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A290923 p-INVERT of the positive integers, where p(S) = 1 - 2*S - 2*S^2. 3

%I #10 Aug 19 2017 14:06:10

%S 2,10,46,208,938,4230,19078,86048,388106,1750490,7895302,35610480,

%T 160615298,724429270,3267420814,14737172032,66469626002,299800475370,

%U 1352201455582,6098885514512,27508034668634,124070532153830,559600027205398,2523985228499040

%N p-INVERT of the positive integers, where p(S) = 1 - 2*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="/A290923/b290923.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 (6, -8, 6, -1)

%F G.f.: (2 (1 - x + x^2))/(1 - 6 x + 8 x^2 - 6 x^3 + x^4).

%F a(n) = 6*a(n-1) - 8*a(n-2) + 6*a(n-3) - a(n-4).

%t z = 60; s = x/(1 - x)^2; p = 1 - 2 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] (* A290923 *)

%t u/2 (* A290924 *)

%Y Cf. A000027, A033453, A290890.

%K nonn,easy

%O 0,1

%A _Clark Kimberling_, Aug 19 2017

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Last modified March 28 08:22 EDT 2024. Contains 371236 sequences. (Running on oeis4.)