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A240970
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Numbers n such that the sum of n consecutive positive cubes is a cube for some initial starting number k.
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4
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1, 3, 4, 20, 25, 49, 64, 99, 125, 153, 288, 343, 512, 1000, 1331, 1849, 2197, 2744, 4096, 4913, 6591, 6859, 8000, 10200, 10648, 12167, 13923, 14161, 15625, 17576, 19220, 21456, 21952, 24389, 25201, 29791, 32768, 33124, 39304, 42875, 49776, 50653, 54872, 63001, 64000, 68921, 79507, 85184, 97336
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OFFSET
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1,2
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COMMENTS
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This sequence gives numbers n such that k^3 + (k+1)^3 + ... + (k+n-1)^3 = y^3 for some nontrivial k. - Derek Orr, Jan 18 2015
For n > 3, n appears not to be squarefree. [This is incorrect, 25201 = 11*29*79 is a member of this sequence. - Derek Orr, Mar 09 2015]
5000 < a(17) <= 6591. - Derek Orr, Jan 17 2015
It is known that 14161 and 63001 are members of this sequence.
Let S(n,k) = Sum_{i=0..n-1} (k+i)^3. The smallest k-values for the given n in the sequence are {1, 3, 11, 3, 6, 291, 6, 11, 34, 213, 273, 213, 406, 1134, 1735, 34228, 3606, 4966, 8790, 11368}.
If n is of the form (2m)^3 for some m = 1, 2, 3, ... there exist only a finite number of k that can make S(n,k) be a cube. For k large enough, S(n,k) - floor(S(n,k)^(1/3))^3 = m^3*(n^2-1)(2k + n - 1). So it will never be a cube.
If n is of the form (2m+1)^3 for some m = 1, 2, 3, ... there also exist only a finite number of k that can make S(n,k) cube. For k large enough, S(n,k) - floor(S(n,k)^(1/3))^3 = f(m)*n(2k + n - 1) where f(m) is a function of m. f(1) = 91, f(2) = 1953, f(3) = 14706, f(4) = 66430, f(5) = 221445, f(6) = 603351, ...
As mentioned above, when k is greater than a certain number, S(n,k) will never be a cube. Let the critical value for k be called g(n), since it depends on n. The function g(n) is only concretely known when n is a cube (mentioned above "for k large enough"). Below are the known g(n) values.
n = 8, g(n) <= 6.
n = 27, g(n) <= 168.
n = 64, g(n) <= 1333.
n = 125, g(n) <= 6447.
n = 216, g(n) <= 23219.
n = 343, g(n) <= 68456.
n = 512, g(n) <= 174506.
n = 729, g(n) <= 398215.
n = 1000, g(n) <= 832832.
n = 1331, g(n) <= 1623264.
n = 1728, g(n) <= 2985119.
n = 2197, g(n) <= 5227943.
...
Example: for n = 8, if k > 6, then S(8,k) - floor(S(8,k)^(1/3))^3 = 63*(2*k+7). So g(n) = 6.
There is a possibility that g(n) could be defined for other values of n besides the cubes and this would give a bound on k for all n. Note that g(n) ~ 0.065*n^(2.4). Graphing this function, one can see that for all n > 4 listed in the sequence, their k-values given in the first comment are well less than 0.065*n^(2.4). For large n, 0.065*n^(2.4) seems to be a large overestimate. Thus, 0.065*n^(2.4) can act as a sufficient bound for k for large n (essentially for n > 4).
If n = v^3 where v is a number not divisible by 3, then k = (v^4-3*v^3-2*v^2+4)/6. Thus there are infinitely many cubes in this sequence. - Robert Israel, Aug 06 2014
The equation S(n,k) = y^3 is a superelliptic curve and thus has only a finite number of integral points.
Using the transformation X = (2*y)/(2k - 1 + n) and Y = 1/(2k - 1 + n), S(n,k) = y^3 transforms to Y^2*(n^3 - n) = X^3 - n, which is in Weierstrass form and can be solved for rationals (X,Y).
Further, the above transformation can be manipulated to arrive at a new Diophantine equation; this one is over the integers: n*X^3 + (n^3 - n)*X = (2y)^3 where X = 2k + n - 1.
The Bilu and Hanrot link gives explicit bounds for the possible values of x for the equation a*y^p = f(x) for p >= 3 and f of degree d >= 2. Here, a = 1, p = 3, and d = 3.
If M is a cubed integer not divisible by 3, there always exists at least one nontrivial solution for the sum of M consecutive cubed integers starting from a^3 and equaling a cubed integer c^3. There are no nontrivial solutions for M = m^3 if m=0(mod 3). For n>=1, for integers m(n) = A001651(n), all nontrivial solutions for M(n) = m^3 = A118719(n+1) are a(n) = (m-1)(m^2(m-2) - 4(m+1))/6 (A253778) and c(n)= m(m^2-1)(m^2+2)/6 (A253779). - Vladimir Pletser, Jan 12 2015
The above relation for the number of terms equal to a cube not 0(mod 3), i.e., a(n) = (m-1)(m^2(m-2) - 4(m+1))/6 (V. Pletser) or k = (v^4-3*v^3-2*v^2+4)/6 (R. Israel), was already given by A. Martin in 1871 and J. Matteson in 1888 (see L. E. Dickson, p. 584). - Vladimir Pletser, Jan 27 2015
Additional terms in this sequence (that are not cubes) are n = 6591, 21456, 176824, 11859210, yielding the triples (n, first term, cubic root of sum) = (6591, 305, 82680), (21456, 266785, 7715220), (176824, 407526, 28127850), (11859210, 21709458, 6398174475). - Vladimir Pletser, Jan 12 2015
Two additional terms in this sequence (that are not cubes) are n = 13923 and 33124 yielding the triples (n, first term, cubic root of sum) = (13923, 3010, 273819) and (33124, 18551, 1205750) (found in K. S. Brown's Mathpages). - Vladimir Pletser, Feb 01 2015
Another solution (when n is not a cube) is (n, first term, cubic root of sum) = (25201, 46690, 1764070). Note that n = 25201 = 11*29*79 is squarefree. This is the first known squarefree solution after n = 3 (given in the Math Overflow link). - Derek Orr, Mar 09 2015
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REFERENCES
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L.E. Dickson, History of the Theory of Numbers, Vol.II: Diophantine Analysis, Dover Publ., Mineola, 2005.
N. P. Smart, The algorithmic resolution of Diophantine equations, New York: Cambridge University Press, 1998.
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LINKS
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EXAMPLE
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11^3 + 12^3 + 13^3 + 14^3 (sum of four consecutive cubes) is a cube (20^3). So 4 is a member of this sequence.
Fermat's Last Theorem says that x^3 + y^3 = z^3 has no nontrivial solutions. Thus k^3 + (k+1)^3 = y^3 has no nontrivial solutions, so 2 is not a member of this sequence.
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PROG
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(PARI) a(n)=for(k=1, n+10^6, /* the limit here only guarantees the terms given in the sequence */ X=2*k+n-1; s=n*X*(X^2+n^2-1); if(ispower(s, 3), return(k)))
n=1; while(n<5001, if(a(n), print1(n, ", ")); n++)
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CROSSREFS
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KEYWORD
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nonn,hard
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AUTHOR
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EXTENSIONS
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STATUS
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approved
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