A155828 - OEIS

%I #19 Aug 23 2017 06:11:31

%S 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,

%T 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,

%U 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

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

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

%C 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

%C 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

%H Antti Karttunen, <a href="/A155828/b155828.txt">Table of n, a(n) for n = 1..10000</a>

%F a(n) = A060594(n) - 1. - after _Zhi-Wei Sun_, _Michel Marcus_, Jul 10 2014

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

%o (PARI) a(n) = sum(k=1, n, issquare(k*n+1)); \\ _Michel Marcus_, Jul 10 2014

%Y Cf. A060594

%K nonn

%O 1,8

%A _John W. Layman_, Jan 28 2009