%I #8 Oct 03 2017 20:54:03
%S 1,3,5,8,22,100,444,1680,5496,16096,43936,117360,323056,946288,
%T 2930320,9287792,29222800,89856944,269619792,795460592,2334102160,
%U 6882700336,20508738256,61728245104,186833742864,565643533232,1706639551568,5125652284144,15338915301264
%N p-INVERT of the odd positive integers, where p(S) = 1 - S + S^2 - S^3.
%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="/A292492/b292492.txt">Table of n, a(n) for n = 0..1000</a>
%H <a href="/index/Rec#order_05">Index entries for linear recurrences with constant coefficients</a>, signature (7, -19, 23, -8, 6)
%F G.f.: -(((1 + x) (1 - 5 x + 8 x^2 - x^3 + x^4))/((-1 + 3 x) (1 - 4 x + 7 x^2 - 2 x^3 + 2 x^4))).
%F a(n) = 7*a(n-1) - 19*a(n-2) + 23*a(n-3) - 8*a(n-4) + 6*a(n-5) for n >= 7.
%t z = 60; s = x (x + 1)/(1 - x)^2; p = 1 - s + s^2 - s^3;
%t Drop[CoefficientList[Series[s, {x, 0, z}], x], 1] (* A005408 *)
%t Drop[CoefficientList[Series[1/p, {x, 0, z}], x], 1] (* A292492 *)
%o (PARI) x='x+O('x^99); Vec(((1+x)*(1-5*x+8*x^2-x^3+x^4))/((1-3*x)*(1-4*x+7*x^2-2*x^3+2*x^4))) \\ _Altug Alkan_, Oct 03 2017
%Y Cf. A005408, A292480.
%K nonn,easy
%O 0,2
%A _Clark Kimberling_, Oct 03 2017