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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.
0
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
1,1
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
EXAMPLE
For n=1, we have 1,1,17,41,19,97,121. So a(1)=6.
CROSSREFS
Sequence in context: A339590 A074288 A156383 * A010726 A084365 A066135
KEYWORD
nonn
AUTHOR
STATUS
approved