OFFSET
1,2
COMMENTS
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).
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-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)
EXAMPLE
MAPLE
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:
CROSSREFS
KEYWORD
sign,easy
AUTHOR
Vladimir Pletser, Jan 12 2015
STATUS
approved