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A240970 Numbers n such that the sum of n consecutive positive cubes is a cube for some initial starting number k. 4
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 (list; graph; refs; listen; history; text; internal format)
OFFSET

1,2

COMMENTS

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 exists 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 exists 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 equations 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 triplets (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 triplets (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

A118719 (excluding its initial term) is a subsequence. - Derek Orr, May 12 2015

REFERENCES

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.

LINKS

Table of n, a(n) for n=1..49.

Yuri F. Bilu and Guillaume Hanrot, Solving superelliptic Diophantine equations by Baker's method, Compositio Mathematica, Volume 112, Issue 03, July 1998, pp 273-312.

K. S. Brown's Mathpages, Sum of Consecutive Nth Powers Equals an Nth Power

Derek Orr (Math Overflow post), Sum of Consecutive Cubes

Ben Vitale, Sum of Cubes Equals a Cube

EXAMPLE

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.

PROG

(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++)

CROSSREFS

Cf. A001032, A253778, A253779, A253780.

Sequence in context: A237884 A256532 A051719 * A047165 A124631 A262033

Adjacent sequences:  A240967 A240968 A240969 * A240971 A240972 A240973

KEYWORD

nonn,hard

AUTHOR

Derek Orr, Aug 05 2014

EXTENSIONS

More terms from Derek Orr, May 12 2015

STATUS

approved

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Last modified September 26 00:10 EDT 2017. Contains 292500 sequences.