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A290902 p-INVERT of the positive integers, where p(S) = 1 - 3*S.

%I A290902 #13 Aug 18 2017 15:49:37
%S A290902 3,15,72,345,1653,7920,37947,181815,871128,4173825,19997997,95816160,
%T A290902 459082803,2199597855,10538906472,50494934505,241935766053,
%U A290902 1159183895760,5553983712747,26610734667975,127499689627128,610887713467665,2926938877711197,14023806675088320
%N A290902 p-INVERT of the positive integers, where p(S) = 1 - 3*S.
%C A290902 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 A290902 See A290890 for a guide to related sequences.
%H A290902 Clark Kimberling, <a href="/A290902/b290902.txt">Table of n, a(n) for n = 0..1000</a>
%H A290902 <a href="/index/Rec#order_02">Index entries for linear recurrences with constant coefficients</a>, signature (5, -1)
%F A290902 G.f.: 3/(1 - 5 x + x^2).
%F A290902 a(n) = 5*a(n-1) - a(n-2).
%F A290902 a(n) = 3*A004254(n+1) for n >= 0.
%e A290902 s = (1,2,3,4,...), p(S) = 1-3*S;
%e A290902 S(x) = x + 2 x^2 + 3 x^3 + ... ;
%e A290902 p(S(x)) = 1 - 3(x + 2 x^2 + 3 x^3 + ...);
%e A290902 1/p(S(x)) = 1 + 3 x + 15 x^2 + 72 x^3 + ... ;
%e A290902 (-p(0) + 1/p(S(x)))/x = 3 + 15 x + 72 x^2 + ... ;
%e A290902 t(s) = (3, 15, 72, ...), with offset 0.
%t A290902 z = 60; s = x/(1 - x)^2; p = 1 - 3 s;
%t A290902 Drop[CoefficientList[Series[s, {x, 0, z}], x], 1] (* A000027 *)
%t A290902 u = Drop[CoefficientList[Series[1/p, {x, 0, z}], x], 1] (* A290902 *)
%t A290902 u/3 (* A004254 shifted *)
%Y A290902 Cf. A000027, A004254, A290890.
%K A290902 nonn,easy
%O A290902 0,1
%A A290902 _Clark Kimberling_, Aug 17 2017