A291014 p-INVERT of (1,1,1,1,1,...), where p(S) = (1 - S^3)^2.

0, 0, 2, 6, 12, 23, 48, 105, 228, 486, 1026, 2161, 4548, 9555, 20026, 41874, 87384, 182043, 378648, 786429, 1631120, 3378750, 6990510, 14447045, 29826156, 61516455, 126761190, 260978922, 536870916, 1103567983, 2266788288, 4652881233, 9544371772, 19565962134

Offset: 0

Author

Clark Kimberling, Aug 23 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 A291000 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 (6,-15,22,-21,12,-4).

Crossrefs

Cf. A000012, A010892, A033453, A099254, A289780, A291000.

Programs

Magma

R:=PowerSeriesRing(Integers(), 50); [0,0] cat Coefficients(R!( x^2*(2-6*x+6*x^2-3*x^3)/((1-2*x)*(1-x+x^2))^2 )); // G. C. Greubel, Jun 05 2023

Mathematica

z = 60; s = x/(1-x); p = (1 - s^3)^2;
Drop[CoefficientList[Series[s, {x, 0, z}], x], 1]  (* A000012 *)
Drop[CoefficientList[Series[1/p, {x, 0, z}], x], 1]  (* A291014 *)
LinearRecurrence[{6,-15,22,-21,12,-4}, {0,0,2,6,12,23}, 50] (* G. C. Greubel, Jun 05 2023 *)

SageMath

def A291014_list(prec):
    P. = PowerSeriesRing(ZZ, prec)
    return P( x^2*(2-6*x+6*x^2-3*x^3)/((1-2*x)*(1-x+x^2))^2 ).list()
A291014_list(50) # G. C. Greubel, Jun 05 2023

Formula

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

a(n) = 6*a(n-1) - 15*a(n-2) + 22*a(n-3) - 21*a(n-4) + 12*a(n-5) - 4*a(n-6) for n >= 7.

a(n) = (1/9)*( 2^(n-1)*(n + 8) - 3*(A099254(n) - A099254(n-1)) - A010892(n) - 5*A010892(n-1) ). - G. C. Greubel, Jun 05 2023