A291236 p-INVERT of (0,1,0,1,0,1,...), where p(S) = (1 - S)(1 - 3 S).

4, 13, 44, 147, 488, 1616, 5344, 17661, 58348, 192739, 636620, 2102688, 6944828, 22937405, 75757420, 250210275, 826389232, 2729379568, 9014530520, 29772975309, 98333463212, 324773375891, 1072653608596, 3542734230336, 11700856345972, 38645303343277

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 A291219 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 (4, -1, -4, -1)

Formula

G.f.: (4 - 3 x - 4 x^2)/(1 - 4 x + x^2 + 4 x^3 + x^4).

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

Mathematica

z = 60; s = x/(1 - x^2); p = (1 - s)(1 - 3s);
Drop[CoefficientList[Series[s, {x, 0, z}], x], 1]  (* A000035 *)
Drop[CoefficientList[Series[1/p, {x, 0, z}], x], 1]  (* A291236 *)

Crossrefs

Cf. A000035, A291219.

Sequence in context: A117882 A257674 A027123 * A339850 A273904 A027125

Adjacent sequences: A291233 A291234 A291235 * A291237 A291238 A291239

Keyword

nonn,easy

Author

Clark Kimberling, Aug 28 2017

Status

approved