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A291027 p-INVERT of the positive integers, where p(S) = 1 - 5*S + S^2. 2

%I

%S 5,34,226,1501,9968,66195,439582,2919134,19385099,128730656,854861845,

%T 5676882210,37698479330,250344342349,1662462010576,11039913707011,

%U 73312769785118,486848208799710,3233013554202907,21469477452590144,142572387761274149,946780646936461346

%N p-INVERT of the positive integers, where p(S) = 1 - 5*S + 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="/A291027/b291027.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 (9, -17, 9, -1)

%F G.f.: (5 - 11 x + 5 x^2)/(1 - 9 x + 17 x^2 - 9 x^3 + x^4).

%F a(n) = 9*a(n-1) - 17*a(n-2) + 9*a(n-3) - a(n-4).

%t z = 60; s = x/(1 - x)^2; p = 1 - 5 s + s^2;

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

%t Drop[CoefficientList[Series[1/p, {x, 0, z}], x], 1] (* A291027 *)

%Y Cf. A000027, A290890.

%K nonn,easy

%O 0,1

%A _Clark Kimberling_, Aug 19 2017

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Last modified January 17 12:18 EST 2020. Contains 330958 sequences. (Running on oeis4.)