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A253779 Numbers c whose cubes are equal to the sum of m^3 consecutive cubes for m^3 not divisible by 3 (A118719).

%I A253779 #28 Mar 03 2015 16:35:27
%S A253779 0,6,180,540,2856,5544,16830,27060,62244,90090,175440,237456,413820,
%T A253779 534660,860706,1074744,1630200,1983150,2872044,3422580,4776480,
%U A253779 5597856,7579110,8760780,11565756,13214994,17077320,19320840,24514644,27500220,34343370,38241456,47098800
%N A253779 Numbers c whose cubes are equal to the sum of m^3 consecutive cubes for m^3 not divisible by 3 (A118719).
%C A253779 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).
%C A253779 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.
%C A253779 There are no nontrivial solutions for M=m^3 if m=0(mod 3).
%C A253779 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 and c(n)= m(m^2-1)(m^2+2)/6.
%H A253779 Vladimir Pletser, <a href="/A253779/b253779.txt">Table of n, a(n) for n = 1..10000</a>
%H A253779 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 A253779 Vladimir Pletser, <a href="/A253780/a253780.txt">File Triplets (M,b,c) for M=m^3</a>
%H A253779 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 A253779 a(n) = m(m^2-1)(m^2+2)/6 where m = A001651(n).
%F A253779 Conjectures from _Colin Barker_, Jan 13 2015: (Start)
%F A253779 a(n) = (81*n^5 - 270*n^4 + 396*n^3 - 312*n^2 + 96*n) / 64 for n even.
%F A253779 a(n) = (81*n^5 - 135*n^4 + 126*n^3 - 66*n^2 - 15*n + 9) / 64 for n odd.
%F A253779 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).
%F A253779 (End)
%e A253779 For n=1, b(1)= 0 and a(1)= 0 for M(1)=1= A118719(n+1) = 1^3= (A001651(n))^3.
%e A253779 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.
%e A253779 For n=3, b(3)=6 and a(3)=180 for M(3)=64= A118719(n+1) = 4^3= (A001651(n))^3.
%e A253779 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].
%p A253779 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:
%Y A253779 Cf. A240970, A253778, A253780.
%K A253779 nonn,easy
%O A253779 1,2
%A A253779 _Vladimir Pletser_, Jan 12 2015