A291016 p-INVERT of (1,1,1,1,1,...), where p(S) = 1 - 4*S + S^2.

4, 19, 90, 426, 2016, 9540, 45144, 213624, 1010880, 4783536, 22635936, 107114400, 506870784, 2398538304, 11350005120, 53708800896, 254152774656, 1202663842560, 5691066407424, 26930415389184, 127436093890560, 603034071008256, 2853587862706176

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,-6).

Crossrefs

Cf. A000012, A033453, A289780, A291000.

Programs

Magma

I:=[4,19]; [n le 2 select I[n] else 6*(Self(n-1)-Self(n-2)): n in [1..40]]; // G. C. Greubel, Jun 06 2023

Mathematica

z = 60; s = x/(1-x); p = 1 - 4 s + s^2;
Drop[CoefficientList[Series[s, {x, 0, z}], x], 1]  (* A000012 *)
Drop[CoefficientList[Series[1/p, {x, 0, z}], x], 1]  (* A291016 *)
LinearRecurrence[{6,-6}, {4,19}, 40] (* G. C. Greubel, Jun 06 2023 *)

SageMath

A291016=BinaryRecurrenceSequence(6,-6,4,19)
[A291016(n) for n in range(51)] # G. C. Greubel, Jun 06 2023

Formula

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

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

From G. C. Greubel, Jun 06 2023: (Start)

a(n) = ((3+sqrt(3))^(n+4) - (3-sqrt(3))^(n+4))/(72*sqrt(3)).

a(n) = 6^(n/2)*(4*ChebyshevU(n, sqrt(3/2)) - (5/sqrt(6))*ChebyshevU(n - 1, sqrt(3/2))).

E.g.f.: exp(3*x)*(4*cosh(sqrt(3)*x) + (7/sqrt(3))*sinh(sqrt(3)*x)). (End)