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%I #29 Mar 03 2015 16:35:34
%S 0,-2,6,34,213,406,1134,1735,3606,4966,8790,11368,18171,22534,33558,
%T 40381,57084,67150,91206,105406,138705,158038,202686,228259,286578,
%U 319606,394134,435940,529431,581446,696870,760633,901176,978334,1147398,1239706,1440909,1550230
%N Numbers b that are the first of m^3 consecutive cubes whose sum is a cube c^3, where m^3 is not divisible by 3 (A118719).
%C A253778 is a subset of A240970.
%C 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).
%C 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 There is no nontrivial solution for M=m^3 if m=0(mod 3).
%C 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 Vladimir Pletser, <a href="/A253778/b253778.txt">Table of n, a(n) for n = 1..10000</a>
%H 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 Vladimir Pletser, <a href="/A253780/a253780.txt">File Triplets (M,b,c) for M=m^3</a>
%H 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 a(n) = (m-1)(m^2 (m-2)-4(m+1))/6 where m = A001651(n).
%F Conjectures from _Colin Barker_, Jan 13 2015: (Start)
%F a(n) = (54*n^4 - 252*n^3 + 312*n^2 - 144*n + 64) / 64 for n even.
%F a(n) = (54*n^4 - 180*n^3 + 96*n^2 - 12*n + 42) / 64 for n odd.
%F 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).
%F (End)
%e For n=1, a(1)= 0 and c(1)= 0 for M(1)=1= A118719(n+1) = 1^3= (A001651(n))^3.
%e For n=2, a(2)=-2 and c(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 For n=3, a(3)=6 and c(3)=180 for M(3)=64= A118719(n+1) = 4^3= (A001651(n))^3.
%e See "File Triplets (M,a,c) for M=m^3" link.
%p 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:
%Y Cf. A240970, A253779, A253780.
%K sign,easy
%O 1,2
%A _Vladimir Pletser_, Jan 12 2015
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