A253779 -id:A253779 - cp's OEIS frontend

A240970 Numbers n such that the sum of n consecutive positive cubes is a cube for some initial starting number k.

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

Offset: 1

Derek Orr, Aug 05 2014

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

			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.

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.

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

Cf. A001032, A253778, A253779, A253780.

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++)

More terms from Derek Orr, May 12 2015

A253778 Numbers b that are the first of m^3 consecutive cubes whose sum is a cube c^3, where m^3 is not divisible by 3 (A118719).

Original entry on oeis.org

0, -2, 6, 34, 213, 406, 1134, 1735, 3606, 4966, 8790, 11368, 18171, 22534, 33558, 40381, 57084, 67150, 91206, 105406, 138705, 158038, 202686, 228259, 286578, 319606, 394134, 435940, 529431, 581446, 696870, 760633, 901176, 978334, 1147398, 1239706, 1440909, 1550230

Offset: 1

Vladimir Pletser, Jan 12 2015

A253778 is a subset of A240970.
Numbers b such that b^3 + (b+1)^3 + ... + (b+M-1)^3 = c^3 has nontrivial solutions over the integers for M equal to a cube not divisible by 3 (A118719).
If M is a cube not divisible by 3, there always exists at least one nontrivial solution for the sum of M consecutive cubes starting from b^3 and equaling a cube c^3.
There is no nontrivial solution 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 b(n) =(m-1)(m^2 (m-2)-4(m+1))/6 and c(n)= m(m^2-1)(m^2+2)/6.

			For n=1, a(1)= 0 and c(1)= 0 for M(1)=1= A118719(n+1) = 1^3= (A001651(n))^3.
For n=2, a(2)=-2 and c(2)=6 for M(2)=8= A118719(n+1) = 2^3= (A001651(n))^3 , which is Euler relation: (-2)^3 + (-1)^3 + 0^3 + 1^3 + 2^3 + 3^3 + 4^3 + 5^3 = 6^3.
For n=3, a(3)=6 and c(3)=180 for M(3)=64= A118719(n+1) = 4^3= (A001651(n))^3.
See "File Triplets (M,a,c) for M=m^3" link.

Vladimir Pletser, Table of n, a(n) for n = 1..10000
K. S. Brown's Mathpages, Sum of Consecutive Nth Powers Equals an Nth Power
Vladimir Pletser, File Triplets (M,b,c) for M=m^3
Ben Vitale, Sum of Cubes Equals a Cube

Cf. A240970, A253779, A253780.

restart: for n from 1 to 15000 do m:=n: if(modp(m,3)>0) then a:=(m-1)*(m^2*(m-2)-4*(m+1))/6: print (a): fi: od:

a(n) = (m-1)(m^2 (m-2)-4(m+1))/6 where m = A001651(n).
Conjectures from Colin Barker, Jan 13 2015: (Start)
a(n) = (54*n^4 - 252*n^3 + 312*n^2 - 144*n + 64) / 64 for n even.
a(n) = (54*n^4 - 180*n^3 + 96*n^2 - 12*n + 42) / 64 for n odd.
G.f.: -x^2*(x^7+5*x^6+60*x^5+69*x^4+147*x^3+36*x^2+8*x-2) / ((x-1)^5*(x+1)^4).
(End)

A253780 Cubes c^3 that are equal to the sum of m^3 consecutive cubes starting at b^3 with b (A253778) for m^3 not divisible by 3 (A118719).

Original entry on oeis.org

0, 216, 5832000, 157464000, 23295638016, 170400029184, 4767078987000, 19814511816000, 241152896222784, 731189187729000, 5399901725184000, 13389040129314816, 70865430394968000, 152838610998696000, 637623759116775816, 1241409566336822784, 4332341335608000000

Offset: 1

Vladimir Pletser, Jan 12 2015

Consider the set of all nontrivial solutions of the equation b^3 + (b+1)^3 + ... + (b+M-1)^3 = c^3 for integers b, M and c, with M equal to a cube not divisible by 3 (A118719). This sequence gives the values of c^3.
If M is a cube not divisible by 3, there always exists at least one nontrivial solution for the sum of M consecutive cubes starting from b^3 and equaling a cube a=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 b(n)=(m-1)(m^2(m-2)-4(m+1))/6, c(n) = m(m^2-1)(m^2+2)/6.

			For n=1, b(1)=0, c(1)=0 and a(1)=C^3=0 for M(1)=1=A118719(n+1) = 1^3 = (A001651(n))^3.
For n=2, b(2)=-2, c(2)=6 and a(2)=c^3=216 for M(2)=8= A118719(n+1) = 2^3 = (A001651(n))^3, which is Euler relation: (-2)^3 + (-1)^3 + 0^3 + 1^3 + 2^3 + 3^3 + 4^3 + 5^3 = 6^3.
For n=3, b(3)=6, c(3)=180 and a(3)=c^3=5832000 for M(3)=64= A118719(n+1) = 4^3 = (A001651(n))^3.
See "File Triplets (M,b,c) for M=m^3" link.

Vladimir Pletser, Table of n, a(n) for n = 1..10000
K. S. Brown's Mathpages, Sum of Consecutive Nth Powers Equals an Nth Power
Vladimir Pletser, File Triplets (M,b,c) for M=m^3
Ben Vitale, Sum of Cubes Equals a Cube

Cf. A240970, A253778, A253779.

restart: for n from 1 to 15000 do m:=n: if(modp(m,3)>0) then a:=(m*(m^2-1)*(m^2+2))/6)^3: print (a): fi: od:

a253780[n_] := (# (#^2 - 1) (#^2 + 2)/6)^3 & /@ Select[Range@ n, Mod[#, 3] != 0 &]; a253780[25] (* Michael De Vlieger, Jan 19 2015 *)

a(n) = m=floor((3*n-1)/2); (m*(m^2-1)*(m^2+2)/6)^3 \\ Colin Barker, Jan 14 2015

a(n) = (m(m^2-1)(m^2+2)/6)^3 where m = A001651(n).

Empirical g.f.: 216*x^2*(x^28 +26999*x^27 +701985*x^26 +106716191*x^25 +670508953*x^24 +19676938356*x^23 +59522167700*x^22 +716736365044*x^21 +1294915592031*x^20 +8429482190425*x^19 +9962263692743*x^18 +39936619145457*x^17 +32146645170615*x^16 +84954433749528*x^15 +47129019463944*x^14 +84954433749528*x^13 +32146645170615*x^12 +39936619145457*x^11 +9962263692743*x^10 +8429482190425*x^9 +1294915592031*x^8 +716736365044*x^7 +59522167700*x^6 +19676938356*x^5 +670508953*x^4 +106716191*x^3 +701985*x^2 +26999*x +1) / ((x-1)^16*(x +1)^15). - Colin Barker, Jan 14 2015

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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.

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A253778 Numbers b that are the first of m^3 consecutive cubes whose sum is a cube c^3, where m^3 is not divisible by 3 (A118719).

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A253780 Cubes c^3 that are equal to the sum of m^3 consecutive cubes starting at b^3 with b (A253778) for m^3 not divisible by 3 (A118719).

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