A153091 a(n) = least positive k such that k*5^n*(5^n-1)+j is prime, with j = -1 or 1 or both.

1, 1, 3, 1, 2, 5, 5, 1, 2, 2, 18, 12, 12, 7, 1, 1, 4, 1, 9, 2, 36, 10, 70, 1, 3, 16, 6, 3, 2, 9, 74, 4, 6, 19, 20, 8, 14, 2, 2, 62, 3, 29, 47, 11, 47, 16, 58, 1, 49, 18, 51, 3, 12, 5, 18, 23, 1, 19, 54, 7, 35, 12, 7, 1, 12, 3, 5, 121, 70, 89, 12, 61, 33, 36, 9, 17, 135, 35, 21, 23, 20, 86, 18

Offset: 1

Author

Pierre CAMI, Dec 18 2008

Keywords

• nonn

Examples

			For n = 1, 1*5^1*(5^1-1)-1 = 19 is prime, so a(1) = 1.
For n = 2, 1*5^2*(5^2-1)-1 = 599 is prime, as well as 1*5^2*(5^2-1)+1 = 601, so a(2) = 1.
For n = 3, k = 3 is the least k satisfying the required condition: 3*5^3*(5^3-1)-1 = 46499 is prime, so a(3) = 3.
		

Links

• Pierre CAMI, Table of n, a(n) for n = 1..1000.

Crossrefs

Cf. A152414, A152090, A152092, A152093, A152094, A152095.

Programs

Mathematica

A153091[n_] := Module[{k = 0}, While[NoneTrue[++k*# + {-1, 1}, PrimeQ]] & [5^n*(5^n - 1)]; k];
Array[A153091, 100] (* Paolo Xausa, Jun 30 2025 *)

Formula

Limit_{n->oo} ( (Sum_{i=1..n} a(i)) / (n*(n+1)/2) ) = 13*log(5)/40.

Limit_{n->oo} ( (Sum_{i=1..n} a(2*i)) / (n*(n+1)) ) = log(5)/4.

Limit_{n->oo} ( (Sum_{i=1..n} a(2*i+1)) / (n*(n+2)) ) = 2*log(5)/5.

Extensions

a(5) corrected by Paolo Xausa, Jun 30 2025