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A291237 p-INVERT of (0,1,0,1,0,1,...), where p(S) = (1 - S)(1 - 2 S)(1 - 4 S). 2

%I

%S 7,35,162,721,3139,13504,57707,245671,1043634,4428053,18774815,

%T 79573152,337178159,1428553243,6052037010,25638260873,108608846171,

%U 460082737472,1948961747155,8255982722783,34973020586946,148148373971341,627567262233463,2658419223345984

%N p-INVERT of (0,1,0,1,0,1,...), where p(S) = (1 - S)(1 - 2 S)(1 - 4 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 A291219 for a guide to related sequences.

%H Clark Kimberling, <a href="/A291237/b291237.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 (7, -11, -6, 11, 7, 1).

%F G.f.: (-7 + 14 x + 6 x^2 - 14 x^3 - 7 x^4)/(-1 + 7 x - 11 x^2 - 6 x^3 + 11 x^4 + 7 x^5 + x^6).

%F a(n) = 7*a(n-1) - 11*a(n-2) - 6*a(n-3) + 11*a(n-4) + 7*a(n-5) + a(n-6) for n >= 7.

%t z = 60; s = x/(1 - x^2); p = (1 - s)(1 - 2s)(1 - 4 s);

%t Drop[CoefficientList[Series[s, {x, 0, z}], x], 1] (* A000035 *)

%t Drop[CoefficientList[Series[1/p, {x, 0, z}], x], 1] (* A291237 *)

%Y Cf. A000035, A291219.

%K nonn,easy

%O 0,1

%A _Clark Kimberling_, Aug 28 2017

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Last modified April 18 18:57 EDT 2021. Contains 343089 sequences. (Running on oeis4.)