

A125211


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


3



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
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OFFSET

1,3


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.


LINKS

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


EXAMPLE

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.


MATHEMATICA

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


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.
Sequence in context: A204933 A240224 A118105 * A139367 A117648 A037222
Adjacent sequences: A125208 A125209 A125210 * A125212 A125213 A125214


KEYWORD

nonn


AUTHOR

Alexander Adamchuk, Nov 23 2006


STATUS

approved



