A290909 p-INVERT of the positive integers, where p(S) = 1 - 5*S^2.

0, 5, 20, 75, 300, 1200, 4780, 19045, 75900, 302475, 1205400, 4803680, 19143320, 76288725, 304020900, 1211564475, 4828248580, 19241224720, 76678887300, 305575754325, 1217760780300, 4852941691355, 19339630115120, 77071046136000, 307138560414000

Offset: 0

Author

Clark Kimberling, Aug 18 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 A290890 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).

Crossrefs

Cf. A000027, A290890, A290910.

Programs

Mathematica

z = 60; s = x/(1 - x)^2; p = 1 - 5 s^2;
Drop[CoefficientList[Series[s, {x, 0, z}], x], 1] (* A000027 *)
u = Drop[CoefficientList[Series[1/p, {x, 0, z}], x], 1] (* A290909 *)
u/5 (* A290910 *)

Formula

G.f.: (5 x)/(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).

a(n) = 5*A290910(n) for n >= 0.