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Primes p such that (2^p + 2^((p+1)/2) + 1)/5 is prime.
4

%I #19 Sep 17 2024 04:45:13

%S 7,17,89,1223,5479,11257,11519,12583,23081,36479,52567,52919,125929,

%T 365689,1127239,1148729,4533073

%N Primes p such that (2^p + 2^((p+1)/2) + 1)/5 is prime.

%C PrimePi[ a(n) ] = {4, 7, 24, 200, 724, 1361, 1389, 1503, 2578, 3868, 5368, 5400, 11814, 31200, ...}.

%C 3 terms found by _David Broadhurst_ in Nov 2006: {36479, 52567, 52919}.

%C Only 2 terms found by Jean Penne in Nov 2006 belong to a(n): {125929, 365689}.

%C 5 other numbers found by Jean Penne in Nov 2006 belong to related sequence of primes p such that (2^p - 2^((p+1)/2) + 1)/5 is prime: {221891, 235099, 305867, 311027, 333227}.

%C All terms belong to A124112 = {5, 7, 9, 11, 13, 17, 29, 43, 53, 89, 283, 557, 563, 613, 691, 1223, 2731, ...} (numbers k such that ((1+i)^k+1)/(2+i) is a Gaussian prime).

%C The terms 1127239 and 1148729 were found by Borys Jaworski in 2006-2007. - _Alexander Adamchuk_, Jun 20 2007

%H Henri Lifchitz and Renaud Lifchitz: <a href="http://www.primenumbers.net/prptop/prptop.php">PRP Records. Probable Primes Top 10000</a>.

%t Do[p=Prime[n];f=(2^p+2^((p+1)/2)+1)/5;If[PrimeQ[f],Print[{n,p}]],{n,1,200}]

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

%K hard,more,nonn

%O 1,1

%A _Alexander Adamchuk_, Dec 02 2006, Dec 04 2006

%E a(17) from _Serge Batalov_, Mar 31 2014