|
| |
|
|
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).
|
|
4
|
|
|
|
0, 216, 5832000, 157464000, 23295638016, 170400029184, 4767078987000, 19814511816000, 241152896222784, 731189187729000, 5399901725184000, 13389040129314816, 70865430394968000, 152838610998696000, 637623759116775816, 1241409566336822784, 4332341335608000000
(list;
graph;
refs;
listen;
history;
text;
internal format)
|
|
|
|
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).
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.
|
|
|
LINKS
|
|
|
|
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
|
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.
|
|
|
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
|
|
|
|
STATUS
|
approved
|
| |
|
|
