A376012 -id:A376012 - cp's OEIS frontend

A277608 Least number of fractions of the form (k+1)/k, for k a positive integer, whose product equals n.

0, 1, 2, 2, 3, 3, 4, 3, 4, 4, 5, 4, 5, 5, 5, 4, 5, 5, 6, 5, 6, 6, 7, 5, 6, 6, 6, 6, 7, 6, 7, 5, 6, 6, 7, 6, 7, 7, 7, 6, 7, 7, 7, 7, 7, 8, 8, 6, 7, 7, 7, 7, 8, 7, 8, 7, 8, 8, 9, 7, 8, 8, 8, 6, 7, 7, 8, 7, 8, 8, 9, 7, 8, 8, 8, 8, 8, 8, 9, 7, 8, 8, 9, 8, 8, 8, 9, 8

Offset: 1

Joseph Myers, Oct 23 2016

nonn

If each intermediate product of the first j of the fractions, for all j < a(n), is also restricted to be an integer, the resulting sequence is A117497. The first n for which a shorter product can be obtained by allowing intermediate non-integer products is 43 = 2/1 * 2/1 * 2/1 * 2/1 * 2/1 * 4/3 * 129/128, a product of 7 fractions, where A117497(43) = 8.

Joseph Myers, Table of n, a(n) for n = 1..10000
United Kingdom Mathematics Trust, Mathematical Olympiad for Girls 2016, problem 5.

Cf. A117497 (restriction to intermediate products being integers), A014701 (always generating n from n-1 for n odd and from n/2 for n even), A376012.

A334127 Number of nonempty sets {p_1, p_2, ..., p_k} such that Product_{i=1..k} p_i divides Product_{i=1..k} (n + p_i), where the p_i are distinct primes.

Original entry on oeis.org

1, 3, 4, 7, 6, 19, 8, 17, 8, 25, 12, 105, 8, 35, 22, 24, 16, 59, 28, 77, 30, 26, 22, 159, 8, 117, 23, 161, 26, 787, 32, 69, 46, 57, 30, 534, 32, 69, 90, 137, 22, 707, 20, 266, 54, 73, 50, 423, 38, 626, 62, 229, 52, 1324, 220, 489, 130, 285, 58, 2943, 24, 119, 274, 171, 202, 12089, 46, 181, 158, 201, 66, 1999, 58, 391, 234, 917, 126, 451, 42, 1767, 73, 1034, 86, 34691, 81, 150, 142, 233, 94, 18319, 226, 477, 70, 477, 114, 4419, 54, 157, 234, 2252

Offset: 1

Jinyuan Wang, May 10 2020

nonn

a(n) is always finite. Proof: let p_1 < p_2 < ... < p_k, we can prove p_k <= 2*n^2 + n. If p_k > 2*n^2 + n, then 2*p_k > p_k + n, thus p_k - n is in the set. If p_k - m*n is in the set and m < n, then 2*(p_k - m*n) > p_k + n, thus p_k - (m+1)*n is in the set. Therefore, p_k - m*n are in the set for 0 <= m <= n. Since p_k - n*n > n + 1, p_k - m*n can be divisible by n + 1 for some m <= n, which is a contradiction to the p_i being primes.

			For n = 3, {3}, {2, 3}, {2, 5} and {2, 3, 5} are such sets, thus a(3) = 4.

Cf. A085098, A118086, A355626, A355627, A355628, A355629, A355630, A355631, A376011, A376012,

a(n, k=primepi(2*n^2+n)) = {my(c=-1, p=primes(k)); forsubset(k, v, if(vecprod(vector(#v, i, p[v[i]]+n))%vecprod(vector(#v, i, p[v[i]])) == 0, c++)); c; }

Terms a(13) onward from Max Alekseyev, Apr 08 2025

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A277608 Least number of fractions of the form (k+1)/k, for k a positive integer, whose product equals n.

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A334127 Number of nonempty sets {p_1, p_2, ..., p_k} such that Product_{i=1..k} p_i divides Product_{i=1..k} (n + p_i), where the p_i are distinct primes.

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