%I #6 Aug 28 2017 20:14:39

%S 5,15,40,100,236,535,1175,2515,5270,10846,21980,43950,86850,169840,

%T 329042,632135,1205205,2281925,4293270,8030558,14940700,27659095,

%U 50968455,93518940,170905555,311159365,564521620,1020800470,1840124050,3307314163,5927828905

%N p-INVERT of (0,1,0,1,0,1,...), where p(S) = (1 - S)^5.

%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 A291219 for a guide to related sequences.

%H Clark Kimberling, <a href="/A291225/b291225.txt">Table of n, a(n) for n = 0..1000</a>

%H <a href="/index/Rec#order_10">Index entries for linear recurrences with constant coefficients</a>, signature (5, -5, -10, 15, 11, -15, -10, 5, 5, 1)

%F a(n) = 5*a(n-1) - 5*a(n-2) - 10*a(n-3) + 15*a(n-4) + 11*a(n-5) - 15*a(n-6) - 10*a(n-7) + 5*a(n-8) + 5*a(n-9) + a(n-10) for n >= 11.

%F G.f.: (5 - 10*x - 10*x^2 + 25*x^3 + 11*x^4 - 25*x^5 - 10*x^6 + 10*x^7 + 5*x^8) / (1 - x - x^2)^5. - _Colin Barker_, Aug 28 2017

%t z = 60; s = x/(1 - x^2); p = (1 - s)^5;

%t Drop[CoefficientList[Series[s, {x, 0, z}], x], 1] (* A000035 *)

%t Drop[CoefficientList[Series[1/p, {x, 0, z}], x], 1] (* A291225 *)

%o (PARI) Vec((5 - 10*x - 10*x^2 + 25*x^3 + 11*x^4 - 25*x^5 - 10*x^6 + 10*x^7 + 5*x^8) / (1 - x - x^2)^5 + O(x^40)) \\ _Colin Barker_, Aug 28 2017

%Y Cf. A000035, A291219.

%K nonn,easy

%O 0,1

%A _Clark Kimberling_, Aug 28 2017