A358527 -id:A358527 - cp's OEIS frontend

A359387 Primes p such that the smallest prime factor of (2^(p-1)-1)/(3*p) is greater than p.

11, 23, 47, 59, 83, 107, 167, 179, 227, 263, 347, 359, 383, 443, 467, 479, 503, 563, 587, 647, 719, 839, 863, 887, 983, 1019, 1187, 1283, 1307, 1319, 1367, 1439, 1487, 1523, 1619, 1823, 1847, 1907, 2027, 2039, 2063, 2099, 2207, 2243, 2447, 2459, 2579, 2687, 2699

Offset: 1

Alain Rocchelli, Dec 29 2022

nonn

This sequence corresponds to the values of p (>7) in A358527 for which p appears in second position in the factorization of 2^(p-1)-1.
All terms are congruent to 11 mod 12, cf. A068231.
It is conjectured that there are infinitely many terms in this sequence, and their estimated asymptotic density n/a(n) ~ C/(log(a(n)))^2 where C is a constant between 0.7 and 0.9.

			7 is not a term since for p=7, (2^(p-1)-1)/(3*p) = (2^6-1)/(3*7) = 3 and 3 is not greater than 7.
11 is a term since for p=11, (2^(p-1)-1)/(3*p) = (2^10-1)/(3*11) = 31, which is greater than 11.
23 is a term since (2^22-1)/(3*23) = 60787 = 89*683 and 89 is greater than 23.

Cf. A068231, A096060, A358527.

q[p_] := AllTrue[Range[p], ! PrimeQ[#] || PowerMod[2, p - 1, 3*p*#] > 1 &]; Select[Prime[Range[4, 400]], q] (* Amiram Eldar, Dec 31 2022 *)

isok(p) = (p%12==11 && isprime(p)) || return(0); forprime(div=5, p-1, if(Mod(2,div)^(p-1)==1, return(0))); 1;

A359395 Least odd prime p in position n in the prime factorization of M(p) = 2^(p - 1) - 1.

Original entry on oeis.org

3, 5, 17, 13, 71, 37, 157, 61, 211, 313, 1289, 241, 337, 181, 577, 601, 541, 1381, 421, 1201, 1009, 1621, 1873, 3433, 4561, 1801, 3301, 2161, 3061, 5281, 3361, 2521, 7393, 6481, 4201, 4621, 8737, 9181, 6301, 19501, 7561, 16633, 12241, 26881, 15601, 9241, 21001, 14281, 12601, 53551

Offset: 1

Jean-Marc Rebert, Dec 31 2022

nonn

			M(5) = 15 = 3*5 and 5 is in second position in the prime factorization of M(5), and no lesser odd prime satisfies this, so a(2) = 5.
M(17) = 65535 = 3*5*17*257 and 17 is in third position in the prime factorization of M(17), and no lesser odd prime satisfies this, so a(3) = 17.

Carlos Rivera, Puzzle 1116. A358527, The Prime Puzzles & Problems Connection.

Cf. A098102, A358527.

f[n_] := Module[{p = Prime[n]}, Count[Prime[Range[n - 1]], ?(PowerMod[2, p - 1, #] == 1 &)] + 1]; f[1] = 0; With[{s = Array[f, 4000]}, ind = TakeWhile[ FirstPosition[s, #] & /@ Range[Max[s]] // Flatten, NumberQ ]; Prime[ind]] (* _Amiram Eldar, Jan 02 2023 *)

isok(p, n) = my(f=factor(2^(p-1) - 1, p+1)); if (#f~ < n, 0, f[n,1] == p);
a(n) = my(p=3); while(!isok(p, n), p=nextprime(p+1)); p; \\ Michel Marcus, Jan 13 2023

cp's OEIS Frontend

A359387 Primes p such that the smallest prime factor of (2^(p-1)-1)/(3*p) is greater than p.

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