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%I #6 Oct 03 2017 20:53:14
%S 1,9,52,261,1323,6814,35077,180261,926348,4761289,24472527,125783886,
%T 646502873,3322895889,17079026852,87782799261,451186103523,
%U 2319006747614,11919233055677,61262485125261,314876977751548,1618404981969089,8318279426249127
%N p-INVERT of the odd positive integers, where p(S) = 1 - S - 5 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="/A292488/b292488.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, -2, 13, 5)
%F G.f.: -(((1 + x) (1 + 3 x + 6 x^2))/(-1 + 5 x - 2 x^2 + 13 x^3 + 5 x^4)).
%F a(n) = 5*a(n-1) - 2*a(n-2) + 13*a(n-3) + 5*a(n-4) for n >= 5.
%t z = 60; s = x (x + 1)/(1 - x)^2; p = 1 - s - 5 s^2;
%t Drop[CoefficientList[Series[s, {x, 0, z}], x], 1] (* A005408 *)
%t Drop[CoefficientList[Series[1/p, {x, 0, z}], x], 1] (* A292488 *)
%o (PARI) x='x+O('x^99); Vec(((1+x)*(1+3*x+6*x^2))/(1-5*x+2*x^2-13*x^3-5*x^4)) \\ _Altug Alkan_, Oct 03 2017
%Y Cf. A005408, A292480.
%K nonn,easy
%O 0,2
%A _Clark Kimberling_, Oct 03 2017
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