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A341051
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a(n) is the largest possible number of n-digit integers that can be in geometric progression with common ratio > 1.
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5
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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, 25, 26, 26, 27, 27, 28, 28, 29, 30, 31, 31, 32, 32, 33, 33, 34, 35, 36, 36, 37, 37, 38, 38, 39, 40, 40, 41, 41, 42, 43, 43, 44, 44, 45, 46, 46, 47, 47, 48, 48, 49, 49
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OFFSET
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1,1
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COMMENTS
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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.
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REFERENCES
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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.
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LINKS
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The IMO Compendium, Problem 10, 4th Canadian Mathematical Olympiad 1972.
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EXAMPLE
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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.
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PROG
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(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
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CROSSREFS
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KEYWORD
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nonn,base
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AUTHOR
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EXTENSIONS
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STATUS
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approved
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