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A234256 Cubes t^3 = (p+q+r)/3 which are the arithmetic mean of three consecutive primes such that p < t^3 < q < r.

%I A234256 #21 Jun 25 2022 17:08:31
%S A234256 5735339,10503459,73560059,253636137,393832837,761048497,791453125,
%T A234256 1064332261,1829276567,2014698447,2487813875,2893640625,4533086375,
%U A234256 7845011803,14437662875,45998156287,55611739513,62429032063,63378025803,72877493233,87115050737,104154702625
%N A234256 Cubes t^3 = (p+q+r)/3 which are the arithmetic mean of three consecutive primes such that p < t^3 < q < r.
%H A234256 K. D. Bajpai, <a href="/A234256/b234256.txt">Table of n, a(n) for n = 1..3680</a>
%e A234256 5735339 is in the sequence because 5735339 = 179^3 = (5735291+5735357+5735369)/3, the arithmetic mean of three consecutive primes.
%e A234256 10503459 is in the sequence because 10503459 = 219^3 = (10503443+10503461+10503473)/3, the arithmetic mean of three consecutive primes.
%p A234256 with(numtheory):KD := proc() local a,b,d,e,f; a:=n^3; b:=prevprime(a); d:=nextprime(a); e:=nextprime(d);  f:=(b+d+e)/3; if  a=f then RETURN (a);  fi;  end: seq(KD(), n=2..10000);
%o A234256 (PARI) list(lim)=my(v=List(), p=2, q=3, t); forprime(r=5, nextprime(nextprime(lim\3+1)+1), t=(p+q+r)/3; if(denominator(t)==1 && ispower(t,3) && t < q, listput(v, t)); p=q; q=r); Vec(v) \\ _Charles R Greathouse IV_, Jan 03 2014
%Y A234256 Cf. A000578 (cubes: n^3).
%Y A234256 Cf. A062703 (squares: sum of two consecutive primes).
%Y A234256 Cf. A069495 (squares: arithmetic mean of two consecutive primes).
%Y A234256 Cf. A234240 (cubes: arithmetic mean of two consecutive primes).
%K A234256 nonn
%O A234256 1,1
%A A234256 _K. D. Bajpai_, Dec 22 2013
%E A234256 Definition corrected by _Michel Marcus_ and _Charles R Greathouse IV_, Jan 03 2014