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
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