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

0,5

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)

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.

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 *)

Crossrefs

Cf. A154272, A291728.

Sequence in context: A078017 A343170 A169816 * A057053 A081850 A247237

Adjacent sequences: A291736 A291737 A291738 * A291740 A291741 A291742

Keyword

nonn,easy

Author

Clark Kimberling, Sep 11 2017

Status

approved