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

1,1

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

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 *)

Prog

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

Crossrefs

Cf. A002586, A292559.

Sequence in context: A230761 A073763 A359750 * A362941 A030021 A269036

Adjacent sequences: A292831 A292832 A292833 * A292835 A292836 A292837

Keyword

hard,nonn

Author

Thomas Ordowski, Sep 24 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

Status

approved