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A089008
Numbers k such that 18*k^2 + 1 is prime.
4
1, 2, 3, 7, 8, 9, 10, 11, 12, 14, 15, 22, 24, 25, 29, 31, 32, 33, 34, 35, 41, 44, 45, 51, 52, 54, 59, 62, 63, 67, 68, 73, 76, 79, 80, 85, 88, 91, 95, 99, 100, 102, 107, 108, 109, 117, 119, 120, 122, 125, 129, 131, 133, 135, 139, 141, 142, 143, 147, 150, 152, 154, 156
OFFSET
1,2
COMMENTS
There are 8 consecutive terms at n=13537 and n=105819293 for n < 10^9. - Jean C. Lambry, Oct 19 2015
Since 18*k^2 + 1 is divisible by 17 for k == 4, 13 (mod 17), the maximum possible number of consecutive terms is 8, in which case the first term must be congruent to 5 modulo 17 and 7 or 8 modulo 11. - Jianing Song, Nov 14 2021
FORMULA
a(n) = A089001(n+1)/3.
MATHEMATICA
Select[Range[200], PrimeQ[18#^2+1]&] (* Harvey P. Dale, Apr 25 2011 *)
PROG
(PARI) for(n=0, 1e3, if(isprime(k=(18*n^2 + 1)), print1(n", "))) \\ Altug Alkan, Oct 19 2015
CROSSREFS
KEYWORD
nonn
AUTHOR
N. J. A. Sloane, Dec 20 2003
STATUS
approved