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A122723 Primes that are the sum of three distinct positive cubes.

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%I A122723 #14 Jan 21 2015 03:34:50
%S A122723 73,197,251,281,307,349,521,547,577,701,757,853,863,881,919,953,1009,
%T A122723 1091,1217,1249,1483,1559,1637,1861,1907,2069,2087,2267,2269,2287,
%U A122723 2339,2477,2521,2729,2753,2843,2927,2953,2969,3067,3257,3413,3457,3527,3529
%N A122723 Primes that are the sum of three distinct positive cubes.
%C A122723 Considering parity, a prime sum of three cubes cannot be the sum of three evens nor two odds and an even, but must be the sum of three odds (such as 1^3 + 3^3 + 9^3 = 757 or 3^3 + 5^3 + 9^3 = 881) or two evens and an odd (such as 1^3 + 2^3 + 10^3 = 1009). Without "distinct" we have solutions such as 1^3 + 1^3 + 3^3 = 29; 2^3 + 2^3 + 3^3 = 43; 1^3 + 1^3 + 5^3 = 127. A subset of the three odds subset is primes which are the sum of the cubes of three distinct primes, such as 3^3 + 5^3 + 11^3 = 1483; or 3^3 + 7^3 + 19^3 = 7229; or 7^3 + 11^3 + 23^3 = 13841; or 3^3 + 5^3 + 41^3 = 69073.
%H A122723 T. D. Noe, <a href="/A122723/b122723.txt">Table of n, a(n) for n=1..1000</a>
%F A122723 Primes in A024975.
%e A122723 a(1) = 73 = 1^3 + 2^3 + 4^3.
%e A122723 a(7) = 521 = 1^3 + 2^3 + 8^3.
%t A122723 lst={};Do[Do[Do[p=n^3+m^3+k^3;If[PrimeQ[p],AppendTo[lst,p]],{n,m+1,4!}],{m,k+1,4!}],{k,4!}];Take[Union[lst],30] (* _Vladimir Joseph Stephan Orlovsky_, May 23 2009 *)
%Y A122723 Cf. A000040, A024975.
%K A122723 easy,nonn
%O A122723 1,1
%A A122723 _Jonathan Vos Post_, Sep 23 2006
%E A122723 Corrected and extended by _Vladimir Joseph Stephan Orlovsky_ and _T. D. Noe_, Jul 16 2010