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Numbers k such that 398*k^2 - 1 is prime.
2

%I #14 Apr 09 2021 01:04:35

%S 1,3,4,5,6,7,8,9,10,11,13,14,17,18,19,20,22,24,25,27,28,29,33,34,37,

%T 38,43,44,46,47,51,52,54,55,58,59,60,67,68,71,73,75,79,80,81,82,83,85,

%U 86,87,89,90,93,94,95,96,97,100,103,106,107,108,110,112,114,116,117,119,121,124,125,128

%N Numbers k such that 398*k^2 - 1 is prime.

%C There are 414 such primes for 1 <= x <= 1000, and 3280 for 1 <= x <= 10000.

%H Robert Israel, <a href="/A338477/b338477.txt">Table of n, a(n) for n = 1..10000</a>

%H V. Granville, <a href="https://mathoverflow.net/questions/375133/quadratic-progressions-with-very-high-prime-density">Quadratic progressions with very high prime density</a>, MathOverflow.

%F a(n) = sqrt(A338476(n) + 1)/398.

%e a(3)=4 is in the sequence because 398*4^2 - 1 = 6367 is prime.

%p select(t -> isprime(398*t^2-1), [$1..1000]);

%Y Cf. A338476.

%Y Cf. A331947, where 398 is a term.

%K nonn

%O 1,2

%A _Robert Israel_, Oct 29 2020