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%I #11 May 21 2021 04:15:46
%S 1,1296,24389,274625,531441,970299,2343750,2515456,4492125,5268024,
%T 5451776,6967871,8000000,18821096,25672375,27270901,32461759,37748736,
%U 41421736,43243551,50653000,64000000,69426531,80062991,81746504,82881856,94818816,100663296
%N a(n) = (p1 + p2)/216 such that p1 >= 5 and p2 = p1 + 2 are twin primes and p1 + p2 is a k-th power with k >= 3.
%C The values of k corresponding to the first terms are: 3, 7, 3, 3, 3, 3, 4, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 5, 3, 3, 3, 3, 3, 3, 3, 3, 3, 4, 3, 3, 4, 3, 3, 3, ...
%H Amiram Eldar, <a href="/A330980/b330980.txt">Table of n, a(n) for n = 1..100</a>
%e a(1) = 1: p1 = 107 and p2 = 109 is the first pair with a sum that is a 3rd power, 216=6^3;
%e a(2) = 1296: p1 = 1296*108 - 1 = 139967, p2 = 1296*108 + 1 = 139969, p1 + p2 = 279936 = 6^7.
%o (PARI) my(pp=5,j); forprime(p=7,10000000000, if(p-pp==2, if(j=ispower(p+pp), if(j>2, print1((p+pp)/216,", ")))); pp=p)
%Y Cf. A076467, A119768, A270231, A330978.
%K nonn
%O 1,2
%A _Hugo Pfoertner_, Jan 05 2020