A341051 - OEIS

%I #59 Aug 26 2022 10:25:50

%S 4,5,6,7,8,9,11,11,11,12,14,14,15,15,16,17,18,19,20,20,21,22,23,24,25,

%T 25,26,26,27,27,28,28,29,30,31,31,32,32,33,33,34,35,36,36,37,37,38,38,

%U 39,40,40,41,41,42,43,43,44,44,45,46,46,47,47,48,48,49,49

%N a(n) is the largest possible number of n-digit integers that can be in geometric progression with common ratio > 1.

%C This sequence is the generalization of the 10th problem proposed during the 4th Canadian Mathematical Olympiad in 1972; the problem asked the question for the set of integers between 100 and 1000 (see Doob reference, link IMO and Example section for a(3)).

%C The corresponding ratios are always of the form b/(b-1) where b (A341053) must satisfy: 10^(-n/(k-1)) < 1/b < 1 - 10^(-1/(k-1)).

%C The successive ratios are 2, 3/2, 3/2, 4/3, then, for a(5) = 8, there exist two possible ratios:

%C -> with ratio = 4/3, the 8 integers go from 10935 to 81920, and,

%C -> with ratio = 5/4, the 8 integers go from 16384 to 78125.

%C a(n) = n iff n = 20, 21, 22, 23, 24, 25.

%D Michael Doob, The Canadian Mathematical Olympiad & L'Olympiade Mathématique du Canada 1969-1993, Canadian Mathematical Society & Société Mathématique du Canada, Problem 10, 1972, page 47, 1993.

%H Michel Marcus, <a href="/A341051/b341051.txt">Table of n, a(n) for n = 1..5000</a>

%H Diophante, <a href="http://www.diophante.fr/problemes-par-themes/arithmetique-et-algebre/a1-pot-pourri/2006-a10219-progressions-maximales">A10219, Progressions maximales</a> (in French).

%H The IMO Compendium, <a href="https://imomath.com/othercomp/Can/CanMO72.pdf">Problem 10</a>, 4th Canadian Mathematical Olympiad 1972.

%H <a href="/index/O#Olympiads">Index to sequences related to Olympiads</a>.

%e a(1) = 4 because (1, 2, 4, 8) is the largest possible string with 1-digit numbers that are in geometric progression; the ratio = 2.

%e a(2) = 5 because (16, 24, 36, 54, 81) is the largest possible string with 2-digit numbers that are in geometric progression; the ratio = 3/2.

%e a(3) = 6 because (128, 192, 288, 432, 648, 972) is the largest possible string with 3-digit numbers that are in geometric progression; the ratio = 3/2.

%o (PARI) a(n) = {my(b=1,m=1, t); while(10^n>t=b^m, while(ceil(10^(n-1)/t)*(b+1)^m<10^n, t=b^m++); b++); m; } \\ _Jinyuan Wang_, Feb 06 2021

%Y Cf. A341052 (number of ratios), A341053 (numerator of ratios).

%K nonn,base

%O 1,1

%A _Bernard Schott_, Feb 04 2021

%E a(8)-a(9) from _Metin Sariyar_, Feb 04 2021

%E More terms from _Jinyuan Wang_, Feb 06 2021