|
%I #4 Oct 03 2017 08:49:06
%S -1,-1,4,9,5,8,63,183,348,745,2061,5456,12991,30831,76660,192137,
%T 472597,1155032,2843007,7024935,17315404,42592489,104847389,258355104,
%U 636507775,1567442143,3859933668,9507231753,23417547813,57675809960,142047927231,349856144791
%N p-INVERT of the odd positive integers, 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 A292480 for a guide to related sequences.
%H Clark Kimberling, <a href="/A292484/b292484.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 (3, -4, 7, -1)
%F G.f.: ((1 + x) (-1 + 3 x))/(1 - 3 x + 4 x^2 - 7 x^3 + x^4).
%F a(n) = 3*a(n-1) - 4*a(n-2) + 7*a(n-3) - a(n-4) for n >= 5.
%t z = 60; s = x (x + 1)/(1 - x)^2; p = 1 + s - s^2;
%t Drop[CoefficientList[Series[s, {x, 0, z}], x], 1] (* A005408 *)
%t Drop[CoefficientList[Series[1/p, {x, 0, z}], x], 1] (* A292484 *)
%Y Cf. A005408, A292480.
%K easy,sign
%O 0,3
%A _Clark Kimberling_, Oct 02 2017
|
