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A155828
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Number of integers k in {1,2,3,..,n} such that kn+1 is a square.
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0
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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
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1,8
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COMMENTS
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Conjecture: All terms a(n) are of the form 2^m-1. This has been verified up to n=1000.
Comment from Zhi-Wei SUN (zwsun(AT)nju.edu.cn), Feb 11 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 as a=1, 2, or a>2.
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. [From John W. Layman, Feb 04 2009]
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LINKS
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Table of n, a(n) for n=1..105.
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EXAMPLE
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1*8+1=9, 3*8+1=25, 6*8+1=49, whereas other values are not square, so a(8)=3.
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CROSSREFS
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A060594 [From John W. Layman, Feb 04 2009]
Sequence in context: A091842 A060901 A087612 * A226203 A051997 A155744
Adjacent sequences: A155825 A155826 A155827 * A155829 A155830 A155831
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KEYWORD
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nonn
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AUTHOR
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John W. Layman, Jan 28 2009
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STATUS
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approved
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