OFFSET
1,2
COMMENTS
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).
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)^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
EXAMPLE
MAPLE
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:
MATHEMATICA
a253780[n_] := (# (#^2 - 1) (#^2 + 2)/6)^3 & /@ Select[Range@ n, Mod[#, 3] != 0 &]; a253780[25] (* Michael De Vlieger, Jan 19 2015 *)
PROG
(PARI) a(n) = m=floor((3*n-1)/2); (m*(m^2-1)*(m^2+2)/6)^3 \\ Colin Barker, Jan 14 2015
CROSSREFS
KEYWORD
nonn,easy
AUTHOR
Vladimir Pletser, Jan 12 2015
STATUS
approved