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. [Any further examples are greater than 5 million. - Charles R Greathouse IV, Apr 12 2020] [Any further example is greater than 10^12. - M. F. Hasler, Jan 10 2021]
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
From James R. Buddenhagen, Oct 16 2008: (Start)
(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 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. (End)
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 the form n*m^3 with n = some previous term of the sequence and m >= 2) are 2, 9, 28, 35, 65, 91, 126, etc. - Zak Seidov, 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
By Fermat's Last Theorem (the special case for exponent 3, proved by Euler, is sufficient), this sequence contains no cubes. - Charles R Greathouse IV, Apr 03 2021
REFERENCES
C. G. J. Jacobi, Gesammelte Werke, vol. 6, 1969, Chelsea, NY, p. 354.
LINKS
N. J. A. Sloane, Table of n, a(n) for n = 1..20000 (first 1000 terms from T. D. Noe)
F. Beukers, The Diophantine equation Ax^p+By^q=Cz^r, Duke Math. J. 91 (1998), 61-88.
Kevin A. Broughan, Characterizing the sum of two cubes, J. Integer Seqs., Vol. 6, 2003.
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.
C. G. J. Jacobi, Gesammelte Werke.
Michael Penn, 1674 is not a perfect cube, 2020 video
N. J. A. Sloane, Table of n, a(n) for n = 1..59562
Eric Weisstein's World of Mathematics, Cubic Number
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 *)
With[{upto=2000}, Select[Total/@Tuples[Range[Ceiling[Surd[upto, 3]]]^3, 2], #<=upto&]]//Union (* Harvey P. Dale, Jun 11 2016 *)
PROG
(PARI) cubes=sum(n=1, 11, x^(n^3), O(x^1400)); v = select(x->x, Vec(cubes^2), 1); vector(#v, k, v[k]+1) \\ edited by Michel Marcus, May 08 2017
(PARI) isA003325(n) = for(k=1, sqrtnint(n\2, 3), ispower(n-k^3, 3) && return(1)) \\ M. F. Hasler, Oct 17 2008, improved upon suggestion of Altug Alkan and Michel Marcus, Feb 16 2016
(PARI) T=thueinit('z^3+1); is(n)=#select(v->min(v[1], v[2])>0, thue(T, n))>0 \\ Charles R Greathouse IV, Nov 29 2014
(PARI) list(lim)=my(v=List()); lim\=1; for(x=1, sqrtnint(lim-1, 3), my(x3=x^3); for(y=1, min(sqrtnint(lim-x3, 3), x), listput(v, x3+y^3))); Set(v) \\ Charles R Greathouse IV, Jan 11 2022
(Haskell)
a003325 n = a003325_list !! (n-1)
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
(Python)
from sympy import integer_nthroot
def aupto(lim):
cubes = [i*i*i for i in range(1, integer_nthroot(lim-1, 3)[0] + 1)]
sum_cubes = sorted([a+b for i, a in enumerate(cubes) for b in cubes[i:]])
return [s for s in sum_cubes if s <= lim]
print(aupto(1343)) # Michael S. Branicky, Feb 09 2021
CROSSREFS
KEYWORD
nonn,easy,nice
AUTHOR
EXTENSIONS
Error in formula line corrected by Zak Seidov, Jul 23 2009
STATUS
approved