cp's OEIS Frontend

A291396 p-INVERT of (1,1,0,0,0,0,...), where p(S) = (1 - S)(1 - 2 S)(1 - 3 S).

%I A291396 #4 Sep 06 2017 21:15:52
%S A291396 6,31,140,596,2440,9751,38344,149147,575794,2211278,8460912,32289105,
%T A291396 122994890,467887343,1778208080,6753481344,25636583768,97283620659,
%U A291396 369070501684,1399909005427,5309251592646,20133801242298,76346423589984,289487843638333
%N A291396 p-INVERT of (1,1,0,0,0,0,...), where p(S) = (1 - S)(1 - 2 S)(1 - 3 S).
%C A291396 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 A291396 See A291382 for a guide to related sequences.
%H A291396 Clark Kimberling, <a href="/A291396/b291396.txt">Table of n, a(n) for n = 0..1000</a>
%H A291396 <a href="/index/Rec#order_06">Index entries for linear recurrences with constant coefficients</a>, signature (6, -5, -16, 7, 18, 6)
%F A291396 G.f.: -(((1 + x) (6 - 11 x - 5 x^2 + 12 x^3 + 6 x^4))/((-1 + x + x^2) (-1 + 2 x + 2 x^2) (-1 + 3 x + 3 x^2))).
%F A291396 a(n) = 6*a(n-1) - 5*a(n-2) - 16*a(n-3) + 7*a(n-4) + 18*a(n-5) + 6*a(n-6) for n >= 7.
%t A291396 z = 60; s = x + x^2; p = (1 - s)(1 - 2s)(1 - 3s);
%t A291396 Drop[CoefficientList[Series[s, {x, 0, z}], x], 1]  (* A019590 *)
%t A291396 u = Drop[CoefficientList[Series[1/p, {x, 0, z}], x], 1]  (* A291396 *)
%Y A291396 Cf. A019590, A291382.
%K A291396 nonn,easy
%O A291396 0,1
%A A291396 _Clark Kimberling_, Sep 06 2017