A125742 Primes p such that (2^p - 2^((p+1)/2) + 1)/5 is prime.

5, 11, 13, 29, 43, 53, 283, 557, 563, 613, 691, 2731, 5147, 5323, 9533, 10771, 221891, 235099, 305867, 311027, 333227, 792061, 1347781, 1669219, 1882787, 2305781

Offset

1,1

Comments

PrimePi[ a(n) ] = {3, 5, 6, 10, 14, 16, 61, 102, 103, 112, 125, 399, 686, 705, 1180, 1312, 19768, 20843, 26482, 26882, 28656, ...}. (2^p - 2^((p+1)/2) + 1) is the Aurifeuillan cofactor of 4^p + 1, where p is odd prime. All a(n) belong to A124112(n) = {5, 7, 9, 11, 13, 17, 29, 43, 53, 89, 283, 557, 563, 613, 691, 1223, 2731, ...} Numbers n such that ((1+I)^n+1)/(2+I) is a Gaussian prime. 5 largest currently known terms found by Jean Penne in Nov 2006: {221891, 235099, 305867, 311027, 333227}.

Links

Table of n, a(n) for n=1..26.

Henri Lifchitz and Renaud Lifchitz: PRP Records. Probable Primes Top 10000

Mathematica

Do[p=Prime[n]; f=(2^p-2^((p+1)/2)+1)/5; If[PrimeQ[f], Print[{PrimePi[p], p}]], {n, 1, 28656}]

Prog

(PARI) is(p)=isprime(p)&&ispseudoprime((2^p - 2^((p+1)/2) + 1)/5) \\ Charles R Greathouse IV, May 15 2013

Crossrefs

Cf. A124165 (primes p such that (2^p + 2^((p+1)/2) + 1)/5 is prime).

Cf. A124112 (numbers n such that ((1+i)^n+1)/(2+i) is a Gaussian prime).

Sequence in context: A051654 A225754 A098973 * A269844 A374719 A116440

Adjacent sequences: A125739 A125740 A125741 * A125743 A125744 A125745

Keyword

hard,more,nonn

Author

Alexander Adamchuk, Dec 04 2006

Extensions

a(23-25) = 1347781, 1669219, 1882787 were found by Borys Jaworski between 2008 and 2012 (see the PRP Records link). - Alexander Adamchuk, Nov 27 2008

a(22) = 792061 was found out-of-sequence by Thomas Ritschel in March of 2014 (see the PRP Records link). - Serge Batalov, Mar 31 2014

a(26) = 2305781 from Serge Batalov, Mar 31 2014

Status

approved