A366193 - OEIS

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

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OFFSET

0,7

COMMENTS

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

LINKS

Table of n, a(n) for n=0..89.

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.

FORMULA

a(n) = 0 for n from A001912.

EXAMPLE

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