A289798 p-INVERT of (-1 + 2^n), where p(S) = 1 - S - S^2.

1, 5, 22, 93, 387, 1602, 6623, 27377, 113174, 467877, 1934315, 7996978, 33061703, 136686153, 565097958, 2336269341, 9658775347, 39932014114, 165089847535, 682526498529, 2821750872886, 11665888441301, 48229967585083, 199395852702354, 824356889826903

Offset: 0

Author

Clark Kimberling, Aug 12 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 A289780 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 (7, -15, 14, -4)

Crossrefs

Cf. A000225, A033453, A289780.

Programs

Mathematica

z = 60; s = x/((1 - 2*x)*(1 - x)); p = 1 - s - s^2;
Drop[CoefficientList[Series[s, {x, 0, z}], x], 1] (* A000225 *)
Drop[CoefficientList[Series[1/p, {x, 0, z}], x], 1] (* A289798 *)

Formula

G.f.: (1 - 2 x + 2 x^2)/(1 - 7 x + 15 x^2 - 14 x^3 + 4 x^4).

a(n) = 7*a(n-1) - 15*a(n-2) + 14*a(n-3) - 4*a(n-4).