A190636 a(n)=(n^3+3*n^7)/4.

1, 98, 1647, 12304, 58625, 210006, 617743, 1572992, 3587409, 7500250, 14615711, 26874288, 47061937, 79060814, 128145375, 201327616, 307755233, 459166482, 670405519, 960002000, 1350818721, 1870771078, 2553622127, 3439857024, 4577640625, 6023862026, 7845269823

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

Sequence in context: A292552 A233373 A221747 * A019563 A285045 A197616

Adjacent sequences: A190633 A190634 A190635 * A190637 A190638 A190639

Keyword

nonn,easy

Author

Rafael Parra Machio, May 15 2011

Status

approved