cp's OEIS Frontend

A247270 Let k == 1 or 5 (mod 6) (A007310). a(n) is the greatest number of the initial values of k such that k^2+6*n-4 divided by the maximal possible power of 3 takes only prime values or 1.

%I A247270 #8 Sep 13 2014 14:45:05
%S A247270 6,10,6,8,2,7,6,3,1,2,4,5,0,1,4,15,2,0,3,2,1,9,3,1,0,3,17,0,1,2,2,4,0,
%T A247270 1,1,7,5,0,0,2,3,1,0,1,2,0,3,0,1,2,6,2,0,1,2,1,1,0,1,0,0,7,0,2,2,2,0,
%U A247270 1,1,1,25,0,0,0,2,5,1,0,1,2,0,3,0,1,0,2
%N A247270 Let k == 1 or 5 (mod 6) (A007310). a(n) is the greatest number of the initial values of k such that k^2+6*n-4 divided by the maximal possible power of 3 takes only prime values or 1.
%C A247270 In the first 2*10^7 terms the numbers 15, 16, 17 and 25 appear only once. Here is the distribution:
%C A247270 0       16206595
%C A247270 1       3157812
%C A247270 2       547566
%C A247270 3       71442
%C A247270 4       12617
%C A247270 5       2848
%C A247270 6       817
%C A247270 7       211
%C A247270 8       53
%C A247270 9       20
%C A247270 10      11
%C A247270 11      2
%C A247270 12      2
%C A247270 13      0
%C A247270 14      0
%C A247270 15      1
%C A247270 16      1
%C A247270 17      1
%C A247270 18      0
%C A247270 19      0
%C A247270 20      0
%C A247270 21      0
%C A247270 22      0
%C A247270 23      0
%C A247270 24      0
%C A247270 25      1
%e A247270 For n=1, we have 1,1,17,41,19,97,121. So a(1)=6.
%K A247270 nonn
%O A247270 1,1
%A A247270 _Vladimir Shevelev_ and _Peter J. C. Moses_, Sep 11 2014