A291218 p-INVERT of (0,1,0,1,0,1,...), where p(S) = 1 - S^5.

0, 0, 0, 0, 1, 0, 5, 0, 15, 1, 35, 10, 70, 55, 127, 220, 225, 715, 450, 2003, 1175, 5025, 3775, 11650, 12630, 25850, 40150, 57475, 118425, 134883, 325075, 345090, 840725, 952195, 2083888, 2722455, 5056055, 7765010, 12293890, 21615771, 30591685, 58293475

Offset: 0

Author

Clark Kimberling, Aug 24 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 A291219 for a guide to related sequences.

Links

• Andrew Howroyd, Table of n, a(n) for n = 0..1000
• Index entries for linear recurrences with constant coefficients, signature (0,5, 0,-10,1,10,0,-5,0,1)

Crossrefs

Cf. A000035, A291000, A291219.

Programs

Magma

I:=[0,0,0,0,1,0,5,0,15,1]; [n le 10 select I[n] else 5*Self(n-2)-10*Self(n-4)+Self(n-5)+10*Self(n-6)-5*Self(n-8)+Self(n-10): n in [1..45]]; // Vincenzo Librandi, Aug 25 2017

Mathematica

z = 60; s = x/(1 - x^2); p = 1 - s^3;
Drop[CoefficientList[Series[s, {x, 0, z}], x], 1]  (* A000035 *)
Drop[CoefficientList[Series[1/p, {x, 0, z}], x], 1]  (* A291218 *)
LinearRecurrence[{0, 5, 0, -10, 1, 10, 0, -5, 0, 1}, {0, 0, 0, 0, 1, 0, 5, 0, 15, 1}, 50] (* Vincenzo Librandi, Aug 25 2017 *)

Formula

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

a(n) = 5*a(n-2) - 10*a(n-4) + a(n-5) + 10* a(n-6) - 5*a(n-8) + a(n-10) for n >= 11.