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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).

%I A253780 #37 Mar 03 2015 16:35:20
%S A253780 0,216,5832000,157464000,23295638016,170400029184,4767078987000,
%T A253780 19814511816000,241152896222784,731189187729000,5399901725184000,
%U A253780 13389040129314816,70865430394968000,152838610998696000,637623759116775816,1241409566336822784,4332341335608000000
%N 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).
%C A253780 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.
%C A253780 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.
%C A253780 There are no nontrivial solutions for M=m^3 if m=0(mod 3).
%C A253780 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.
%H A253780 Vladimir Pletser, <a href="/A253780/b253780.txt">Table of n, a(n) for n = 1..10000</a>
%H A253780 K. S. Brown's Mathpages, <a href="http://www.mathpages.com/home/kmath147.htm">Sum of Consecutive Nth Powers Equals an Nth Power</a>
%H A253780 Vladimir Pletser, <a href="/A253780/a253780.txt">File Triplets (M,b,c) for M=m^3</a>
%H A253780 Ben Vitale, <a href="http://benvitalenum3ers.wordpress.com/2014/08/06/when-sum-of-cubes-equals-a-cube/">Sum of Cubes Equals a Cube</a>
%F A253780 a(n) = (m(m^2-1)(m^2+2)/6)^3 where m = A001651(n).
%F A253780 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
%e A253780 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.
%e A253780 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.
%e A253780 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.
%e A253780 See "File Triplets (M,b,c) for M=m^3" link.
%p A253780 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:
%t A253780 a253780[n_] := (# (#^2 - 1) (#^2 + 2)/6)^3 & /@ Select[Range@ n, Mod[#, 3] != 0 &]; a253780[25] (* _Michael De Vlieger_, Jan 19 2015 *)
%o A253780 (PARI) a(n) = m=floor((3*n-1)/2); (m*(m^2-1)*(m^2+2)/6)^3 \\ _Colin Barker_, Jan 14 2015
%Y A253780 Cf. A240970, A253778, A253779.
%K A253780 nonn,easy
%O A253780 1,2
%A A253780 _Vladimir Pletser_, Jan 12 2015