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%I #6 Feb 20 2020 12:11:22
%S 0,0,1,4,7,8,12,31,71,125,201,367,749,1471,2679,4814,9014,17304,32739,
%T 60683,112444,210938,397800,746347,1392898,2601701,4876692,9149911,
%U 17138518,32060349,60002060,112404852,210600344,394370928,738281497,1382360598
%N p-INVERT of (1,1,0,0,0,0,...), where p(S) = 1 - S^3 - S^4.
%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 A291382 for a guide to related sequences.
%H <a href="/index/Rec#order_08">Index entries for linear recurrences with constant coefficients</a>, signature (0, 0, 1, 4, 7, 7, 4, 1)
%F G.f.: -((x^2 (1 + x)^3 (1 + x + x^2))/(-1 + x^3 + 4 x^4 + 7 x^5 + 7 x^6 + 4 x^7 + x^8)).
%F a(n) = a(n-3) + 4*a(n-4) + 7*a(n-5) + 7*a(n-6) + 4*a(n-7) + a(n-8) for n >= 9.
%t z = 60; s = x + x^2; p = 1 - s^3 - s^4;
%t Drop[CoefficientList[Series[s, {x, 0, z}], x], 1] (* A019590 *)
%t u = Drop[CoefficientList[Series[1/p, {x, 0, z}], x], 1] (* A291402 *)
%t LinearRecurrence[{0,0,1,4,7,7,4,1},{0,0,1,4,7,8,12,31},40] (* _Harvey P. Dale_, Feb 20 2020 *)
%Y Cf. A019590, A291382.
%K nonn,easy
%O 0,4
%A _Clark Kimberling_, Sep 06 2017
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