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A262807 a(n) = (Product_{k=1..n} prime(k+1)) mod (Sum_{k=1..n} prime(k+1)) where prime(k) is the k-th prime number.

%I A262807 #21 Oct 11 2015 11:25:37
%S A262807 0,7,0,11,0,7,45,91,24,55,0,113,93,175,308,153,414,395,273,355,609,
%T A262807 779,558,23,0,843,962,185,0,547,1634,21,170,1149,1455,2483,1830,2275,
%U A262807 2865,1989,0,1515,1211,2013,1105,403,2733,819,0,4011,0,1457,4278,1155,391,1717,2596,2163,0,5985
%N A262807 a(n) = (Product_{k=1..n} prime(k+1)) mod (Sum_{k=1..n} prime(k+1)) where prime(k) is the k-th prime number.
%C A262807 Remainder when product of first n odd primes is divided by sum of first n odd primes.
%C A262807 Obviously a(2n) cannot be 0. Does 0 appear in the sequence infinitely often?
%F A262807 a(n) = A070826(n+1) mod A071148(n).
%e A262807 a(1) = prime(2) mod prime(2) = 3 mod 3 = 0.
%e A262807 a(2) = (prime(2) * prime(3)) mod (prime(2) + prime(3)) = 15 mod 8 = 7.
%e A262807 a(3) = (prime(2) * prime(3) * prime(4)) mod (prime(2) + prime(3) + prime(4)) = 105 mod 15 = 0.
%e A262807 a(4) = (prime(2) * prime(3) * prime(4) * prime(5)) mod (prime(2) + prime(3) + prime(4) + prime(5)) = 1155 mod 26 = 11.
%t A262807 Table[Mod[Product[Prime[k + 1], {k, n}], Sum[Prime[k + 1], {k, n}]], {n, 60}] (* _Michael De Vlieger_, Oct 02 2015 *)
%o A262807 (PARI) a(n) = prod(k=1, n, prime(k+1)) % sum(k=1, n, prime(k+1));
%o A262807 vector(60, n, a(n))
%Y A262807 Cf. A065091, A070826, A071089, A071148, A259643.
%K A262807 nonn,easy
%O A262807 1,2
%A A262807 _Altug Alkan_, Oct 02 2015