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%I #14 Mar 20 2020 10:35:37
%S 0,0,0,2,3,5,6,7,13,4,10,9,2,12,8,14,6,8,16,8,9,8,19,10,15,18,17,8,10,
%T 14,9,13,10,15,14,11,15,10,13,20,15,13,14,16,16,15,19,17,14,18,13,13,
%U 15,15,7,14,16,21,12,11,13,20,7,19,18,13,8,19
%N a(n) is the number of consecutive primes x-3,x+3 such that x=j*(p(n)#/3)/p(k), where 1 <= j < p(n+1) and 3 <= k <= n and p(k) doesn't divide j.
%C p(n) is the n-th prime; # denotes primorial (A002110).
%C a(n) seems to grow like 2*log(p(n)).
%H Jinyuan Wang, <a href="/A088968/b088968.txt">Table of n, a(n) for n = 1..100</a>
%e a(5) = 3 because for (j,k) = (1,3),(10,4),(8,5), j*(11#/3)/p(k)+-3 are consecutive primes.
%o (PARI) a(n) = {my(p=vector(n, i, prime(i)), x, y=2*prod(i=3, n, p[i])); sum(j=1, prime(n+1)-1, sum(k=3, n, j%p[k]>0 && ispseudoprime(x=j*y/p[k]-3) && nextprime(x+1)==x+6)); } \\ _Jinyuan Wang_, Mar 20 2020
%Y Cf. A002110, A087859, A087941.
%K nonn
%O 1,4
%A _Pierre CAMI_, Oct 29 2003
%E Edited by _Don Reble_, Nov 16 2005
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