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 A290925 p-INVERT of the positive integers, where p(S) = 1 - 3*S - 2*S^2. 2

%I

%S 3,17,92,495,2661,14304,76891,413329,2221860,11943663,64203453,

%T 345127232,1855239875,9972887313,53609499612,288179176047,

%U 1549114207525,8327301302176,44763611772699,240627889663761,1293501104827044,6953246818258415,37377348295412093

%N p-INVERT of the positive integers, where p(S) = 1 - 3*S - 2*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 A290890 for a guide to related sequences.

%H Clark Kimberling, <a href="/A290925/b290925.txt">Table of n, a(n) for n = 0..1000</a>

%H <a href="/index/Rec#order_04">Index entries for linear recurrences with constant coefficients</a>, signature (7, -10, 7, -1)

%F G.f.: (3 - 4 x + 3 x^2)/(1 - 7 x + 10 x^2 - 7 x^3 + x^4).

%F a(n) = 7*a(n-1) - 10*a(n-2) + 7*a(n-3) - a(n-4).

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

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

%t u = Drop[CoefficientList[Series[1/p, {x, 0, z}], x], 1] (* A290925 *)

%Y Cf. A000027, A290890.

%K nonn,easy

%O 0,1

%A _Clark Kimberling_, Aug 19 2017

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Last modified May 12 23:01 EDT 2021. Contains 343829 sequences. (Running on oeis4.)