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A292559
Composite numbers m such that lpf(2^m - 1) == 1 (mod m).
3
169, 221, 323, 611, 779, 793, 923, 1121, 1159, 1271, 1273, 1343, 1349, 1513, 1717, 1829, 1919, 2033, 2077, 2201, 2413, 2533, 2603, 2759, 2951, 3097, 3131, 3173, 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
OFFSET
1,1
COMMENTS
All terms are coprime to 2, 3, 5, 7, 11. - Robert Israel, Sep 20 2017
If p = lpf(2^m - 1) and A002326((p-1)/2) = m composite, then m is in this sequence. - Thomas Ordowski, Sep 20 2017
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
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
If lpf(2^n - 1) == 1 (mod n), then gpf(2^n - 1) == 1 (mod n). Cf. A291855. - Thomas Ordowski, Oct 20 2017
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
FORMULA
A049479(m) == 1 (mod m).
MATHEMATICA
searchMax = 1000; Complement[Select[Range[searchMax], Mod[FactorInteger[2^# - 1][[1, 1]], #] == 1 &], Prime[Range[PrimePi[searchMax]]]] (* Alonso del Arte, Sep 19 2017 *)
PROG
(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
CROSSREFS
Subsequence of A236769.
Sequence in context: A350705 A141075 A124979 * A322568 A350381 A018820
KEYWORD
nonn,hard
AUTHOR
EXTENSIONS
a(10)-a(54) from Charles R Greathouse IV, Sep 19 2017
STATUS
approved