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A260407
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Numbers n such that (n-1)^2+1 divides 2^(n-1)-1.
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2
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OFFSET
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1,2
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COMMENTS
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a(7) = 1382401 is the first composite number of this sequence (which makes it different from A260072).
The Fermat numbers 2^(2^k)+1 = A000215(k) with k>1 are a subsequence of this sequence. I conjecture that they are equal to the intersection of this and A260406 (except for the conventional 1).
Conjecture: also numbers n such that ((2^k)^(n-1)-1) == 0 mod ((n-1)^2+1) for all k >= 1. - Jaroslav Krizek, Jun 02 2016
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LINKS
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FORMULA
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MATHEMATICA
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Join [{1}, Select[Range[43*10^8], PowerMod[2, #-1, (#-1)^2+1]==1&]] (* Harvey P. Dale, Sep 07 2018 *)
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PROG
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(PARI) forstep(n=1, 1e7, 2, Mod(2, (n-1)^2+1)^(n-1)==1&&print1(n", "))
(Magma) [n: n in [1..10^6] | (2^(n-1)-1) mod ((n-1)^2+1) eq 0 ]; // Vincenzo Librandi, Jul 25 2015
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CROSSREFS
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KEYWORD
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nonn,more
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AUTHOR
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STATUS
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approved
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