A330382 -id:A330382 - cp's OEIS frontend

A014741 Numbers k such that k divides 2^(k+1) - 2.

1, 2, 6, 18, 42, 54, 126, 162, 294, 342, 378, 486, 882, 1026, 1134, 1314, 1458, 1806, 2058, 2394, 2646, 3078, 3402, 3942, 4374, 5334, 5418, 6174, 6498, 7182, 7938, 9198, 9234, 10206, 11826, 12642, 13122, 14154, 14406, 16002, 16254

Offset: 1

N. J. A. Sloane, Olivier Gérard

nonn

Also, numbers k such that k divides Eulerian number A000295(k+1) = 2^(k+1) - k - 2.
Also, numbers k such that k divides A086787(k) = Sum_{i=1..k} Sum_{j=1..k} i^j.
All terms greater than 1 are even; for a proof, see comment in A036236. - Max Alekseyev, Feb 03 2012
If k>1 is a term, then 3*k is also a term. - Alexander Adamchuk, Nov 03 2006
Prime numbers of the form a(m)+1 are given by A069051. - Max Alekseyev, Nov 14 2012
The number 2^m - 2 is a term of this sequence if and only if m - 1 is a term. - Thomas Ordowski, Jul 01 2024

Amiram Eldar, Table of n, a(n) for n = 1..10000

Cf. A000295, A086787, A015919, A006517, A330382.

Join[{1,2},Select[Range[17000],PowerMod[2,#+1,#]==2&]] (* Harvey P. Dale, Feb 11 2015 *)

is(n)=Mod(2,n)^(n+1)==2 \\ Charles R Greathouse IV, Nov 03 2016

For n > 1, a(n) = 2*A014945(n-1). - Max Alekseyev, Nov 14 2012

A069051 Primes p such that p-1 divides 2^p-2.

Original entry on oeis.org

2, 3, 7, 19, 43, 127, 163, 379, 487, 883, 1459, 2647, 3079, 3943, 5419, 9199, 11827, 14407, 16759, 18523, 24967, 26407, 37339, 39367, 42463, 71443, 77659, 95923, 99079, 113779, 117307, 143263, 174763, 175447, 184843, 265483, 304039, 308827

Offset: 1

Benoit Cloitre, Apr 03 2002

nonn

These are the prime values of n such that 2^n == 2 (mod n*(n-1)). - V. Raman, Sep 17 2012
These are the prime values p such that n^(2^(p-1)) is congruent to n or -n (mod p) for all n in Z/pZ, the commutative ring associated with each term. This results follows from Fermat's little theorem. - Philip A. Hoskins, Feb 08 2013
A prime p is in this sequence iff p-1 belongs to A014741. For p>2, this is equivalent to (p-1)/2 belonging to A014945. - Max Alekseyev, Aug 31 2016
From Thomas Ordowski, Nov 20 2018: (Start)
Conjecture: if n-1 divides 2^n-2, then (2^n-2)/(n-1) is squarefree.
Numbers n such that b^n == b (mod (n-1)*n) for every integer b are 2, 3, 7, and 43; i.e., only prime numbers of the form A014117(k) + 1. (End)
These are primes p such that p^2 divides b^(2^p-2) - 1 for every b coprime to p. - Thomas Ordowski, Jul 01 2024

Charles R Greathouse IV, Table of n, a(n) for n = 1..3314 (First 553 terms from V. Raman)
Mersenne Forum, Prime Conjecture

Cf. A216822, A217468, A211203, A211349, A330382.

a(n)-1 form subsequence of A014741; (a(n)-1)/2 for n>1 forms a subsequence of A014945.

Filtered([1..350000],p->IsPrime(p) and (2^p-2) mod (p-1)=0); # Muniru A Asiru, Dec 03 2018

[p : p in PrimesUpTo(310000) | IsZero((2^p-2) mod (p-1))]; // Vincenzo Librandi, Dec 03 2018

Select[Prime[Range[10000]], Mod[2^# - 2, # - 1] == 0 &] (* T. D. Noe, Sep 19 2012 *)
Join[{2,3},Select[Prime[Range[30000]],PowerMod[2,#,#-1]==2&]] (* Harvey P. Dale, Apr 17 2022 *)

isA069051(p)=Mod(2,p-1)^p==2 && isprime(p); \\ Charles R Greathouse IV, Sep 19 2012

from sympy import prime
for n in range(1,350000):
    if (2**prime(n)-2) % (prime(n)-1)==0:
        print(prime(n)) # Stefano Spezia, Dec 07 2018

a(1) added by Charles R Greathouse IV, Sep 19 2012

cp's OEIS Frontend

A014741 Numbers k such that k divides 2^(k+1) - 2.

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