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%I #9 Aug 13 2017 23:00:56
%S 1,2,5,13,29,73,168,410,962,2317,5483,13131,31193,74509,177311,423025,
%T 1007505,2402354,5723761,13644587,32514730,77501115,184698088,
%U 440216833,1049148789,2500520812,5959478837,14203542282,33851496564,80679640434,192285583548
%N p-INVERT of (1,0,2,0,3,0,4,0,5,...) (A027656), where p(S) = 1 - 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 A289780 for a guide to related sequences.
%H Clark Kimberling, <a href="/A289843/b289843.txt">Table of n, a(n) for n = 0..1000</a>
%H <a href="/index/Rec#order_08">Index entries for linear recurrences with constant coefficients</a>, signature (1, 5, -2, -6, 1, 4, 0, -1)
%F G.f.: (1 + x - 2 x^2 + x^4)/(1 - x - 5 x^2 + 2 x^3 + 6 x^4 - x^5 - 4 x^6 + x^8).
%F a(n) = a(n-1) + 5*a(n-2) - 2*a(n-3) - 6*a(n-4) + a(n-5) + 4*a(n-6) - a(n-8).
%t z = 60; s = x/(1 - x^2)^2; p = 1 - s - s^2;
%t Drop[CoefficientList[Series[s, {x, 0, z}], x], 1] (* A027656 *)
%t Drop[CoefficientList[Series[1/p, {x, 0, z}], x], 1] (* A289843 *)
%Y Cf. A027656, A289780.
%K nonn,easy
%O 0,2
%A _Clark Kimberling_, Aug 12 2017
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