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%I #40 May 26 2019 18:51:59
%S 1,1,1,1,3,1,4,0,1,1,5,1,3,0,1,1,6,1,7,0,1,1,6,0,1,0,1,1,6,1,9,0,0,0,
%T 3,1,11,0,1,1,9,1,5,0,1,1,10,0,2,0,1,1,9,0,2,0,1,1,10,1,9,0,0,0,3,1,9,
%U 0,1,1,8,1,9,0,0,0,5,1,9,0,1,1,11,0,1,0,1,1,8,0,3,0,0,0,2,1,10,0,1,1,10,1
%N a(n) is the number of primes of the form k! + n, 1 <= k <= n.
%C Note the triples of consecutive zeros in a(n) for n = {{32,33,34}, {62,63,64}, {74,75,76}, {92,93,94}, {116,117,118}, {122,123,124}, {140,141,142}, {152,153,154}, {158,159,160}, {182,183,184}, {200,201,202}, {206,207,208}, {212,213,214}, {218,219,220}, {242,243,244}, {272,273,274}, {284,285,286}, ...}. The middle index of most zero triples is a multiple of 3. See A125164.
%C The first consecutive quintuple of zeros has indices n = {294,295,296,297,298}, where the odd zero index n = 295 is not a multiple of 3.
%C Also for n >= 2, a(n) is the number of primes of the form k! + n for all k, since n divides k! + n for k >= n. Note that it is not known whether there are infinitely many primes of the form k! + 1; see A088332 for such primes and A002981 for the indices k. - _Jianing Song_, Jul 28 2018
%H Michel Marcus, <a href="/A125162/a125162.txt">Example table</a>
%e a(n) is the length of n-th row in the table of numbers k such that k! + n is a prime, 1 <= k <= n.
%e n: numbers k
%e -------------
%e 1: {1},
%e 2: {1},
%e 3: {2},
%e 4: {1},
%e 5: {2, 3, 4},
%e Thus a(1)-a(4) = 1, a(5) = 3.
%e See Example table link for more rows.
%t Table[Length[Select[Range[n],PrimeQ[ #!+n]&]],{n,1,300}]
%o (PARI) a(n)=c=0;for(k=1,n,if(ispseudoprime(k!+n),c++));c
%o vector(100,n,a(n)) \\ _Derek Orr_, Oct 15 2014
%Y Cf. A125163 (indices of 0), A125164 (triples).
%K nonn
%O 1,5
%A _Alexander Adamchuk_, Nov 21 2006
%E Name clarified by _Jianing Song_, Jul 28 2018
%E Edited by _Michel Marcus_, Jul 29 2018
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