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p-INVERT of the positive integers, where p(S) = 1 - 3*S.
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%I #13 Aug 18 2017 15:49:37

%S 3,15,72,345,1653,7920,37947,181815,871128,4173825,19997997,95816160,

%T 459082803,2199597855,10538906472,50494934505,241935766053,

%U 1159183895760,5553983712747,26610734667975,127499689627128,610887713467665,2926938877711197,14023806675088320

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

%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 A290890 for a guide to related sequences.

%H Clark Kimberling, <a href="/A290902/b290902.txt">Table of n, a(n) for n = 0..1000</a>

%H <a href="/index/Rec#order_02">Index entries for linear recurrences with constant coefficients</a>, signature (5, -1)

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

%F a(n) = 5*a(n-1) - a(n-2).

%F a(n) = 3*A004254(n+1) for n >= 0.

%e s = (1,2,3,4,...), p(S) = 1-3*S;

%e S(x) = x + 2 x^2 + 3 x^3 + ... ;

%e p(S(x)) = 1 - 3(x + 2 x^2 + 3 x^3 + ...);

%e 1/p(S(x)) = 1 + 3 x + 15 x^2 + 72 x^3 + ... ;

%e (-p(0) + 1/p(S(x)))/x = 3 + 15 x + 72 x^2 + ... ;

%e t(s) = (3, 15, 72, ...), with offset 0.

%t z = 60; s = x/(1 - x)^2; p = 1 - 3 s;

%t Drop[CoefficientList[Series[s, {x, 0, z}], x], 1] (* A000027 *)

%t u = Drop[CoefficientList[Series[1/p, {x, 0, z}], x], 1] (* A290902 *)

%t u/3 (* A004254 shifted *)

%Y Cf. A000027, A004254, A290890.

%K nonn,easy

%O 0,1

%A _Clark Kimberling_, Aug 17 2017