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

0, 6, 180, 540, 2856, 5544, 16830, 27060, 62244, 90090, 175440, 237456, 413820, 534660, 860706, 1074744, 1630200, 1983150, 2872044, 3422580, 4776480, 5597856, 7579110, 8760780, 11565756, 13214994, 17077320, 19320840, 24514644, 27500220, 34343370, 38241456, 47098800

Offset: 1

Author

Vladimir Pletser, Jan 12 2015

Keywords

• nonn
• easy

Comments

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).

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 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 and c(n)= m(m^2-1)(m^2+2)/6.

Examples

			For n=1, b(1)= 0 and a(1)= 0 for M(1)=1= A118719(n+1) = 1^3= (A001651(n))^3.
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.
For n=3, b(3)=6 and a(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, [where in this File, M is the number of term, a the first term and c the square root of the sum].
		

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, A253778, A253780.

Programs

Maple

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:

Formula

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

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

a(n) = (81*n^5 - 270*n^4 + 396*n^3 - 312*n^2 + 96*n) / 64 for n even.

a(n) = (81*n^5 - 135*n^4 + 126*n^3 - 66*n^2 - 15*n + 9) / 64 for n odd.

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).

(End)