A289785 - OEIS

%I #25 Jul 08 2022 12:58:06

%S 1,7,48,325,2183,14588,97161,645719,4285240,28411789,188257719,

%T 1246893028,8256349457,54659946215,361825274112,2394939574997,

%U 15851402375719,104912178457996,694343294142105,4595323060281271,30412598132972936,201274210714545437

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

%C 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).

%C See A289780 for a guide to related sequences.

%H Clark Kimberling, <a href="/A289785/b289785.txt">Table of n, a(n) for n = 0..1000</a>

%H <a href="/index/Rec#order_02">Index entries for linear recurrences with constant coefficients</a>, signature (11, -29)

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

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

%F 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

%F a(n) = A081575(n+1)-4*A081575(n). - _R. J. Mathar_, Jul 08 2022

%t z = 60; s = x/(1 - 5*x); p = 1 - s - s^2;

%t Drop[CoefficientList[Series[s, {x, 0, z}], x], 1] (* A000351 *)

%t Drop[CoefficientList[Series[1/p, {x, 0, z}], x], 1] (* A289785 *)

%o (PARI) Vec(x*(1 - 4*x) / (1 - 11*x + 29*x^2) + O(x^30)) \\ _Colin Barker_, Aug 11 2017

%Y Cf. A000351, A289780.

%K nonn,easy

%O 0,2

%A _Clark Kimberling_, Aug 10 2017