A253778 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).

0, -2, 6, 34, 213, 406, 1134, 1735, 3606, 4966, 8790, 11368, 18171, 22534, 33558, 40381, 57084, 67150, 91206, 105406, 138705, 158038, 202686, 228259, 286578, 319606, 394134, 435940, 529431, 581446, 696870, 760633, 901176, 978334, 1147398, 1239706, 1440909, 1550230

Offset: 1

Author

Vladimir Pletser, Jan 12 2015

Keywords

• sign
• easy

Comments

A253778 is a subset of A240970.

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).

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.

There is no nontrivial solution 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 and c(n)= m(m^2-1)(m^2+2)/6.

Examples

			For n=1, a(1)= 0 and c(1)= 0 for M(1)=1= A118719(n+1) = 1^3= (A001651(n))^3.
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.
For n=3, a(3)=6 and c(3)=180 for M(3)=64= A118719(n+1) = 4^3= (A001651(n))^3.
See "File Triplets (M,a,c) for M=m^3" link.
		

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

Crossrefs

Cf. A240970, A253779, A253780.

Programs

Maple

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:

Formula

a(n) = (m-1)(m^2 (m-2)-4(m+1))/6 where m = A001651(n).

Conjectures from Colin Barker, Jan 13 2015: (Start)

a(n) = (54*n^4 - 252*n^3 + 312*n^2 - 144*n + 64) / 64 for n even.

a(n) = (54*n^4 - 180*n^3 + 96*n^2 - 12*n + 42) / 64 for n odd.

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).

(End)