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%I #111 Jan 19 2019 04:14:59
%S 1,3,4,20,25,49,64,99,125,153,288,343,512,1000,1331,1849,2197,2744,
%T 4096,4913,6591,6859,8000,10200,10648,12167,13923,14161,15625,17576,
%U 19220,21456,21952,24389,25201,29791,32768,33124,39304,42875,49776,50653,54872,63001,64000,68921,79507,85184,97336
%N Numbers n such that the sum of n consecutive positive cubes is a cube for some initial starting number k.
%C 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
%C 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]
%C 5000 < a(17) <= 6591. - _Derek Orr_, Jan 17 2015
%C It is known that 14161 and 63001 are members of this sequence.
%C 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}.
%C 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.
%C 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, ...
%C 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.
%C n = 8, g(n) <= 6.
%C n = 27, g(n) <= 168.
%C n = 64, g(n) <= 1333.
%C n = 125, g(n) <= 6447.
%C n = 216, g(n) <= 23219.
%C n = 343, g(n) <= 68456.
%C n = 512, g(n) <= 174506.
%C n = 729, g(n) <= 398215.
%C n = 1000, g(n) <= 832832.
%C n = 1331, g(n) <= 1623264.
%C n = 1728, g(n) <= 2985119.
%C n = 2197, g(n) <= 5227943.
%C ...
%C 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.
%C 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).
%C 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
%C The equation S(n,k) = y^3 is a superelliptic curve and thus has only a finite number of integral points.
%C 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).
%C 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.
%C 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.
%C 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
%C 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
%C 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
%C 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
%C 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
%C A118719 (excluding its initial term) is a subsequence. - _Derek Orr_, May 12 2015
%D L.E. Dickson, History of the Theory of Numbers, Vol.II: Diophantine Analysis, Dover Publ., Mineola, 2005.
%D N. P. Smart, The algorithmic resolution of Diophantine equations, New York: Cambridge University Press, 1998.
%H Yuri F. Bilu and Guillaume Hanrot, <a href="http://dx.doi.org/10.1023/A:1000305028888">Solving superelliptic Diophantine equations by Baker's method</a>, Compositio Mathematica, Volume 112, Issue 03, July 1998, pp 273-312.
%H K. S. Brown's Mathpages, <a href="http://www.mathpages.com/home/kmath147.htm">Sum of Consecutive Nth Powers Equals an Nth Power</a>
%H Derek Orr (Math Overflow post), <a href="http://mathoverflow.net/questions/199297/sum-of-consecutive-cubes/199396#199396">Sum of Consecutive Cubes</a>
%H Ben Vitale, <a href="http://benvitalenum3ers.wordpress.com/2014/08/06/when-sum-of-cubes-equals-a-cube/">Sum of Cubes Equals a Cube</a>
%e 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.
%e 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.
%o (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)))
%o n=1;while(n<5001,if(a(n),print1(n,", "));n++)
%Y Cf. A001032, A253778, A253779, A253780.
%K nonn,hard
%O 1,2
%A _Derek Orr_, Aug 05 2014
%E More terms from _Derek Orr_, May 12 2015
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