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%I #5 Sep 03 2017 21:42:13
%S 4,14,48,159,512,1618,5036,15491,47192,142624,428144,1277884,3795208,
%T 11222716,33060072,97060033,284095940,829298422,2414859016,7016265637,
%U 20344112608,58879534286,170117201548,490736173432,1413562889020,4066259673834,11682314946048
%N p-INVERT of (0,1,0,1,0,1, ...), where p(S) = (1 - 2 S - S^2)^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 A291219 for a guide to related sequences.
%H Clark Kimberling, <a href="/A291254/b291254.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 (4, 2, -16, -3, 16, 2, -4, -1)
%F G.f.: (4 - 2 x - 16 x^2 + 3 x^3 + 16 x^4 - 2 x^5 - 4 x^6)/(1 - 2 x - 3 x^2 + 2 x^3 + x^4)^2.
%F a(n) = 4*a(n-1) + 2*a(n-2) - 16*a(n-3) - 3*a(n-4) + 16*a(n-5) + 2*a(n-6) - 4*a(n-7) - a(n-8) for n >= 9.
%t z = 60; s = x/(1 - x^2); p = (1 - 2 s - s^2)^2;
%t Drop[CoefficientList[Series[s, {x, 0, z}], x], 1] (* A000035 *)
%t Drop[CoefficientList[Series[1/p, {x, 0, z}], x], 1] (* A291254 *)
%Y Cf. A000035, A291219.
%K nonn,easy
%O 0,1
%A _Clark Kimberling_, Sep 01 2017
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