cp's OEIS Frontend

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

%I A155828 #21 Jun 18 2025 11:59:09
%S A155828 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 A155828 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 A155828 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 A155828 Number of integers k in {1,2,3,..,n} such that kn+1 is a square.
%C A155828 Conjecture: All terms a(n) are of the form 2^m - 1. This has been verified up to n = 1000.
%C A155828 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 A155828 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 A155828 Antti Karttunen, <a href="/A155828/b155828.txt">Table of n, a(n) for n = 1..10000</a>
%F A155828 a(n) = A060594(n) - 1. - after _Zhi-Wei Sun_, _Michel Marcus_, Jul 10 2014
%e A155828 1*8 + 1 = 9, 3*8 + 1 = 25, 6*8 + 1 = 49, whereas other values are not square, so a(8) = 3.
%o A155828 (PARI) a(n) = sum(k=1, n, issquare(k*n+1)); \\ _Michel Marcus_, Jul 10 2014
%o A155828 (PARI) a(n)=if(n>2, 2^#znstar(n)[3]-1) \\ _Charles R Greathouse IV_, Jun 18 2025
%Y A155828 Cf. A060594.
%K A155828 nonn
%O A155828 1,8
%A A155828 _John W. Layman_, Jan 28 2009