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A069051
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Primes p such that p-1 divides 2^p-2.
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10
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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
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OFFSET
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1,1
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COMMENTS
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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
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. How to prove that this set is complete? (End)
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LINKS
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MATHEMATICA
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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 *)
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PROG
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(Magma) [p : p in PrimesUpTo(310000) | IsZero((2^p-2) mod (p-1))]; // Vincenzo Librandi, Dec 03 2018
(GAP) Filtered([1..350000], p->IsPrime(p) and (2^p-2) mod (p-1)=0); # Muniru A Asiru, Dec 03 2018
(Python)
from sympy import prime
for n in range(1, 350000):
if (2**prime(n)-2) % (prime(n)-1)==0:
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CROSSREFS
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a(n)-1 form subsequence of A014741; (a(n)-1)/2 for n>1 forms a subsequence of A014945.
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KEYWORD
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nonn
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AUTHOR
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EXTENSIONS
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STATUS
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approved
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