A206568 Expand 1/(8 - 8 x + 3 x^3 - 2 x^4) in powers of x, then multiply coefficient of x^n by 8^(1 + floor(n/3)) to get integers.

1, 1, 1, 5, 4, 3, 25, 23, 22, 149, 130, 110, 785, 693, 623, 4389, 3880, 3397, 23977, 21115, 18684, 131893, 116502, 102680, 724705, 638985, 563949, 3980357, 3512812, 3098935, 21873593, 19295871, 17024690

Offset: 0

Author

Roger L. Bagula, Feb 09 2012

Keywords

• nonn
• easy

Comments

Bob Hanlon (hanlonr(AT)cox.net) helped convert the expansion to a recursion.

Links

• Alois P. Heinz, Table of n, a(n) for n = 0..1000

Crossrefs

Cf. A202907, A167602, A167602, A117791, A107293, A204631, A185357, A205961.

Programs

Maple

a:= n-> (<<0|1|0|0>, <0|0|1|0>, <0|0|0|1>, <64|69|21|-1>>^ iquo(n, 3, 'r'). `if`(r=0, <<1, 5, 25, 149>>, `if`(r=1, <<1, 4, 23, 130>>, <<1, 3, 22, 110>>)))[1, 1]: seq (a(n), n=0..40); # Alois P. Heinz, Feb 11 2012

Mathematica

(* expansion*)
Table[8^(1 + Floor[n/3])*SeriesCoefficient[Series[1/(8 - 8 x + 3 x^3 - 2 x^4), {x, 0, 50}], n], {n, 0,50}]
(*recursion*)
a[1] = 1; a[2] = 1; a[3] = 1; a[4] = 5; a[5] = 4; a[6] = 3;
a[7] = 25; a[8] = 23; a[9] = 22; a[10] = 149; a[11] = 130;
a[12] = 110;
a[n_Integer?Positive] := a[n] = 64*a[-12 + n] + 69*a[-9 + n] + 21*a[-6 +n] - a[-3 + n]
Table[a[n], {n, 1, 50}]

Formula

G.f.: (-4*x^8-6*x^7-9*x^6-4*x^5-5*x^4-6*x^3-x^2-x-1) / (64*x^12 +69*x^9 +21*x^6 -x^3-1).