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 A260407 Numbers n such that (n-1)^2+1 divides 2^(n-1)-1. 2
 1, 17, 257, 8209, 65537, 649801, 1382401, 4294967297 (list; graph; refs; listen; history; text; internal format)
 OFFSET 1,2 COMMENTS 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 LINKS FORMULA a(n) = A247165(n)+1. MATHEMATICA Join [{1}, Select[Range[43*10^8], PowerMod[2, #-1, (#-1)^2+1]==1&]] (* Harvey P. Dale, Sep 07 2018 *) PROG (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 CROSSREFS Cf. A000215, A081762, A247165, A260072, A260406. Sequence in context: A090457 A174408 A260072 * A193329 A256499 A217796 Adjacent sequences:  A260404 A260405 A260406 * A260408 A260409 A260410 KEYWORD nonn,more AUTHOR M. F. Hasler, Jul 24 2015 STATUS approved

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Last modified July 6 08:51 EDT 2020. Contains 335476 sequences. (Running on oeis4.)