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%I #7 Oct 19 2017 03:14:16
%S 1,1,1,0,3,3,4,4,6,2,1,10,5,3,9,6,6,4,5,8,6,7,19,25,11,2,1,3,9,23,7,7,
%T 39,5,7,2,1,5,78,2,1,15,19,12,17,6,3,14,8,21,23,17,14,40,16,6,8,13,15,
%U 5,15,82,46,51,39,43,6,11,61,57,16,2,1,26,54,2,1,13,4,62,31,69,27,155,21
%N Smallest k such that k*(k+1)*(k+2)*...*(k+n-1) + 1 is prime, or 0 if no such number exists.
%C The product of four consecutive integers + 1 is always composite (a square), so a(4) = 0. Are there any more zeros in the sequence?
%C Since rather large numbers (up to 193 digits) are encountered in the computation, the Pocklington-Lehmer "P-1" primality test is used, as implemented in PARI 2.1.3.
%e 1*2*3*4*5 + 1 = 121 = 11*11 and 2*3*4*5*6 + 1 = 721 = 7*103 are composite, but 3*4*5*6*7 + 1 = 2521 is prime, so a(5) = 3.
%o (PARI) m=1000; for(n=1,85,b=0; k=1; while(b<1&&k<m,if(!isprime(prod(j=k,k+n-1,j)+1,1),k++,b=1)); print1(if(k<m,k,0),","))
%Y Cf. A083520, A083521.
%K nonn
%O 1,5
%A _Amarnath Murthy_ and Meenakshi Srikanth (menakan_s(AT)yahoo.com), May 05 2003
%E Edited and extended by _Klaus Brockhaus_ and _Don Reble_, May 06 2003
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