%I A003325
%S A003325 2,9,16,28,35,54,65,72,91,126,128,133,152,189,217,224,243,250,280,341,
%T A003325 344,351,370,407,432,468,513,520,539,559,576,637,686,728,730,737,756,793,
%U A003325 854,855,945,1001,1008,1024,1027,1064,1072,1125,1216,1241,1332,1339,1343
%N A003325 Numbers that are the sum of 2 positive cubes.
%C A003325 It is conjectured that this sequence and A052276 have infinitely many 
               numbers in common, although only one example (128) is known.
%C A003325 A119976 is a subsequence; if m is a term then m+k^3 is a term of A003072 
               for all k>0. - Reinhard Zumkeller (reinhard.zumkeller(AT)gmail.com), 
               Jun 03 2006
%D A003325 F. Beukers, The Diophantine equation Ax^p+By^q=Cz^r, Duke Math. J. 91 
               (1998), 61-88.
%D A003325 Nils Bruin, On powers as sums of two cubes, in Algorithmic number theory 
               (Leiden, 2000), 169-184, Lecture Notes in Comput. Sci., 1838, Springer, 
               Berlin, 2000.
%D A003325 C. G. J. Jacobi, Gesammelte Werke, vol. 6, 1969, Chelsea, NY, p. 354.
%H A003325 T. D. Noe, <a href="b003325.txt">Table of n, a(n) for n = 1..1000</a>
%H A003325 C. G. J. Jacobi, <a href="http://www.hti.umich.edu/cgi/t/text/text-idx?c=umhistmath;
               idno=ABR8803">Gesammelte Werke</a>.
%H A003325 D. Tournes, <a href="http://www.reunion.iufm.fr/dep/mathematiques/Seminaires/
               ActesPDF/Tournes53.pdf">A Glance on Indian Mathematician Srinivasa 
               Ramanujan(1887-1920). [Text in French]</a>
%H A003325 Eric Weisstein's World of Mathematics, <a href="http://mathworld.wolfram.com/
               CubicNumber.html">Link to a section of The World of Mathematics.</
               a>
%H A003325 <a href="Sindx_Su.html#ssq">Index entries for sequences related to sums 
               of cubes</a>
%F A003325 Comment from James Buddenhagen, Oct 16 2008: (i) N and N+1 are both the 
               sum of two positive cubes if N=2*(2*n^2+4*n+1)*(4*n^4+16*n^3+23*n^2+14*n+4), 
               n=1,2,.... (ii) For integer n >= 2, let N = 16*n^6-12*n^4+6*n^2-2, 
               so N+1 = 16*n^6-12*n^4+6*n^2-1. Then the identities 16*n^6-12*n^4+6*n^2-2 
               = (2*n^2-n-1)^3 + (2*n^2+n-1)^3 16*n^6-12*n^4+6*n^2-1 = (2*n^2)^3 
               + (2*n^2-1)^3 show that N, N+1 are in the sequence.
%t A003325 Union[Flatten[Table[Table[n^3 + m^3, {m, 1, n}], {n, 1, 15}]]] - Roger 
               L. Bagula, Oct 16 2008
%o A003325 (PARI) cubes=sum(n=1,11,x^(n^3),O(x^1400)); print(cubes^2)
%o A003325 (PARI) isA003325(n) = for( k=1,sqrtn(n\2,3), round(sqrtn(n-k^3,3))^3+k^3==n 
               & return(1)) [From M. F. Hasler (MHasler(AT)univ-ag.fr), Oct 17 2008]
%Y A003325 Cf. A003072, A001235, A011541, A003826.
%Y A003325 Cf. A085323 (n such that a(n+1)=a(n)+1). [From M. F. Hasler (MHasler(AT)univ-ag.fr), 
               Oct 17 2008]
%Y A003325 Sequence in context: A011193 A085960 A051386 this_sequence A101420 A097965 
               A075645
%Y A003325 Adjacent sequences: A003322 A003323 A003324 this_sequence A003326 A003327 
               A003328
%K A003325 nonn,easy,nice
%O A003325 1,1
%A A003325 N. J. A. Sloane (njas(AT)research.att.com).
%E A003325 Error in formula line corrected by Zak Seidov, Jul 23 2009