A291181 p-INVERT of the positive integers, where p(S) = 1 - 8*S.

8, 80, 792, 7840, 77608, 768240, 7604792, 75279680, 745192008, 7376640400, 73021211992, 722835479520, 7155333583208, 70830500352560, 701149669942392, 6940666199071360, 68705512320771208, 680114457008640720, 6732439057765635992, 66644276120647719200

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 (10, -1)

Crossrefs

Cf. A000027, A290890.

Programs

Mathematica

z = 60; s = x/(1 - x)^2; p = 1 - 8 s;
Drop[CoefficientList[Series[s, {x, 0, z}], x], 1] (* A000027 *)
Drop[CoefficientList[Series[1/p, {x, 0, z}], x], 1]  (* A291181 *)
LinearRecurrence[{10,-1},{8,80},30] (* Harvey P. Dale, Jul 31 2023 *)

Formula

G.f.: 8/(1 - 10 x + x^2).

a(n) = 10*a(n-1) - a(n-2).

a(n) = 8*A004189(n+1) for n >= 0.