A376011 -id:A376011 - cp's OEIS frontend

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.

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

cp's OEIS Frontend

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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