A098749 Let f[n_]=((n^4-n^3-1)/ (n^3-n-1))^2; then a(n) = Floor[f[n]].

1, 1, 1, 5, 10, 17, 26, 37, 50, 65, 82, 101, 122, 145, 170, 197, 226, 257, 290, 325, 362, 401, 442, 485, 530, 577, 626, 677, 730, 785, 842, 901, 962, 1025, 1090, 1157, 1226, 1297, 1370, 1445, 1522, 1601, 1682, 1765, 1850, 1937, 2026, 2117, 2210, 2305, 2402

Offset: 0

Author

Roger L. Bagula, Oct 01 2004

Keywords

• nonn
• easy
• less

Crossrefs

Cf. A002522.

Programs

Mathematica

(* polynomial sequence with Theta1 to Theta0 pattern*) digits=200 f[n_]=((n^4-n^3-1)/ (n^3-n-1))^2 a=Table[Floor[f[n]], {n, 0, digits}]

Formula

It is easy to show that Floor[((n^4-n^3-1)/ (n^3-n-1))^2] = (n-1)^2 + 1 for n >= 3. So this is essentially the same sequence as A002522. - Juan Jose Alba Gonzalez, Nov 09 2006.