A098749 - OEIS

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

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OFFSET

0,4

LINKS

Table of n, a(n) for n=0..50.

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

MATHEMATICA

(* polynomial sequence with Theta1 to Theta0 pattern *) f[n_]=((n^4-n^3-1)/(n^3-n-1))^2; a=Table[Floor[f[n]], {n, 0, 50}]

CROSSREFS

Cf. A002522.

Sequence in context: A277186 A193053 A340047 * A034676 A076598 A374539

Adjacent sequences: A098746 A098747 A098748 * A098750 A098751 A098752

KEYWORD

nonn,easy,less

AUTHOR

Roger L. Bagula, Oct 01 2004

STATUS

approved