained close. The rhetorical question of the title suggests that her glamorous lifestyle might not have brought Marie-Claire happiness or contentment.
Even though Sarstedt himself was not French, the song benefited from the contemporary awareness in Britain of such French and Belgian singers as Serge Gainsbourg and Jacques Brel (Belgium-born of Flemish descent).
The lyrics contain a large number of contemporary and other references:
Marlene Dietrich: German American actress and singer
Zizi Jeanmaire: French ballerina
Pierre Balmain: French designer of elegant fashions
Boulevard Saint-Michel: street in the Latin Quarter of Paris
The Rolling Stones: popular British rock and roll band
Sacha Distel: French singer and musician
Sorbonne: University of Paris
Picasso: Spanish pioneer of modern art
Juan-les-Pins: fashionable beach resort on the French Riviera
Topless swimsuit: first conceived by Austrian-American fashion designer Rudi Gernreich in 1964
Saint Moritz: fashionable ski resort in the Engadin, Swiss Alps
Napoleon brandy: a blended brandy in which the youngest brandy of the blend has been aged for at least six years
Aga Khan: world-travelling Islamic leader and racehorse owner
The version on the album Peter Sarstedt is longer than the radio edit version released as a single, having extra stanzas beginning "You go to the embassy parties..." and "You're in between twenty and thirty...".
Inspiration[edit]
It is often suspected that the name Marie-Claire is inspired by Marie Claire magazine, a women's fashion weekly that began in 1937 in France. One theory says that the song is about the Italian actress Sophia Loren, who was abandoned by her father and had a poverty-stricken life in Naples. Another theory has the song being inspired by Danish singer and actress Nina van Pallandt.
According to Alan Cooper, "Sarstedt insisted it was not written with actress Sophia Loren in mind. 'Yes, it's a portrait of a poor-born girl who becomes a member of the European jet-set. And yes, there's reference to her growing up on the 'back streets of Naples', so I can see why people may think it was written with Sophia Loren in mind. But that's just a coincidence. I really wasn't thinking of anyone specific.'"[5]
The song was written in Copenhagen.[6][7]
In 2009, Sarstedt spoke to a gossip columnist for the Daily Express. He admitted he had lied about the song being about a socialite who died in a fire. He said that the song was about his girlfriend at the time, whom he later married and then divorced. According to Mark Steyn, "Anita is now a dentist in Copenhagen. Peter Sarstedt has spent 40 years singing about wanting to look inside her head. And for most of that time Anita has made a living by looking inside yours."[8]
Reception[edit]
The song was a hit far exceeding Sarstedt's other work, although he is not a one-hit wonder. In 1998 he was earning £60,000 annually in royalties from it.[9]
DJ John Peel repeatedly stated that the song was one of his least favourites. On his show on BFBS radio on 1 July 1989, he said: 'It's a terrible, smug, self-satisfied, hideous record. Really have hated it ever since I first heard it'.[citation needed]
The song was used as a soundtrack in Wes Anderson's 2007 film The Darjeeling Limited, and in Jennifer Saunders' 2016 film Absolutely Fabulous: The Movie.
Sequel[edit]
In 1997 Sarstedt recorded a sequel, "The Last of the Breed (Lovely 2)" on his CD England's Lane.[10] This picks up the story of Marie Claire twenty years on, living now in London. It names more people and places, including Belgravia, Ballets Russes, Cape Town, Claridge's, Gstaad, John Galliano, Harrods, Jerusalem, Long Island, Milan, Rudolf Nureyev, Palm Beach, Rio de Janeiro, and Isabella Rossellini.[11] In recent years,[when?] Sarstedt and a co-writer were working on a further sequel, "Farewell Marie-Claire", in which the story was brought to a conclusion. The song was to feature the same waltz feel as the original. But Sarstedt's retirement from the music industry meant that the track was abandoned.[citation needed]
References[edit]
Jump up ^ Roberts, David (2006). British Hit Singles & Albums (19th ed.). London: Guinness World Records Limited. pp. 229–30. ISBN 1-904994-10-5.
Jump up ^ Lister, David (27 May 1994). "Pop ballads bite back in lyrical fashion". The Independent. Retrieved 26 June 2017.
Jump up ^ Eder, Bruce. "Peter Sarstedt: biography". Allmusic. Retrieved 7 October 2012.
Jump up ^ "Record Details". 45cat. Retrieved 26 June 2017.
Jump up ^ "60's chart topper Peter Sarstedt visits the KZN Midlands". Treeroutes.co.za. Archived from the original on 2014-07-22. Retrieved 2014-03-26.
Jump up ^ Confirmation of this can be found on the CD cover of The Best of Peter Sarstedt, EMI, nr. 8297622, Australian CD.
Jump up ^ "One of those songs by Sarstedt". Sydney Morning Herald. 27 June 1982. Retrieved 2014-05-11.
Jump up ^ Steyn, Mark. "Where Do You Go To My Lovely". Steyn Online. Mark Steyn. Retrieved 10 January 2017.
Jump up ^ Murphy, Peter (27 May 1998). "So Where Did You Go To, My Lovely?". Hot Press. Dublin. Retrieved 2 September 2016.
Jump up ^ England's Lane released as Round Tower RTM CD89
Jump up ^ "Untitled" (PDF). Archived from the original (PDF) on 2 May 2006. Retrieved 26 June 2017.

hide
v
t
e
Ivor Novello Awards – Best Song Musically and Lyrically
1960s
"She's Leaving Home" (1968)
1970s
"Where Do You Go To (My Lovely)?" (1970)
"Something" (1971)
"Don't Let It Die" (1972)
"Without You" (1973)
"Daniel" (1974)
"Streets of London"* (1975)
"Don't Cry for Me Argentina" (1978)
"Baker Street" (1979)
1980s
"The Logical Song" (1980)
"Woman in Love" (1981)
"Memory" (1982)
"Have You Ever Been in Love" (1983)
"Every Breath You Take" (1984)
"Against All Odds" (1985)
"Nikita" (1986)
"Don't Give Up" (1987)
"Something Inside So Strong" (1988)
"They Dance Alone" (1989)
1990s
"The Living Years" (1990)
"Sacrifice" (1991)
"The Whole of the Moon" (1992)
"Why" (1993)
"If I Ever Lose My Faith in You" (1994)
"Think Twice" (1995)
"Common People" (1996)
"Too Much Love Will Kill You" (1997)
"Paranoid Android" (1998)
"Believe" (1999)
2000s
"Strong" (2000)
"Babylon" (2001)
"Walk On" (2002)
"The Other Side" (2003)
"Leave Right Now" (2004)
"Dry Your Eyes" (2005)
"Suddenly I See" (2006)
"Elusive" (2007)
"Love Is a Losing Game" (2008)
"One Day Like This" (2009)
2010s
"The Fear" (2010)
"Becoming a Jackal" (2011)
"The A Team" (2012)
"Next to Me" (2013)
"Strong" (2014)
"Take Me to Church" (2015)
"Wasn't Expecting That" (2016)
"Black Man in a White World" (2017)
*Awarded as The Best Song
Retrieved from "https://en.wikipedia.org/w/index.php?title=Where_Do_You_Go_To_(My_Lovely)%3F&oldid=853149281"
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