A291387 p-INVERT of (1,1,0,0,0,0,...), where p(S) = (1 - 4 S)^2.

8, 56, 352, 2096, 12032, 67328, 369664, 2000128, 10696704, 56666112, 297836544, 1555066880, 8073379840, 41709076480, 214558048256, 1099562549248, 5616171483136, 28599668703232, 145249047412736, 735884541427712, 3720035809886208, 18767645931208704

Offset: 0

Author

Clark Kimberling, Sep 04 2017

Keywords

• nonn
• easy

Comments

Suppose s = (c(0), c(1), c(2), ...) is a sequence and p(S) is a polynomial. Let S(x) = c(0)*x + c(1)*x^2 + c(2)*x^3 + ... and T(x) = (-p(0) + 1/p(S(x)))/x. The p-INVERT of s is the sequence t(s) of coefficients in the Maclaurin series for T(x). Taking p(S) = 1 - S gives the "INVERT" transform of s, so that p-INVERT is a generalization of the "INVERT" transform (e.g., A033453).

See A291382 for a guide to related sequences.

All terms = 0 mod 8. - Muniru A Asiru, Sep 07 2017

Links

• Clark Kimberling, Table of n, a(n) for n = 0..1000
• Index entries for linear recurrences with constant coefficients, signature (8, -8, -32, -16)

Crossrefs

Cf. A019590, A291382, A291388.

Programs

GAP

a:=8*[1,7,44,262];; for n in [5..10^2] do a[n]:=8*a[n-1]-8*a[n-2]-32*a[n-3]-16*a[n-4]; od; a;  # Muniru A Asiru, Sep 07 2017

Mathematica

z = 60; s = x + x^2; p = (1 - 4 s)^2;
Drop[CoefficientList[Series[s, {x, 0, z}], x], 1]  (* A019590 *)
u = Drop[CoefficientList[Series[1/p, {x, 0, z}], x], 1]  (* A291387 *)
u / 8  (* A291388 *)

Formula

G.f.: -((8 (1 + x) (-1 + 2 x + 2 x^2))/(-1 + 4 x + 4 x^2)^2).

a(n) = 8*a(n-1) - 8*a(n-2) - 32*a(n-3) - 16*a(n-4) for n >= 5.