

A003325


Numbers that are the sum of 2 positive cubes.


50



2, 9, 16, 28, 35, 54, 65, 72, 91, 126, 128, 133, 152, 189, 217, 224, 243, 250, 280, 341, 344, 351, 370, 407, 432, 468, 513, 520, 539, 559, 576, 637, 686, 728, 730, 737, 756, 793, 854, 855, 945, 1001, 1008, 1024, 1027, 1064, 1072, 1125, 1216, 1241, 1332, 1339, 1343
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OFFSET

1,1


COMMENTS

It is conjectured that this sequence and A052276 have infinitely many numbers in common, although only one example (128) is known.
A113958 is a subsequence; if m is a term then m+k^3 is a term of A003072 for all k>0.  Reinhard Zumkeller, Jun 03 2006
If n is a term then n*m^3 (m>=2) is also a term, e.g., 2m^3, 9m^3, 28m^3, and 35m^3 are all terms of the sequence. "Primitive" terms (not of form n*m^3 with n = some previous term of the sequence and m>=2) are 2,9,28,35,65,91,126 etc.  Moshe Levin, Oct 12 2011
This is an infinite sequence in which the first term is prime but thereafter all terms are composite.  Ant King May 09 2013


REFERENCES

F. Beukers, The Diophantine equation Ax^p+By^q=Cz^r, Duke Math. J. 91 (1998), 6188.
Nils Bruin, On powers as sums of two cubes, in Algorithmic number theory (Leiden, 2000), 169184, Lecture Notes in Comput. Sci., 1838, Springer, Berlin, 2000.
C. G. J. Jacobi, Gesammelte Werke, vol. 6, 1969, Chelsea, NY, p. 354.


LINKS

T. D. Noe, Table of n, a(n) for n = 1..1000
Kevin A. Broughan, Characterizing the Sum of Two Cubes, J. Integer Seqs., Vol. 6, 2003.
C. G. J. Jacobi, Gesammelte Werke.
D. Tournes, A Glance on Indian Mathematician Srinivasa Ramanujan(18871920). [Text in French]
Eric Weisstein's World of Mathematics, Cubic Number
Index entries for sequences related to sums of cubes


FORMULA

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^612*n^4+6*n^22, so N+1 = 16*n^612*n^4+6*n^21. Then the identities 16*n^612*n^4+6*n^22 = (2*n^2n1)^3 + (2*n^2+n1)^3 16*n^612*n^4+6*n^21 = (2*n^2)^3 + (2*n^21)^3 show that N, N+1 are in the sequence.


MATHEMATICA

nn = 2*20^3; Union[Flatten[Table[x^3 + y^3, {x, nn^(1/3)}, {y, x, (nn  x^3)^(1/3)}]]] (* T. D. Noe, Oct 12 2011 *)


PROG

(PARI) cubes=sum(n=1, 11, x^(n^3), O(x^1400)); print(cubes^2)
(PARI) isA003325(n) = for( k=1, sqrtn(n\2, 3), round(sqrtn(nk^3, 3))^3+k^3==n & return(1)) [From M. F. Hasler, Oct 17 2008]
(Haskell)
a003325 n = a003325_list !! (n1)
a003325_list = filter c2 [1..] where
c2 x = any (== 1) $ map (a010057 . fromInteger) $
takeWhile (> 0) $ map (x ) $ tail a000578_list
 Reinhard Zumkeller, Mar 24 2012


CROSSREFS

Cf. A003072, A001235, A011541, A003826.
Cf. A085323 (n such that a(n+1)=a(n)+1). [From M. F. Hasler, Oct 17 2008]
Cf. A010057, A000578.
Cf. A027750, A010052, subsequence of A045980, A004999.
Sequence in context: A011193 A085960 A051386 * A101420 A248434 A213389
Adjacent sequences: A003322 A003323 A003324 * A003326 A003327 A003328


KEYWORD

nonn,easy,nice


AUTHOR

N. J. A. Sloane.


EXTENSIONS

Error in formula line corrected by Zak Seidov, Jul 23 2009


STATUS

approved



