A343240 The number of solutions x from {0, 1, ..., A343238(n)-1} of the congruence x^2 + 5 == 0 (mod A343238(n)) is given by a(n).

1, 1, 2, 1, 2, 2, 2, 1, 2, 2, 2, 4, 2, 2, 2, 2, 2, 2, 4, 2, 2, 2, 2, 2, 2, 2, 2, 4, 2, 4, 2, 2, 2, 2, 2, 4, 2, 2, 2, 2, 2, 2, 4, 2, 2, 2, 2, 4, 4, 2, 4, 2, 2, 4, 4, 2, 4, 2, 4, 2, 2, 2, 2, 4, 2, 2, 4, 4, 4, 2, 4, 2, 2, 4

Offset: 1

Author

Wolfdieter Lang, May 16 2021

Keywords

• nonn
• easy

Comments

Row length of irregular triangle A343239.

Examples

			a(19) = 4 because A343238(19) = 42 = 2*3*7 has 2^(1+1) = 4 solutions from the primes 3 and 7.
		

Crossrefs

Cf. A343238, A343239.

Programs

PARI

isok(k) = issquare(Mod(-5, k)); \\ A343238
lista(nn) = my(list = List()); for (n=1, nn, if (issquare(Mod(-5, n)), listput(list, sum(i=0, n-1, Mod(i,n)^2 + 5 == 0)););); Vec(list); \\ Michel Marcus, Sep 17 2023

Formula

a(n) = row length of A343239(n), for n >= 1.

a(1) = a(2) = a(4) = a(8) = 1, and otherwise a(n) = 2^{number of distinct primes from A139513}, that is, primes congruent to {1, 3, 7, 9} (mod 20), appearing in the prime factorization of A343238(n).