A291724 p-INVERT of (1,0,1,0,0,0,0,...), where p(S) = 1 - S^5.

0, 0, 0, 0, 1, 0, 5, 0, 10, 1, 10, 10, 5, 45, 2, 120, 15, 210, 105, 253, 455, 230, 1365, 310, 3004, 1185, 5030, 4855, 6735, 15506, 8735, 38790, 17655, 77955, 56134, 130030, 178500, 195365, 481750, 327263, 1088225, 761775, 2095350, 2162550, 3593394, 5940325

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 (0, 0, 0, 0, 1, 0, 5, 0, 10, 0, 10, 0, 5, 0, 1)

Crossrefs

Cf. A154272, A291728.

Programs

Mathematica

z = 60; s = x + x^3; p = 1 - s^5;
Drop[CoefficientList[Series[s, {x, 0, z}], x], 1]  (* A154272 *)
Drop[CoefficientList[Series[1/p, {x, 0, z}], x], 1]  (* A291724 *)
LinearRecurrence[{0, 0, 0, 0, 1, 0, 5, 0, 10, 0, 10, 0, 5, 0, 1}, {0, 0, 0, 0, 1, 0, 5, 0, 10, 1, 10, 10, 5, 45, 2}, 50] (* Vincenzo Librandi, Sep 10 2017 *)

Formula

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

a(n) = a(n-5) + 5*a(n-7) + 10*a(n-9) + 10*a(n-11) + 5*a(n-13) + a(n-15) for n >= 15.