A291727 p-INVERT of (1,0,1,0,0,0,0,...), where p(S) = (1 - S)^4.

4, 10, 24, 55, 116, 234, 460, 879, 1640, 3006, 5424, 9650, 16964, 29510, 50852, 86893, 147360, 248198, 415440, 691428, 1144772, 1886270, 3094292, 5055140, 8227084, 13341756, 21564360, 34746331, 55823080, 89439056, 142928424, 227851285, 362396564, 575135150

Offset: 0

Author

Clark Kimberling, Sep 08 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 (4, -6, 8, -13, 12, -10, 12, -6, 4, -4, 0, -1)

Crossrefs

Cf. A154272, A291728.

Programs

Mathematica

z = 60; s = x + x^3; p = (1 - s)^4;
Drop[CoefficientList[Series[s, {x, 0, z}], x], 1]  (* A154272 *)
Drop[CoefficientList[Series[1/p, {x, 0, z}], x], 1]  (* A291727 *)
LinearRecurrence[{4,-6,8,-13,12,-10,12,-6,4,-4,0,-1},{4,10,24,55,116,234,460,879,1640,3006,5424,9650},40] (* Harvey P. Dale, Mar 13 2025 *)

Formula

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

a(n) = 4*a(n-1) - 6*a(n-2) + 8*a(n-3) - 13*a(n-4) + 12*a(n-5) - 10*a(n-6) + 12*a(n-7) - 6*a(n-8) +4*a(n-9) - 4*a(n-10) - a(n-12) for n >= 13.