We Love Commas


« In music, the schisma is the interval between a Pythagorean comma and a syntonic comma. » (Wikipedia)

Here is my latest post to the Math-Fun mailing list (May 3rd, 2020):

Hello Math-Fun,
S = 2,5,26,159,1447,10274,45206,280278,2298281,...
If we frame any comma above with its two closest digits we get 2.5 for the first comma, 5.2 for the second comma, 6.1 for the third one, etc.

 Look now: 
the division of 5 by 2 starts with 2.5
the division of 26 by 5 starts with 5.2
the division of 159 by 26 starts with 6.1
the division of 1447 by 159 starts with 9.1, etc.

 [In other words, a(n+1)/a(n) starts with [x.y] with x = the rightmost digit of a(n) and y = the leftmost digit of a(n+1)].

 As usual, we want S to be the lexicographically earliest seq of distinct terms with this property... but!

 Here comes the (Math-)fun part...

 I thought that S would increase monotonically for ever – when I bumped into a(9) = 2298280 [and not, as above, a(9) = 2298281].
I realized then that S could decrease at some point! Indeed, the following example works: 120,24,... as the result of 24/120 starts with 0.2

 But this "zero comma" trick doesn't systematically work with integers ending in 0 (it works rarely, in reality). If it doesn't when you need it, you'll have to backtrack – which consists most of the time in increasing by 1 the term ending in "0" that would stop the sequence; this is what I did above with a(9).

 The question remains: will the "+1" trick always work? Or will S stop at some point? Enter into a loop? Extend for ever? 
[I've noticed that the ratio a(n+1)/a(n) gets closer at every step to [x.y], (the divisor), but I don't know if it means something for the future of S].
Best,
É.

 ____________________
Update, 19:00 Brussels time, same day

 Jean-Marc Falcoz was quick to compute the hereunder 62-term list: merci beaucoup, Jean-Marc !

 S = 2, 5, 26, 159, 1447, 10274, 45206, 280278, 2298281, 2757938, 22615092, 56537732, 118729239, 1080436075, 5942398412, 12479036665, 69882605324, 293506942361, 381559025069, 3548498933141, 4967898506397, 36265659096698, 297378404592924, 1219251458830989, 11095188275362001, 12204707102898201, 13425177813188022, 29535391189013652, 79745556210336864, 342905891704448512, 994427085942900864, 4375479178148764161, 7438314602852899842, 15620460665991090177, 110905270728536752128, 987056909483977080832, 2171525200864749813761, 2605830241037700300801, 3387579313349010653184, 13889075184730944569344, 63889745849762352988162, 134168466284500953858048, 1086764576904457704439808, 9672204734449674992222208, 85115401663157146803503104, 365996227151575708921233408, 3037768685358078631436353536, 18530388980684281191078035456, 113035372782174132198055084032, 248677820120783118983218855936, 1516934702736777191164183838721, 1668628173010454910280602222592, 3837844797924046969185329217536, 23794637747129090871179069095936, 145147290257487474130030681915392, 333838767592221246343705947799552, 968132426017441700865860094132224, 4259782674476743944978386256920576, 26410652581755813957663950781808641, 34333848356282558144963136016351232, 99568160233219433838956955257798656, 657149857539248344502789829023498241, ...

« The earliest instance of this anecdote known to QI appeared on May 8, 1884 in “The Daily Graphic: An Illustrated Evening Newspaper” of New York City under the title “The Casual Observer”. »



Commentaires

Enregistrer un commentaire

Posts les plus consultés de ce blog

A square for three (chess)

Image
(English translation after the French text) Voici cinq problèmes d’échecs disjoints : a ) combien faut-il de coups au minimum pour que trois pions soient capturés sur la même case ? b ) trois tours c ) trois cavaliers d ) trois fous e ) trois dames Here are five disjointed chess problems: a) how many moves are required at least for three pawns to be captured on the same square? b) three rooks c) three knights d) three bishops e) three queens Solutions a ) (corrigé grâce à Orion , ici ) 3,5 coups pour 3 pions capturés en b5 (4 moves for White and 3 moves for Black   — 3 pawns  captured on b5)  b ) 5  coups pour 3 tours capturées en a3 (ci-dessous – merci aussi à Orion !-) (5 moves for White and 5 moves for Black  — 3 rooks  captured on a3  hereunder )  c ) 5  coups pour 3 cavaliers capturés en d5  ( ci-dessous) (5 moves for White and 5 moves for Black   — 3 knights  captured on d5  hereunder )  (merci Orion !-) d ) 7  coups pour 3 fous capturés en h6  ( ci-dessous) (7 moves for Whit
Lire la suite

Beautés ?

Image
Beau comme un clavier ? Un deuxième ? (celui de Modotti)   Beau comme le dos d ’ un timbre proposé par Delcampe ? Beau comme une série ukrainienne où les avions râpent le Kremlin ? Belle comme Chargois ? ... belle comme Germaine Richier ? Bel atelier ? Beau comme une pièce de 1 santim ? ... ou comme une pièce de 50  santeem ? ... Byrrh , beau comme un anus ? Beau comme la collection de Marin Karmitz à Beaubourg ? Belles Ménines ? Beau comme Marcel Duchamp ? Beau comme Jean-Luc Godard ? Beau comme la réouverture de l ’ Einsteinturm à Berlin ? Beau comme un anneau dudit ? Beau comme Eisenstein ? Beau comme les fleurs de Pascale hier matin ? Beau comme on dirait la planète Mars ? Beau comme des recherches personnelles ? Beau comme une marelle (vers le ciel) ? Beau comme un triple selfie (au Man Ray ) ? Beau comme le ciel de Sacha , mardi, en face du restaurant La Montagne ? (qui a rouvert) ... beau comme ce bleu-là de Lo ? Beau comme le froid ? Beau comme une Vespa ? Beau comme Cloon
Lire la suite

Underline, reproduce

Image
Dall.e creation If we underline the... « first, second, third, fourth, fifth, sixth, seventh, eighth, ninth, tenth, eleventh,… » letter of this sentence S 1 , we see that the successive underlined letters simply reproduce S 1 . No big deal. If we decide not to start S 1   with "first" but with "sixth", we can reproduce the new S 6  quite easily: S 6 = sixth , second, third, fourth, fifth, first , seventh, eighth, ninth, tenth, eleventh,… S 6  was reproduced in a slightly less mechanical manner than  S 1 :  just two words of  S 1   swapped here ("first" and "sixth", in red, above). If we start S 8  with "eighth" we just swap again a pair of words of S 1  ("first" and "eighth", in red, hereunder): S 8  = eighth , second, third, fourth, fifth, sixth, seventh, first , ninth, tenth, eleventh,… What about starting S 9  with "ninth"? How many swaps do we need to reproduce S 9 ?  This is more tricky! (Indeed, we
Lire la suite