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%I #8 Aug 18 2017 18:46:22
%S 0,3,12,39,132,456,1572,5409,18612,64053,220440,758640,2610840,
%T 8985147,30922188,106418031,366235308,1260390744,4337606988,
%U 14927778921,51373622388,176801189997,608457401520,2093992746720,7206429919920,24800769855603,85351303248012
%N p-INVERT of the positive integers, where p(S) = 1 - 3*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 <a href="/index/Rec#order_04">Index entries for linear recurrences with constant coefficients</a>, signature (4, -3, 4, -1)
%F G.f.: (3 x)/(1 - 4 x + 3 x^2 - 4 x^3 + x^4).
%F a(n) = 4*a(n-1) - 3*a(n-2) + 4*a(n-3) - a(n-4).
%F a(n) = 3*A290907(n) for n >= 0.
%t z = 60; s = x/(1 - x)^2; p = 1 - 3 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] (* A290906 *)
%t u/3 (* A290907 *)
%Y Cf. A000027, A290890, A290907.
%K nonn,easy
%O 0,2
%A _Clark Kimberling_, Aug 17 2017
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