A289785 p-INVERT of the (5^n), where p(S) = 1 - S - S^2.

1, 7, 48, 325, 2183, 14588, 97161, 645719, 4285240, 28411789, 188257719, 1246893028, 8256349457, 54659946215, 361825274112, 2394939574997, 15851402375719, 104912178457996, 694343294142105, 4595323060281271, 30412598132972936, 201274210714545437

Offset

0,2

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 A289780 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 (11, -29)

Formula

G.f.: (1 - 4 x)/(1 - 11 x + 29 x^2).

a(n) = 11*a(n-1) - 29*a(n-2).

a(n) = (2^(-n-1)*((11-sqrt(5))^(n+1)*(-7+2*sqrt(5)) + (11+sqrt(5))^(n+1)*(7+2*sqrt(5)))) / (29*sqrt(5)). - Colin Barker, Aug 11 2017

a(n) = A081575(n+1)-4*A081575(n). - R. J. Mathar, Jul 08 2022

Mathematica

z = 60; s = x/(1 - 5*x); p = 1 - s - s^2;
Drop[CoefficientList[Series[s, {x, 0, z}], x], 1] (* A000351 *)
Drop[CoefficientList[Series[1/p, {x, 0, z}], x], 1] (* A289785 *)

Prog

(PARI) Vec(x*(1 - 4*x) / (1 - 11*x + 29*x^2) + O(x^30)) \\ Colin Barker, Aug 11 2017

Crossrefs

Cf. A000351, A289780.

Sequence in context: A186161 A370037 A081106 * A036829 A164591 A242630

Adjacent sequences: A289782 A289783 A289784 * A289786 A289787 A289788

Keyword

nonn,easy

Author

Clark Kimberling, Aug 10 2017

Status

approved