A155828 Number of integers k in {1,2,3,..,n} such that kn+1 is a square.

0, 0, 1, 1, 1, 1, 1, 3, 1, 1, 1, 3, 1, 1, 3, 3, 1, 1, 1, 3, 3, 1, 1, 7, 1, 1, 1, 3, 1, 3, 1, 3, 3, 1, 3, 3, 1, 1, 3, 7, 1, 3, 1, 3, 3, 1, 1, 7, 1, 1, 3, 3, 1, 1, 3, 7, 3, 1, 1, 7, 1, 1, 3, 3, 3, 3, 1, 3, 3, 3, 1, 7, 1, 1, 3, 3, 3, 3, 1, 7, 1, 1, 1, 7, 3, 1, 3, 7, 1, 3, 3, 3, 3, 1, 3, 7, 1, 1, 3, 3, 1, 3, 1, 7, 7

Offset: 1

Author

John W. Layman, Jan 28 2009

Keywords

• nonn

Comments

Conjecture: All terms a(n) are of the form 2^m - 1. This has been verified up to n = 1000.

It appears that the terms of this sequence are exactly one less than those of A060594, indicating that the terms are related to the square roots of unity mod n. - John W. Layman, Feb 04 2009

The conjecture that a(n) is of the form 2^m - 1 holds trivially. In fact, the conjecture can be restated as follows: For any positive integer n, the solutions of the congruence x^2 = 1 (mod n) with 1 <= x <= n is a power of two. By the Chinese remainder theorem, this reduces to the case when n is a prime power. It is well known that the solution of x^2 = 1 (mod p^a) is two when p is an odd prime (one may obtain this by induction on a). As for the number of solutions of the congruence x^2 = 1 (mod 2^a), it equals 1, 2, 4 according to whether a=1, 2, or a>2. - Zhi-Wei Sun, Feb 11 2009

Examples

			1*8 + 1 = 9, 3*8 + 1 = 25, 6*8 + 1 = 49, whereas other values are not square, so a(8) = 3.
		

Links

• Antti Karttunen, Table of n, a(n) for n = 1..10000

Crossrefs

Cf. A060594.

Programs

PARI

a(n) = sum(k=1, n, issquare(k*n+1)); \\ Michel Marcus, Jul 10 2014

PARI

a(n)=if(n>2, 2^#znstar(n)[3]-1) \\ Charles R Greathouse IV, Jun 18 2025

Formula

a(n) = A060594(n) - 1. - after Zhi-Wei Sun, Michel Marcus, Jul 10 2014