A309498 Least number k > 0 such that 4*p^2*k^2 + 1 is prime, where p = prime(n) is the n-th prime.

1, 1, 1, 1, 3, 1, 5, 15, 4, 2, 2, 1, 13, 5, 1, 6, 2, 13, 1, 2, 1, 3, 9, 5, 10, 5, 1, 5, 2, 9, 6, 8, 4, 2, 7, 3, 1, 1, 10, 11, 2, 7, 2, 1, 4, 5, 13, 4, 4, 3, 1, 3, 7, 2, 4, 6, 3, 7, 5, 2, 20, 6, 4, 2, 6, 1, 2, 1, 4, 3, 4, 3, 5, 5, 5, 16, 2, 14, 3, 3, 2, 2, 5, 5

Offset: 1

Author

Amiram Eldar, Aug 05 2019

Keywords

• nonn
• easy

Comments

Gagola calculated the first 669 terms of this sequence (for all the primes p < 5000) using an HP 9830 in 1981. She found that the largest value of k was only 45 and that 84% of the values of k were less than or equal to 10.

The Generalized Dickson Conjecture implies that the sequence contains each positive integer infinitely many times. - Robert Israel, Aug 05 2019

Examples

			a(1) = 1 since 4*1^2*prime(1)^2 + 1 = 4*1*2^2 + 1 = 17 is prime.
		

Links

• Jason Yuen, Table of n, a(n) for n = 1..10000
• Gloria Gagola, Progress on primes, News and Letters, Mathematics Magazine, Vol. 54, No. 1 (1981), p. 43.

Crossrefs

Cf. A002496, A005574, A035092, A052291.

Programs

Maple

f:= proc(n) local q,k;
  q:= 4*ithprime(n)^2;
  for k from 1 do
     if isprime(q*k^2+1) then return k fi
  od
end proc:
map(f, [$1..100]); # Robert Israel, Aug 05 2019

Mathematica

a[n_] := Module[{k = 1, p = Prime[n]}, While[!PrimeQ[4 * k^2 * p^2 + 1], k++]; k]; Array[a, 100]

PARI

a(n) = my(k=1, p=prime(n)); while (!isprime(4*p^2*k^2 + 1), k++); k; \\ Michel Marcus, Aug 05 2019