A292834 Numbers m, not powers of 2, such that the least prime factor of 2^m + 1 is congruent to 1 (mod m).

24, 48, 112, 160, 192, 272, 448, 496, 656, 688, 832, 896, 1088, 1152, 1168, 1328, 1360, 1408, 1472, 1520, 1664, 1744, 1920, 1984, 2176, 2304, 2432, 2560, 2688, 2752, 2816, 2944, 2960, 3056, 3072, 3200, 3328, 3520, 3664, 3712, 3776, 4672, 4864, 4928, 5120, 5376, 5552, 5888, 6144

Offset: 1

Author

Thomas Ordowski, Sep 24 2017

Keywords

• hard
• nonn

Comments

Problem: are there infinitely many such numbers?

Theorem: there are no numbers m in the sequence such that, for each prime factor p of 2^m + 1, p == 1 (mod m).

Proof: if all prime factors p of 2^m + 1 are p == 1 (mod m), then 2^m + 1 == 1 (mod m), thus 2^m == 0 (mod m), so m = 2^k.

From Theorem in A002586, all terms are == 0 (mod 8). - Robert G. Wilson v, Jan 02 2018

Links

• Robert G. Wilson v, Table of n, a(n) for n = 1..71

Crossrefs

Cf. A002586, A292559.

Programs

Mathematica

Select[Range[200], And[! IntegerQ @ Log2 @ #, Mod[FactorInteger[2^# + 1][[1, 1]], #] == 1] &] (* Michael De Vlieger, Sep 24 2017 *)
fQ[n_] := If[ OddQ@ n || IntegerQ@ Log2@ n || PrimeQ[2^n +1], False, Block[{p = 3}, While[PowerMod[2, n, p] +1 != p, p = NextPrime@ p]; Mod[p, n] == 1]] (* Robert G. Wilson v, Jan 01 2018 *)

PARI

isok(n) = my(e = valuation(n, 2)); (2^e != n) && ((vecmin(factor(2^n+1)[,1]) % n) == 1); \\ Michel Marcus, Nov 13 2017

Extensions

a(9)-a(15) from Robert G. Wilson v, Jan 01 2018

a(16)-a(49) from Robert G. Wilson v, Jan 02 2018