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Average of twin primes p3 such that p1^2 + p2^3=p3 and p1^3 + p2^2 = p4, p3 and p4 are average of twin primes. p1 and p2 consecutive primes, p1 < p2.
1

%I #10 Jan 09 2020 03:13:13

%S 24918,9270440598450720,1151315644373474442978,1166412457712602408182,

%T 1408820228836430919078,1611036311504881881342,1839287439769397002278,

%U 1876396650678820877442,2541675503832771774858,3760334521638661478022,13232238501319295512260,19086564229432581494760,30269637404459759488308

%N Average of twin primes p3 such that p1^2 + p2^3=p3 and p1^3 + p2^2 = p4, p3 and p4 are average of twin primes. p1 and p2 consecutive primes, p1 < p2.

%H Amiram Eldar, <a href="/A139776/b139776.txt">Table of n, a(n) for n = 1..120</a>

%e 24918 is a term since p1 = 23 and p2 = 29 are consecutive primes such that p1^2 + p2^3 = 24918 and p1^3 + p2^2 = 13008 are averages of twin primes.

%t a={};Do[p1=Prime[n];p2=Prime[n+1];p3=p1^2+p2^3;p4=p1^3+p2^2;If[PrimeQ[p3-1]&&PrimeQ[p3+1]&&PrimeQ[p4-1]&&PrimeQ[p4+1],AppendTo[a,p3]],{n,13^5}];Print[a];

%o (PARI) testA139776(p,q)={my(p3=p^2+q^3,p4=p^3+q^2);ispseudoprime(p3-1)&&ispseudoprime(p3+1)&&ispseudoprime(p4-1)&&ispseudoprime(p4+1)} p=3;forprime(q=5,1e7,if(testA139776(p,q),print1(p^2+q^3","));p=q)

%Y Cf. A001097, A001359, A006512, A014574, A138763.

%K nonn

%O 1,1

%A _Vladimir Joseph Stephan Orlovsky_, May 16 2008

%E Program and more terms from Charles R Greathouse IV Jul 27 2009