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Cubes k^3 such that k^3 + (k+1)^3 is semiprime.
5

%I #12 Dec 15 2015 11:55:48

%S 1,8,27,125,216,512,2744,3375,8000,9261,35937,68921,125000,157464,

%T 328509,421875,474552,704969,729000,970299,1157625,1367631,1685159,

%U 2248091,2628072,2803221,3581577,3723875,4741632,5177717,5451776,6751269,7301384,9129329,9938375

%N Cubes k^3 such that k^3 + (k+1)^3 is semiprime.

%H K. D. Bajpai, <a href="/A240859/b240859.txt">Table of n, a(n) for n = 1..3689</a>

%e a(2) = 8 = 2^3: 2^3 + 3^3 = 35 = 5*7 which is a semiprime.

%e a(4) = 125 = 5^3: 5^3 + 6^3 = 341 = 11*31 which is a semiprime.

%p with(numtheory):KD:= proc() local a,b; a:=n^3+(n+1)^3;b:=bigomega(a); if b=2 then RETURN (n^3); fi; end: seq(KD(), n=1..500);

%t Transpose[Select[Partition[Range[250]^3,2,1],PrimeOmega[Total[#]] == 2&]] [[1]] (* _Harvey P. Dale_, Dec 15 2015 *)

%o (PARI) forprime(p=3,1e4,if(isprime((p^2+3)/4),print1((p^3 - 3*p^2 + 3*p - 1)/8", "))) \\ _Charles R Greathouse IV_, Aug 25 2014

%Y Cf. A001358, A005898.

%K nonn,easy

%O 1,2

%A _K. D. Bajpai_, Apr 13 2014