%I #11 Aug 20 2024 14:32:24

%S 1,4,13,42,133,424,1348,4291,13653,43449,138261,439979,1400101,

%T 4455420,14178073,45117606,143573662,456881476,1453892534,4626590576,

%U 14722780217,46850970327,149089600359,474434334814,1509749422360,4804338875098,15288412556740

%N p-INVERT of (1,2,2,3,3,4,4,...) (A080513), where p(S) = 1 - S - 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 A289780 for a guide to related sequences.

%H Clark Kimberling, <a href="/A289807/b289807.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 (3, 2, -5, 1, 2, -1)

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

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

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

%t Drop[CoefficientList[Series[s, {x, 0, z}], x], 1] (* A080513 *)

%t Drop[CoefficientList[Series[1/p, {x, 0, z}], x], 1] (* A289807 *)

%t LinearRecurrence[{3,2,-5,1,2,-1},{1,4,13,42,133,424},30] (* _Harvey P. Dale_, Aug 20 2024 *)

%Y Cf. A080513, A289780.

%K nonn,easy

%O 0,2

%A _Clark Kimberling_, Aug 12 2017