A291735 p-INVERT of (1,0,1,0,0,0,0,...), where p(S) = 1 - S - S^3.

1, 1, 3, 5, 10, 19, 35, 67, 124, 234, 441, 827, 1558, 2927, 5503, 10349, 19453, 36580, 68774, 129304, 243119, 457093, 859415, 1615837, 3038024, 5711986, 10739431, 20191855, 37963921, 71378219, 134202491, 252322113, 474405911, 891958973, 1677025407

Offset: 0

Author

Clark Kimberling, Sep 11 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 A291728 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 (1, 0, 2, 0, 3, 0, 3, 0, 1)

Crossrefs

Cf. A154272, A291728.

Programs

Mathematica

z = 60; s = x + x^3; p = 1 - s - s^3;
Drop[CoefficientList[Series[s, {x, 0, z}], x], 1]  (* A154272 *)
Drop[CoefficientList[Series[1/p, {x, 0, z}], x], 1]  (* A291735 *)
LinearRecurrence[{1,0,2,0,3,0,3,0,1},{1,1,3,5,10,19,35,67,124},40] (* Harvey P. Dale, Aug 25 2024 *)

Formula

G.f.: -(((1 + x^2) (1 + x^2 + 2 x^4 + x^6))/(-1 + x + 2 x^3 + 3 x^5 + 3 x^7 + x^9)).

a(n) = a(n-1) + 2*a(n-3) + 3*a(n-5) + 3*a(n-7) + a(n-9) for n >= 10.