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

1,2

Comments

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.

Links

Table of n, a(n) for n=1..100.

Example

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

Prog

(PARI) 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; }

Crossrefs

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

Sequence in context: A069213 A130700 A117134 * A095001 A377378 A344458

Adjacent sequences: A334124 A334125 A334126 * A334128 A334129 A334130

Keyword

nonn

Author

Jinyuan Wang, May 10 2020

Extensions

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

Status

approved