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Average k of twin primes such that k = p1^2 + p2^3, where p1 and p2 are consecutive primes, and p1 < p2.
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%I #9 Aug 16 2021 22:34:32

%S 24918,1235652,3250531482,4199242278,13709099778,23871182760,

%T 86428949742,126606734382,168135540408

%N Average k of twin primes such that k = p1^2 + p2^3, where p1 and p2 are consecutive primes, and p1 < p2.

%e Numbers 24917 and 24919 are primes, making 24918 the average of twin primes. Also, 24918 = 23^2 + 29^3. Thus, 24918 is in this sequence. - _Tanya Khovanova_, Aug 14 2021

%t a={};Do[p1=Prime[n];p2=Prime[n+1];pp=p1^2+p2^3;If[PrimeQ[pp-1]&&PrimeQ[pp+1],AppendTo[a,pp]],{n,16^3}];Print[a];

%K nonn,more

%O 1,1

%A _Vladimir Joseph Stephan Orlovsky_, May 15 2008