

A155828


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


3



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
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OFFSET

1,8


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.  ZhiWei Sun, Feb 11 2009


LINKS

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


FORMULA

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


EXAMPLE

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


PROG

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


CROSSREFS

Cf. A060594
Sequence in context: A060901 A087612 A260626 * A226203 A051997 A155744
Adjacent sequences: A155825 A155826 A155827 * A155829 A155830 A155831


KEYWORD

nonn


AUTHOR

John W. Layman, Jan 28 2009


STATUS

approved



