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a(n) = (2^prime(n) + 1)/3.
16

%I #32 Sep 08 2022 08:45:29

%S 3,11,43,683,2731,43691,174763,2796203,178956971,715827883,

%T 45812984491,733007751851,2932031007403,46912496118443,

%U 3002399751580331,192153584101141163,768614336404564651,49191317529892137643

%N a(n) = (2^prime(n) + 1)/3.

%C If p - 1 is squarefree, the multiplicative order of 2 modulo a(n) is 2p. - _Vladimir Shevelev_, Jul 15 2008

%C The prime numbers in this sequence are the Wagstaff primes (A000979). - _Omar E. Pol_, Nov 05 2013

%H Vincenzo Librandi, <a href="/A126614/b126614.txt">Table of n, a(n) for n = 2..200</a>

%e a(2) = (2^prime(2) + 1)/3 = (2^3 + 1)/3 = 9/3 = 3.

%e a(3) = (2^prime(3) + 1)/3 = (2^5 + 1)/3 = 33/3 = 11.

%e a(4) = (2^prime(4) + 1)/3 = (2^7 + 1)/3 = 129/3 = 43.

%t Table[(2^Prime[n] + 1)/3, {n, 2, 20}]

%o (Magma) [(2^NthPrime(n)+1)/3: n in [2..20]]; // _Vincenzo Librandi_, Mar 29 2012

%o (PARI) a(n)=2^prime(n)\/3 \\ _Charles R Greathouse IV_, Mar 29 2012

%o (PARI) vecextract(apply(p->2^p\/3,primes(100)),"2..") \\ _Charles R Greathouse IV_, Mar 29 2012

%Y Cf. A000040, A000979, A000978, A124400.

%K nonn

%O 2,1

%A _Artur Jasinski_, Feb 09 2007