OFFSET
1,2
COMMENTS
Each term is the difference of two cubes because ((n^3+n)/2)^3-((n^3-n)/2)^3=(n^3+3*n^7)/4. More generally, ((n^s+n)/2)^3-((n^s-n)/2)^3 = (n^3+3*n^(2s+1))/4 for any s; in this case, s=3.
LINKS
Nathaniel Johnston, Table of n, a(n) for n = 1..1000
Rafael Parra Machío, Ecuaciones Diofanticas, p. 24
Rafael Parra Machío, Ecuaciones Cuarticas, p. 11
FORMULA
a(n) = ((n^3+n)/2)^3 - ((n^3-n)/2)^3.
G.f.: (z^7+90*z^6+891*z^5+1816*z^4+891*z^3+90*z^2+z)/(z-1)^8.
EXAMPLE
58625=65^3-60^3=(5^3+3*5^7)/4; 47061937=1105^3-1092^3=(13^3 + 3*13^7)/4
MATHEMATICA
Table[((n^3+n)/2)^3 - ((n^3-n)/2)^3, {n, 1, 20}]
Table[1/4*(n^3+3 n^7), {n, 1, 20}]
CROSSREFS
KEYWORD
nonn,easy
AUTHOR
Rafael Parra Machio, May 15 2011
STATUS
approved