A291026 p-INVERT of the positive integers, where p(S) = 1 - 4*S + S^2.

4, 23, 128, 711, 3948, 21920, 121700, 675673, 3751296, 20826953, 115629868, 641969344, 3564171060, 19788040311, 109861881472, 609945846247, 3386378699324, 18800948912352, 104381615697460, 579519775642745, 3217455182279552, 17863096800262569

Offset: 0

Author

Clark Kimberling, Aug 19 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 (8,-15,8,-1).

Crossrefs

Cf. A000027, A033453, A290890.

Programs

Mathematica

z = 60; s = x/(1 - x)^2; p = 1 - 4 s + s^2;
Drop[CoefficientList[Series[s, {x, 0, z}], x], 1] (* A000027 *)
u = Drop[CoefficientList[Series[1/p, {x, 0, z}], x], 1]  (* A291026 *)
LinearRecurrence[{8,-15,8,-1},{4,23,128,711},30] (* Harvey P. Dale, May 18 2024 *)

Formula

G.f.: (4 - 9 x + 4 x^2)/(1 - 8 x + 15 x^2 - 8 x^3 + x^4).

a(n) = 8*a(n-1) - 15*a(n-2) + 8*a(n-3) - a(n-4).