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Numbers m, not powers of 2, such that the least prime factor of 2^m + 1 is congruent to 1 (mod m).
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%I #34 Jun 13 2018 03:26:31

%S 24,48,112,160,192,272,448,496,656,688,832,896,1088,1152,1168,1328,

%T 1360,1408,1472,1520,1664,1744,1920,1984,2176,2304,2432,2560,2688,

%U 2752,2816,2944,2960,3056,3072,3200,3328,3520,3664,3712,3776,4672,4864,4928,5120,5376,5552,5888,6144

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

%C Problem: are there infinitely many such numbers?

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

%C 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.

%C From Theorem in A002586, all terms are == 0 (mod 8). - _Robert G. Wilson v_, Jan 02 2018

%H Robert G. Wilson v, <a href="/A292834/b292834.txt">Table of n, a(n) for n = 1..71</a>

%t Select[Range[200], And[! IntegerQ @ Log2 @ #, Mod[FactorInteger[2^# + 1][[1, 1]], #] == 1] &] (* _Michael De Vlieger_, Sep 24 2017 *)

%t 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 *)

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

%Y Cf. A002586, A292559.

%K hard,nonn

%O 1,1

%A _Thomas Ordowski_, Sep 24 2017

%E a(9)-a(15) from _Robert G. Wilson v_, Jan 01 2018

%E a(16)-a(49) from _Robert G. Wilson v_, Jan 02 2018