cp's OEIS Frontend

This is a front-end for the Online Encyclopedia of Integer Sequences, made by Christian Perfect. The idea is to provide OEIS entries in non-ancient HTML, and then to think about how they're presented visually. The source code is on GitHub.

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A260406 Numbers n such that (n-1)^2-1 divides 2^(n-1)-1.

Original entry on oeis.org

1, 3, 5, 17, 37, 257, 457, 1297, 2557, 4357, 6481, 8009, 11953, 26321, 44101, 47521, 47881, 49681, 57241, 65537, 74449, 84421, 97813, 141157, 157081, 165601, 225457, 278497, 310591, 333433, 365941, 403901, 419711, 476737, 557041, 560737, 576721, 647089, 1011961
Offset: 1

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Author

M. F. Hasler, Jul 24 2015

Keywords

Comments

The initial 1 is conventional.
647089 is the smallest composite number of this sequence (which makes it different from A081762).
The next composite number in this sequence is a(1000) = F_5 = 4294967297. - Robert G. Wilson v, Jul 25 2015
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 A260407 (apart from the conventional 1), i.e., the numbers such that (n-1)^4-1 divides 2^(n-1)-1.

Crossrefs

Programs

  • Magma
    [n: n in [3..6*10^5] | (2^(n-1)-1) mod ((n-1)^2-1) eq 0]; // Vincenzo Librandi, Jul 26 2015
  • Mathematica
    fQ[n_] := PowerMod[2, n - 1, (n - 1)^2 - 1] == 1; Select[ Range[3, 1200000], fQ] (* Robert G. Wilson v, Jul 25 2015 *)
  • PARI
    forstep(n=1,1e7,2,Mod(2,(n-1)^2-1)^(n-1)==1&&print1(n","))
    

A260407 Numbers n such that (n-1)^2+1 divides 2^(n-1)-1.

Original entry on oeis.org

1, 17, 257, 8209, 65537, 649801, 1382401, 4294967297
Offset: 1

Views

Author

M. F. Hasler, Jul 24 2015

Keywords

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

Crossrefs

Programs

  • Magma
    [n: n in [1..10^6] | (2^(n-1)-1) mod ((n-1)^2+1) eq 0 ]; // Vincenzo Librandi, Jul 25 2015
  • Mathematica
    Join [{1},Select[Range[43*10^8],PowerMod[2,#-1,(#-1)^2+1]==1&]] (* Harvey P. Dale, Sep 07 2018 *)
  • PARI
    forstep(n=1,1e7,2,Mod(2,(n-1)^2+1)^(n-1)==1&&print1(n","))
    

Formula

a(n) = A247165(n)+1.
Showing 1-2 of 2 results.