A291242 p-INVERT of (0,1,0,1,0,1,...), where p(S) = 1 - 2 S - S^2 + S^3.

2, 5, 13, 35, 91, 241, 631, 1662, 4362, 11470, 30127, 79179, 208023, 546633, 1436257, 3773939, 9916134, 26055432, 68461966, 179888381, 472667065, 1241962303, 3263330095, 8574599917, 22530279167, 59199680826, 155550750026, 408719050346, 1073934109927

Offset: 0

Author

Clark Kimberling, Aug 28 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

• Clark Kimberling, Table of n, a(n) for n = 0..1000
• Index entries for linear recurrences with constant coefficients, signature (2,4,-5,-4,2,1)

Crossrefs

Cf. A000035, A291219.

Programs

Magma

I:=[2,5,13,35,91,241]; [n le 6 select I[n] else 2*Self(n-1)+4*Self(n-2)-5*Self(n-3)-4*Self(n-4)+2*Self(n-5)+Self(n-6): n in [1..30]]; // Vincenzo Librandi, Aug 29 2017

Mathematica

z = 60; s = x/(1 - x^2); p = 1 - 2 s - s^2 + s^3;
Drop[CoefficientList[Series[s, {x, 0, z}], x], 1]  (* A000035 *)
Drop[CoefficientList[Series[1/p, {x, 0, z}], x], 1]  (* A291242 *)
LinearRecurrence[{2, 4, -5, -4, 2, 1}, {2, 5, 13, 35, 91, 241}, 30] (* Vincenzo Librandi, Aug 29 2017 *)

Formula

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

a(n) = 2*a(n-1) + 4*a(n-2) - 5*a(n-3) - 4*a(n-4) + 2*a(n-5) + a(n-6) for n >= 7.