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A003325
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Numbers that are the sum of 2 positive cubes.
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136
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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
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1,1
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COMMENTS
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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]
(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
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REFERENCES
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C. G. J. Jacobi, Gesammelte Werke, vol. 6, 1969, Chelsea, NY, p. 354.
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LINKS
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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.
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MATHEMATICA
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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 *)
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PROG
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(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
(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]
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CROSSREFS
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KEYWORD
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nonn,easy,nice
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AUTHOR
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EXTENSIONS
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Error in formula line corrected by Zak Seidov, Jul 23 2009
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STATUS
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approved
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