A292489 - OEIS

%I #6 Oct 03 2017 20:53:25

%S 1,10,60,312,1656,8928,48024,257904,1385352,7442784,39985272,

%T 214811280,1154025000,6199749504,33306803352,178933509936,

%U 961281138888,5164272731808,27743925989304,149048175357648,800728728609384,4301739993919680,23110157427289560

%N p-INVERT of the odd positive integers, where p(S) = 1 - S - 6 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 A292480 for a guide to related sequences.

%H Clark Kimberling, <a href="/A292489/b292489.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, 15, 6)

%F G.f.: -(((1 + x) (1 + 4 x + 7 x^2))/((-1 + 5 x + 2 x^2) (1 + 3 x^2))).

%F a(n) = 5*a(n-1) - a(n-2) + 16*a(n-3) + 6*a(n-4)  for n >= 5.

%t z = 60; s = x (x + 1)/(1 - x)^2; p = 1 - s - 6 s^2;

%t Drop[CoefficientList[Series[s, {x, 0, z}], x], 1] (* A005408 *)

%t Drop[CoefficientList[Series[1/p, {x, 0, z}], x], 1]  (* A292489 *)

%o (PARI) x='x+O('x^99); Vec(((1+x)*(1+4*x+7*x^2))/((1-5*x-2*x^2)*(1+3*x^2))) \\ _Altug Alkan_, Oct 03 2017

%Y Cf. A005408, A292480.

%K nonn,easy

%O 0,2

%A _Clark Kimberling_, Oct 03 2017