A125211 a(n) = total number of primes of the form |k! - n|.

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, 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, 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

Offset: 1

Author

Alexander Adamchuk, Nov 23 2006

Keywords

• nonn

Comments

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.

Examples

			a(4) = 3 because there are 3 primes of the form |k! - 4|:
1! - 4 = -3, 2! - 4 = -2, 3! - 4 = 2.
k! - 4 is composite for all k>3 because it is divisible by 4.
		

Links

• Amiram Eldar, Table of n, a(n) for n = 1..2500

Crossrefs

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.

Programs

Mathematica

Table[Length[Select[Range[n],PrimeQ[ #!-n]&]],{n,1,300}]