A291739 p-INVERT of (1,0,1,0,0,0,0,...), where p(S) = 1 - S^3 - S^6.

0, 0, 1, 0, 3, 2, 3, 12, 4, 30, 27, 45, 108, 90, 260, 342, 498, 1115, 1218, 2709, 3913, 5949, 11469, 15262, 28461, 44556, 68028, 123243, 178650, 311337, 498114, 777996, 1340603, 2052765, 3435906, 5569902, 8800392, 14783823, 23242761, 38249550, 62156709

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 (0, 0, 1, 0, 3, 1, 3, 6, 1, 15, 0, 20, 0, 15, 0, 6, 0, 1)

Crossrefs

Cf. A154272, A291728.

Programs

Mathematica

z = 60; s = x + x^3; p = 1 - s^3 - s^6;
Drop[CoefficientList[Series[s, {x, 0, z}], x], 1]  (* A154272 *)
Drop[CoefficientList[Series[1/p, {x, 0, z}], x], 1]  (* A291739 *)

Formula

G.f.: -((x^2 (1 + x^2)^3 (1 + x + x^2) (1 + x + x^3) (1 - 2 x + 2 x^2 - x^3 + x^4))/(-1 + x^3 + 3 x^5 + x^6 + 3 x^7 + 6 x^8 + x^9 + 15 x^10 + 20 x^12 + 15 x^14 + 6 x^16 + x^18)).

a(n) = a(n-3) + 3*a(n-5) + a(n-6) + 3*a(n-7) + 6*a(n-8) + a(n-9) + 15*a(n-10) + 20 *a(n-12) + 15*a(n-14) + 6*a(n-16) + a(n-18) for n >= 19.