A291391 p-INVERT of (1,1,0,0,0,0,...), where p(S) = (1 - 6 S)^2.

12, 120, 1080, 9180, 75168, 599616, 4691520, 36164880, 275503680, 2078711424, 15559682688, 115688917440, 855249269760, 6291326453760, 46080184338432, 336227628720384, 2445042642140160, 17726787591690240, 128173151784867840, 924487654349822976

Offset

0,1

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.

Links

Clark Kimberling, Table of n, a(n) for n = 0..1000

Index entries for linear recurrences with constant coefficients, signature (12, -24, -72, -36)

Formula

G.f.: -((12 (1 + x) (-1 + 3 x + 3 x^2))/(-1 + 6 x + 6 x^2)^2).

a(n) = 12*a(n-1) - 24*a(n-2) - 72*a(n-3) - 36*a(n-4) for n >= 5.

Mathematica

z = 60; s = x + x^2; p = (1 - 6 s)^2;
Drop[CoefficientList[Series[s, {x, 0, z}], x], 1]  (* A019590 *)
u = Drop[CoefficientList[Series[1/p, {x, 0, z}], x], 1]  (* A291391 *)
u / 12  (* A291392 *)

Crossrefs

Cf. A019590, A291382, A291392.

Sequence in context: A009140 A012273 A008465 * A115902 A277491 A004332

Adjacent sequences: A291388 A291389 A291390 * A291392 A291393 A291394

Keyword

nonn,easy

Author

Clark Kimberling, Sep 06 2017

Status

approved