OFFSET
1,1
COMMENTS
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)).
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)).
The successive ratios are 2, 3/2, 3/2, 4/3, then, for a(5) = 8, there exist two possible ratios:
-> with ratio = 4/3, the 8 integers go from 10935 to 81920, and,
-> with ratio = 5/4, the 8 integers go from 16384 to 78125.
a(n) = n iff n = 20, 21, 22, 23, 24, 25.
REFERENCES
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.
LINKS
Michel Marcus, Table of n, a(n) for n = 1..5000
Diophante, A10219, Progressions maximales (in French).
The IMO Compendium, Problem 10, 4th Canadian Mathematical Olympiad 1972.
EXAMPLE
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.
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.
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.
PROG
(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
CROSSREFS
KEYWORD
nonn,base
AUTHOR
Bernard Schott, Feb 04 2021
EXTENSIONS
a(8)-a(9) from Metin Sariyar, Feb 04 2021
More terms from Jinyuan Wang, Feb 06 2021
STATUS
approved