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A292834
Numbers m, not powers of 2, such that the least prime factor of 2^m + 1 is congruent to 1 (mod m).
2
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
Sequence in context: A230761 A073763 A359750 * A362941 A030021 A269036
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