|
%I #8 Aug 18 2017 18:46:55
%S 0,2,8,24,72,222,688,2128,6576,20322,62808,194120,599960,1854270,
%T 5730912,17712288,54742624,169190722,522910632,1616137848,4994929128,
%U 15437616926,47712391952,147462678768,455756685840,1408587979170,4353463496440,13455066133672
%N p-INVERT of the positive integers, where p(S) = 1 - 2*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 A290890 for a guide to related sequences.
%H Clark Kimberling, <a href="/A290904/b290904.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 (4, -4, 4, -1)
%F G.f.: (2 x)/(1 - 4 x + 4 x^2 - 4 x^3 + x^4).
%F a(n) = 4*a(n-1) - 4*a(n-2) + 4*a(n-3) - a(n-4).
%F a(n) = 2*A290905(n) for n >= 0.
%t z = 60; s = x/(1 - x)^2; p = 1 - 2 s^2;
%t Drop[CoefficientList[Series[s, {x, 0, z}], x], 1] (* A000027 *)
%t u = Drop[CoefficientList[Series[1/p, {x, 0, z}], x], 1] (* A290903 *)
%t u/2 (* A290905 *)
%Y Cf. A000027, A290890, A290905.
%K nonn,easy
%O 0,2
%A _Clark Kimberling_, Aug 17 2017
|
