A291241 p-INVERT of (0,1,0,1,0,1,...), where p(S) = 1 - S - S^2 + S^3.

1, 2, 3, 7, 10, 22, 32, 67, 99, 200, 299, 588, 887, 1708, 2595, 4913, 7508, 14018, 21526, 39725, 61251, 111922, 173173, 313752, 486925, 875702, 1362627, 2434747, 3797374, 6746350, 10543724, 18636343, 29180067, 51340988, 80521055, 141089508, 221610563

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 (1,4,-3,-4,1,1)

Crossrefs

Cf. A000035, A291219.

Programs

Magma

I:=[1,2,3,7,10,22]; [n le 6 select I[n] else Self(n-1)+4*Self(n-2)-3*Self(n-3)-4*Self(n-4)+Self(n-5)+Self(n-6): n in [1..40]]; // Vincenzo Librandi, Aug 29 2017

Mathematica

z = 60; s = x/(1 - x^2); p = 1 - 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]  (* A291241 *)
LinearRecurrence[{1, 4, -3, -4, 1, 1}, {1, 2, 3, 7, 10, 22}, 40] (* Vincenzo Librandi, Aug 29 2017 *)

Formula

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

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