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.

%I A334127 #17 Apr 08 2025 17:50:29
%S A334127 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,
%T A334127 117,23,161,26,787,32,69,46,57,30,534,32,69,90,137,22,707,20,266,54,
%U A334127 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
%N 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.
%C A334127 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.
%e A334127 For n = 3, {3}, {2, 3}, {2, 5} and {2, 3, 5} are such sets, thus a(3) = 4.
%o A334127 (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; }
%Y A334127 Cf. A085098, A118086, A355626, A355627, A355628, A355629, A355630, A355631, A376011, A376012,
%K A334127 nonn
%O A334127 1,2
%A A334127 _Jinyuan Wang_, May 10 2020
%E A334127 Terms a(13) onward from _Max Alekseyev_, Apr 08 2025