cp's OEIS Frontend

A217993 Smallest k such that k^(2^n) + 1 and (k+2)^(2^n) + 1 are both prime.

%I A217993 #18 May 03 2018 03:04:14
%S A217993 2,2,2,2,74,112,2162,63738,13220,54808,3656570,6992032,125440,
%T A217993 103859114,56414914,87888966
%N A217993 Smallest k such that k^(2^n) + 1 and (k+2)^(2^n) + 1 are both prime.
%C A217993 a(15)=87888966 but a(14) is unknown. - _Jeppe Stig Nielsen_, Mar 17 2018
%C A217993 The prime pair related to a(14) was found four days ago, and today double checking has proved that they are indeed the first occurrence for n=14. - _Jeppe Stig Nielsen_, May 02 2018
%F A217993 a(n) = A118539(n)-1. - _Jeppe Stig Nielsen_, Feb 27 2016
%e A217993 a(0) = 2 because 2^1+1 = 3 and 4^1+1 = 5 are prime;
%e A217993 a(1) = 2 because 2^2+1 = 5  and 4^2+1 = 17 are prime;
%e A217993 a(2) = 2 because 2^4+1 = 17  and 4^4+1 = 257 are prime;
%e A217993 a(3) = 2 because  2^8+1 = 257 and 4^8+1 = 65537 are prime.
%p A217993 for n from 0  to 5 do:ii:=0:for k from 2 by 2 to 10000 while(ii=0) do:if type(k^(2^n)+1,prime)=true and type((k+2)^(2^n)+1,prime)=true then ii:=1: printf ( "%d %d \n",n,k):else fi:od:od:
%Y A217993 Cf. A006313, A006314, A006315, A006316, A056994, A056995, A057465, A057002, A088361, A088362, A118539.
%K A217993 nonn,hard,more
%O A217993 0,1
%A A217993 _Michel Lagneau_, Oct 17 2012
%E A217993 a(13) from _Jeppe Stig Nielsen_, Mar 17 2018
%E A217993 a(14) and a(15) from _Jeppe Stig Nielsen_, May 02 2018