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p-INVERT of (1,1,0,0,0,0,...), where p(S) = 1 - S - 2 S^2 + S^3.
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%I #7 Sep 27 2017 09:25:05

%S 1,4,10,24,62,156,391,987,2484,6252,15744,39636,99788,251237,632525,

%T 1592480,4009326,10094104,25413498,63982496,161086011,405559431,

%U 1021059816,2570679048,6472089792,16294506424,41023988824,103284359545,260034658537,654678248796

%N p-INVERT of (1,1,0,0,0,0,...), where p(S) = 1 - S - 2 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 A291382 for a guide to related sequences.

%H Clark Kimberling, <a href="/A291412/b291412.txt">Table of n, a(n) for n = 0..1000</a>

%H <a href="/index/Rec#order_06">Index entries for linear recurrences with constant coefficients</a>, signature (1, 3, 3, -1, -3, -1)

%F G.f.: -(((1 + x) (-1 - 2 x - x^2 + 2 x^3 + x^4))/(1 - x - 3 x^2 - 3 x^3 + x^4 + 3 x^5 + x^6)).

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

%t z = 60; s = x + x^2; p = 1 - s - 2 s^2 + s^3;

%t Drop[CoefficientList[Series[s, {x, 0, z}], x], 1] (* A019590 *)

%t Drop[CoefficientList[Series[1/p, {x, 0, z}], x], 1] (* A291412 *)

%o (GAP)

%o a:=[1,4,10,24,62,156];; for n in [7..10^2] do a[n]:=a[n-1]+3*a[n-2]+3*a[n-3]-a[n-4]-3*a[n-5]-a[n-6]; od; a; # _Muniru A Asiru_, Sep 12 2017

%Y Cf. A019590, A291382.

%K nonn,easy

%O 0,2

%A _Clark Kimberling_, Sep 07 2017