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%I #68 Jan 28 2018 13:10:20
%S 169,221,323,611,779,793,923,1121,1159,1271,1273,1343,1349,1513,1717,
%T 1829,1919,2033,2077,2201,2413,2533,2603,2759,2951,3097,3131,3173,
%U 3193,3281,3379,3599,3721,3743,3791,3937,3953,4043,4223,4309,4331,4369,4607,4619,4811,4867,4883,4981,5111,5177,5263,5429,5567,5699
%N Composite numbers m such that lpf(2^m - 1) == 1 (mod m).
%C All terms are coprime to 2, 3, 5, 7, 11. - _Robert Israel_, Sep 20 2017
%C If p = lpf(2^m - 1) and A002326((p-1)/2) = m composite, then m is in this sequence. - _Thomas Ordowski_, Sep 20 2017
%C Conjecture: there are no numbers k in this sequence such that, for each prime factor q of 2^k - 1, q == 1 (mod k). - _Thomas Ordowski_, Sep 20 2017
%C Note: if all prime factors q of 2^k - 1 are q == 1 (mod k), then 2^k - 1 == 1 (mod k), thus 2^k == 2 (mod k), so k is a pseudoprime. The pseudoprime k = a(42) = 4369 = 17*257 is not a counterexample to this conjecture. A pseudoprime k = P*Q such that both 2^P - 1 and 2^Q - 1 are primes would be a counterexample, but the known Mersenne primes do not give such k. - _Thomas Ordowski_, Oct 02 2017
%C If lpf(2^n - 1) == 1 (mod n), then gpf(2^n - 1) == 1 (mod n). Cf. A291855. - _Thomas Ordowski_, Oct 20 2017
%C Composites m such that lpf(2^m - 1)*gpf(2^m - 1) is a Fermat pseudoprime to base 2, i.e., is in A214305. - _Thomas Ordowski_, Oct 29 2017
%F A049479(m) == 1 (mod m).
%t searchMax = 1000; Complement[Select[Range[searchMax], Mod[FactorInteger[2^# - 1][[1, 1]], #] == 1 &], Prime[Range[PrimePi[searchMax]]]] (* _Alonso del Arte_, Sep 19 2017 *)
%o (PARI) lista(nn) = forcomposite(n=1, nn, sp = factor(2^n-1)[1,1]; if ((sp % n) == 1, print1(n, ", "))); \\ _Michel Marcus_, Sep 19 2017
%Y Subsequence of A236769.
%Y Cf. A002326, A049479, A292834, A298076.
%K nonn,hard
%O 1,1
%A _Michel Marcus_ and _Thomas Ordowski_, Sep 19 2017
%E a(10)-a(54) from _Charles R Greathouse IV_, Sep 19 2017
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