A291412 p-INVERT of (1,1,0,0,0,0,...), where p(S) = 1 - S - 2 S^2 + S^3.

1, 4, 10, 24, 62, 156, 391, 987, 2484, 6252, 15744, 39636, 99788, 251237, 632525, 1592480, 4009326, 10094104, 25413498, 63982496, 161086011, 405559431, 1021059816, 2570679048, 6472089792, 16294506424, 41023988824, 103284359545, 260034658537, 654678248796

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, 3, 3, -1, -3, -1)

Crossrefs

Cf. A019590, A291382.

Programs

GAP

a:=[1,4,10,24,62,156];;  for n in [7..10^2] do a[n]:=a[n-1]+3*a[n-2]+3*a[n-3]-a[n-4]-3*a[n-5]-a[n-6]; od; a; # Muniru A Asiru, Sep 12 2017

Mathematica

z = 60; s = x + x^2; p = 1 - s - 2 s^2 + s^3;
Drop[CoefficientList[Series[s, {x, 0, z}], x], 1]  (* A019590 *)
Drop[CoefficientList[Series[1/p, {x, 0, z}], x], 1]  (* A291412 *)

Formula

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

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