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.

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, 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, 1, 1, 1, 25, 0, 0, 0, 2, 5, 1, 0, 1, 2, 0, 3, 0, 1, 0, 2

Offset: 1

Author

Vladimir Shevelev and Peter J. C. Moses, Sep 11 2014

Keywords

• nonn

Comments

In the first 2*10^7 terms the numbers 15, 16, 17 and 25 appear only once. Here is the distribution:

0 16206595

1 3157812

2 547566

3 71442

4 12617

5 2848

6 817

7 211

8 53

9 20

10 11

11 2

12 2

13 0

14 0

15 1

16 1

17 1

18 0

19 0

20 0

21 0

22 0

23 0

24 0

25 1

Examples

			For n=1, we have 1,1,17,41,19,97,121. So a(1)=6.