OFFSET
1,2
COMMENTS
Numbers c 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 are no nontrivial solutions for M=m^3 if m=0(mod 3).
LINKS
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
FORMULA
a(n) = m(m^2-1)(m^2+2)/6 where m = A001651(n).
Conjectures from Colin Barker, Jan 13 2015: (Start)
a(n) = (81*n^5 - 270*n^4 + 396*n^3 - 312*n^2 + 96*n) / 64 for n even.
a(n) = (81*n^5 - 135*n^4 + 126*n^3 - 66*n^2 - 15*n + 9) / 64 for n odd.
G.f.: 6*x^2*(x^8+29*x^7+55*x^6+241*x^5+158*x^4+241*x^3+55*x^2+29*x+1) / ((x-1)^6*(x+1)^5).
(End)
EXAMPLE
For n=2, b(2)=-2 and a(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.
See "File Triplets (M,a,c) for M=m^3" link, [where in this File, M is the number of term, a the first term and c the square root of the sum].
MAPLE
restart: for n from 1 to 15000 do m:=n: if(modp(m, 3)>0) then c:=m*(m^2-1)*(m^2+2))/6: print (c): fi: od:
CROSSREFS
KEYWORD
nonn,easy
AUTHOR
Vladimir Pletser, Jan 12 2015
STATUS
approved