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A291380 p-INVERT of (1,1,0,0,0,0,...), where p(S) = 1 - S^5.

%I A291380 #6 Sep 04 2017 19:30:29
%S A291380 0,0,0,0,1,5,10,10,5,2,10,45,120,210,253,225,225,500,1375,3005,5025,
%T A291380 6625,7575,9850,18508,40150,78275,128375,180625,237888,345090,607105,
%U A291380 1163155,2109140,3426771,5056055,7237835,11059960,18816930,33638409,58293475
%N A291380 p-INVERT of (1,1,0,0,0,0,...), where p(S) = 1 - S^5.
%C A291380 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 A291380 See A291382 for a guide to related sequences.
%H A291380 Clark Kimberling, <a href="/A291380/b291380.txt">Table of n, a(n) for n = 0..1000</a>
%H A291380 <a href="/index/Rec#order_10">Index entries for linear recurrences with constant coefficients</a>, signature (0, 0, 0, 0, 1, 5, 10, 10, 5, 1)
%F A291380 G.f.: -((x^4 (1 + x)^5)/((-1 + x + x^2) (1 + x + 2 x^2 + 3 x^3 + 5 x^4 + 7 x^5 + 7 x^6 + 4 x^7 + x^8))).
%F A291380 a(n) = a(n-5) + 5*a(n-6) + 10*a(n-7) + 10*a(n-8) + 5*a(n-9) + a(n-10) for n >= 11.
%t A291380 z = 60; s = x + x^2; p = (1 - s)^5;
%t A291380 Drop[CoefficientList[Series[s, {x, 0, z}], x], 1]  (* A019590 *)
%t A291380 Drop[CoefficientList[Series[1/p, {x, 0, z}], x], 1]  (* A291380 *)
%Y A291380 Cf. A019590, A291382.
%K A291380 nonn,easy
%O A291380 0,6
%A A291380 _Clark Kimberling_, Sep 04 2017