

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.


0



1, 3, 4, 7, 6, 19, 8, 17, 8, 25, 12, 105
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OFFSET

1,2


COMMENTS

a(n) always exists. 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_i + 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_i + 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



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



KEYWORD

nonn,more


AUTHOR



STATUS

approved



