A291245 - OEIS

%I #9 Aug 29 2017 03:30:14

%S 5,24,120,599,2990,14925,74500,371876,1856265,9265776,46251265,

%T 230868900,1152410620,5752399899,28713814350,143328549649,

%U 715442152480,3571217840400,17826174791885,88981552487776,444162405876285,2217091490069376,11066885918992400

%N p-INVERT of (0,1,0,1,0,1,...), where p(S) = 1 - 5 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 A291219 for a guide to related sequences.

%H Clark Kimberling, <a href="/A291245/b291245.txt">Table of n, a(n) for n = 0..1000</a>

%H <a href="/index/Rec#order_04">Index entries for linear recurrences with constant coefficients</a>, signature (5,1,-5,-1)

%F G.f.: (5 - x - 5*x^2)/(1 - 5*x - x^2 + 5*x^3 + x^4).

%F a(n) = 5*a(n-1) + a(n-2) - 5*a(n-3) - a(n-4) for n >= 5.

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

%t Drop[CoefficientList[Series[s, {x, 0, z}], x], 1]  (* A000035 *)

%t Drop[CoefficientList[Series[1/p, {x, 0, z}], x], 1]  (* A291245 *)

%Y Cf. A000035, A291219.

%K nonn,easy

%O 0,1

%A _Clark Kimberling_, Aug 28 2017