A291015 p-INVERT of (1,1,1,1,1,...), where p(S) = (1 - S^3)^2.

2, 7, 23, 75, 244, 793, 2576, 8366, 27167, 88215, 286439, 930072, 3019941, 9805712, 31838986, 103380599, 335674791, 1089929347, 3538978588, 11490991649, 37311016064, 121148109014, 393365440335, 1277249563655, 4147203285279, 13465884484800, 43723452275981

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 (5,-6,1).

Crossrefs

Cf. A000012, A033453, A289780, A291000.

Programs

Magma

I:=[2,7,23]; [n le 3 select I[n] else 5*Self(n-1) -6*Self(n-2) +Self(n-3): n in [1..50]]; // G. C. Greubel, Jun 06 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]  (* A291015 *)
LinearRecurrence[{5,-6,1}, {2,7,23}, 50] (* G. C. Greubel, Jun 06 2023 *)

SageMath

@CachedFunction
def a(n): # a = A291015
    if (n<3): return (2,7,23)[n]
    else: return 5*a(n-1) - 6*a(n-2) + a(n-3)
[a(n) for n in range(51)] # G. C. Greubel, Jun 06 2023

Formula

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

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