%I #9 Apr 16 2021 04:36:08
%S 0,0,2,3,2,1,3,2,2,0,5,1,7,1,1,0,9,1,6,1,2,1,4,1,2,1,1,0,5,1,8,1,1,0,
%T 2,0,10,1,1,0,6,1,10,1,1,0,10,1,3,0,0,0,7,1,2,0,0,0,7,1,11,1,1,0,2,0,
%U 9,1,1,0,9,1,11,1,1,0,4,0,11,1,1,0,8,1,3,0,0,0,14,1,3,0,0,0,2,0,11,1,1,0,9
%N a(n) = total number of primes of the form |k! - n|.
%C Numbers n such that a(n) = 0 are listed in A125212(n) = {1,2,10,16,28,34,36,40,46,50,51,52,56,57,58,64,66,70,76,78,82,86,87,88,92,93,94,96,100,...} Numbers n such that no prime exists of the form k! - n. Note the triples of consecutive zeros in a(n) for n = {{50,51,52}, {56,57,58}, {86,87,88}, {92,93,94}, ...}. Most zeros in a(n) have even indices. The middle index of most consecutive zero triples is odd and is a multiple of 3. Numbers n such that no prime exists of the form (k! - 3n - 1), (k! - 3n), (k! - 3n + 1) are listed in A125213(n) = {17,19,29,31,45,49,57,59,63,69,73,79,83,85,87,89,97,99,...}. The first pair of odd middle indices of zero triples that are not divisible by 3 is n = 325 and n = 329. They belong to the first septuplet of consecutive zeros in a(n): a(324)-a(330) = 0.
%H Amiram Eldar, <a href="/A125211/b125211.txt">Table of n, a(n) for n = 1..2500</a>
%e a(4) = 3 because there are 3 primes of the form |k! - 4|:
%e 1! - 4 = -3, 2! - 4 = -2, 3! - 4 = 2.
%e k! - 4 is composite for all k>3 because it is divisible by 4.
%t Table[Length[Select[Range[n],PrimeQ[ #!-n]&]],{n,1,300}]
%Y Cf. A125162 = number of primes of the form k! + n. Cf. A125163 = numbers n such that no prime exists of the form k! + n. Cf. A125164 = numbers n such that no prime exists of the form (k! + 3n - 1), (k! + 3n), (k! + 3n + 1). Cf. A125212, A125213.
%K nonn
%O 1,3
%A _Alexander Adamchuk_, Nov 23 2006
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