A366193 For n >= 0, a(n) is the least x >= 0 such that x^2 + (x + 2*n)^2 + 1 = p, p prime number (A000040).

1, 0, 0, 0, 1, 0, 9, 0, 0, 6, 0, 6, 0, 0, 3, 15, 1, 2, 0, 1, 0, 6, 1, 2, 6, 3, 9, 0, 0, 6, 15, 4, 5, 0, 3, 2, 6, 0, 2, 3, 1, 9, 0, 4, 3, 0, 7, 0, 3, 1, 6, 6, 1, 5, 6, 0, 2, 6, 0, 6, 0, 1, 0, 0, 13, 0, 6, 0, 6, 3, 4, 11, 12, 0, 3, 0, 9, 3, 0, 3, 0, 21, 9, 2, 3, 0, 6, 18, 0, 3

Offset: 0

Author

Ctibor O. Zizka, Oct 03 2023

Keywords

• nonn

Comments

For a(n) = 0 the resulting primes p >= 5 see in A002496.

Examples

			n = 0: x^2 + x^2 + 1 = p is valid for the least x = 1, p = 3, thus a(0) = 1.
n = 6: x^2 + (x + 12)^2 + 1 = p is valid for the least x = 9, p = 523, thus a(6) = 9.
		

Links

• Stoyan I. Dimitrov, A ternary diophantine inequality by primes with one of the form p = x^2 + y^2 + 1, arXiv:2011.03967 [math.NT], 2020.

Crossrefs

Cf. A000040, A001912, A002496.

Programs

PARI

a(n) = my(x=0); while (!isprime(x^2 + (x + 2*n)^2 + 1), x++); x; \\ Michel Marcus, Oct 03 2023

Formula

a(n) = 0 for n from A001912.

Extensions

More terms from Michel Marcus, Oct 03 2023