A236769 -id:A236769 - cp's OEIS frontend

A049479 Smallest prime dividing 2^n - 1.

3, 7, 3, 31, 3, 127, 3, 7, 3, 23, 3, 8191, 3, 7, 3, 131071, 3, 524287, 3, 7, 3, 47, 3, 31, 3, 7, 3, 233, 3, 2147483647, 3, 7, 3, 31, 3, 223, 3, 7, 3, 13367, 3, 431, 3, 7, 3, 2351, 3, 127, 3, 7, 3, 6361, 3, 23, 3, 7, 3, 179951, 3, 2305843009213693951, 3, 7, 3, 31

Offset: 2

Olivier Gérard

nonn

If p is prime then a(p) == 1 (mod p). Are there composite numbers k such that a(k) == 1 (mod k)? - Thomas Ordowski, Jan 27 2014
Yes, up to 1200, the following composites have the desired property: 169, 221, 323, 611, 779, 793, 923, 1121, 1159. - Michel Marcus, Jan 28 2014
a(n) <= a(lpf(n)) for every n, where lpf(n) = A020639(n). For which n is a(n) < a(lpf(n))? See A236769. - Thomas Ordowski, Jan 30 2014

			a(6)=3 since 2^6 - 1 = 63 = 3^2*7.

Max Alekseyev, Table of n, a(n) for n = 2..1236
Eric Weisstein's World of Mathematics, Mersenne Number.

Cf. A005420.

a = {}; Do[w = 2^n - 1; c = FactorInteger[w]; b = c[[1]][[1]]; AppendTo[a, b], {n, 2, 65}]; a (* Artur Jasinski, Dec 11 2007 *)
FactorInteger[#][[1,1]]&/@(2^Range[2,70]-1) (* Harvey P. Dale, Nov 18 2019 *)

from sympy import factorint
def A049479(n):
    return min(factorint(2**n-1)) # Chai Wah Wu, Jun 03 2019

a(n) > lpf(n) while a(2k) = 3 and a(2k+1) > 2*lpf(2k+1), where lpf(m) = A020639(m). - Thomas Ordowski, Jan 29 2014

For k >= 1, a(2k) = 3, a(6k-3)=7. - Zak Seidov, Mar 21 2014

More terms from Michael Lugo (mlugo(AT)thelabelguy.com), Dec 22 1999
Terms to a(500) in b-file from  T. D. Noe, Dec 06 2006
a(501)-a(1060) in b-file from Michel Marcus, Sep 15 2017
a(1061)-a(1236) in b-file added at the suggestion of Eric Chen by Max Alekseyev, Apr 25 2022

A292559 Composite numbers m such that lpf(2^m - 1) == 1 (mod m).

Original entry on oeis.org

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

Michel Marcus and Thomas Ordowski, Sep 19 2017

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

Subsequence of A236769.

Cf. A002326, A049479, A292834, A298076.

searchMax = 1000; Complement[Select[Range[searchMax], Mod[FactorInteger[2^# - 1][[1, 1]], #] == 1 &], Prime[Range[PrimePi[searchMax]]]] (* Alonso del Arte, Sep 19 2017 *)

lista(nn) = forcomposite(n=1, nn, sp = factor(2^n-1)[1,1]; if ((sp % n) == 1, print1(n, ", "))); \\ Michel Marcus, Sep 19 2017

A049479(m) == 1 (mod m).

a(10)-a(54) from Charles R Greathouse IV, Sep 19 2017

cp's OEIS Frontend

A049479 Smallest prime dividing 2^n - 1.

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