A291408 p-INVERT of (1,1,0,0,0,0,...), where p(S) = (1 - S)(1 - S^2).

1, 3, 6, 11, 21, 39, 70, 126, 224, 394, 690, 1201, 2079, 3585, 6158, 10541, 17991, 30623, 51996, 88092, 148944, 251364, 423492, 712369, 1196557, 2007135, 3362598, 5626847, 9405465, 15705447, 26200066, 43667802, 72719312, 121000846, 201185334, 334265089

Offset: 0

Author

Clark Kimberling, Sep 07 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 A291382 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 (1,2,1,-2,-3,-1)

Crossrefs

Cf. A019590, A275439, A291382.

Programs

GAP

a:=[1,3,6,11,21,39];;
for n in [7..10^2] do a[n]:=a[n-1]+2*a[n-2]+a[n-3]-2*a[n-4]-3*a[n-5]- a[n-6]; od; a; # Muniru A Asiru, Sep 10 2017

Mathematica

z = 60; s = x + x^2; p = (1 - s)(1 - s^2);
Drop[CoefficientList[Series[s, {x, 0, z}], x], 1]  (* A019590 *)
Drop[CoefficientList[Series[1/p, {x, 0, z}], x], 1]  (* A291408 *)

Formula

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

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

a(n) = (1/2) * A275439(n+4). - Alois P. Heinz, May 20 2025