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Numbers k such that 18*k^2 + 1 is prime.
4

%I #28 Nov 15 2021 01:59:57

%S 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,

%T 54,59,62,63,67,68,73,76,79,80,85,88,91,95,99,100,102,107,108,109,117,

%U 119,120,122,125,129,131,133,135,139,141,142,143,147,150,152,154,156

%N Numbers k such that 18*k^2 + 1 is prime.

%C There are 8 consecutive terms at n=13537 and n=105819293 for n < 10^9. - _Jean C. Lambry_, Oct 19 2015

%C 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

%H Zak Seidov, <a href="/A089008/b089008.txt">Table of n, a(n) for n = 1..1000</a>

%F a(n) = A089001(n+1)/3.

%t Select[Range[200],PrimeQ[18#^2+1]&] (* _Harvey P. Dale_, Apr 25 2011 *)

%o (PARI) for(n=0, 1e3, if(isprime(k=(18*n^2 + 1)), print1(n", "))) \\ _Altug Alkan_, Oct 19 2015

%Y Cf. A089001, A090612, A090698.

%K nonn

%O 1,2

%A _N. J. A. Sloane_, Dec 20 2003