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

%I A153091 #38 Jul 04 2025 01:27:16
%S A153091 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,
%T A153091 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,
%U A153091 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
%N A153091 a(n) = least positive k such that k*5^n*(5^n-1)+j is prime, with j = -1 or 1 or both.
%H A153091 Pierre CAMI, <a href="/A153091/b153091.txt">Table of n, a(n) for n = 1..1000</a>.
%F A153091 Limit_{n->oo} ( (Sum_{i=1..n} a(i)) / (n*(n+1)/2) ) = 13*log(5)/40.
%F A153091 Limit_{n->oo} ( (Sum_{i=1..n} a(2*i)) / (n*(n+1)) ) = log(5)/4.
%F A153091 Limit_{n->oo} ( (Sum_{i=1..n} a(2*i+1)) / (n*(n+2)) ) = 2*log(5)/5.
%e A153091 For n = 1, 1*5^1*(5^1-1)-1 = 19 is prime, so a(1) = 1.
%e A153091 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.
%e A153091 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.
%t A153091 A153091[n_] := Module[{k = 0}, While[NoneTrue[++k*# + {-1, 1}, PrimeQ]] & [5^n*(5^n - 1)]; k];
%t A153091 Array[A153091, 100] (* _Paolo Xausa_, Jun 30 2025 *)
%Y A153091 Cf. A152414, A152090, A152092, A152093, A152094, A152095.
%K A153091 nonn
%O A153091 1,3
%A A153091 _Pierre CAMI_, Dec 18 2008
%E A153091 a(5) corrected by _Paolo Xausa_, Jun 30 2025