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A291389 p-INVERT of (1,1,0,0,0,0,...), where p(S) = (1 - 5 S)^2.

%I A291389 #7 Sep 09 2017 21:43:16
%S A291389 10,85,650,4700,32750,222375,1481250,9721875,63062500,405175000,
%T A291389 2582687500,16353078125,102955156250,644991640625,4023367968750,
%U A291389 25002220312500,154848222656250,956155732421875,5888138769531250,36171585068359375,221714776953125000
%N A291389 p-INVERT of (1,1,0,0,0,0,...), where p(S) = (1 - 5 S)^2.
%C A291389 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 A291389 See A291382 for a guide to related sequences.
%C A291389 All terms = 0 mod 5. - _Muniru A Asiru_, Sep 07 2017
%H A291389 Clark Kimberling, <a href="/A291389/b291389.txt">Table of n, a(n) for n = 0..1000</a>
%H A291389 <a href="/index/Rec#order_04">Index entries for linear recurrences with constant coefficients</a>, signature (10, -15, -50, -25)
%F A291389 G.f.: -((5 (1 + x) (-2 + 5 x + 5 x^2))/(-1 + 5 x + 5 x^2)^2).
%F A291389 a(n) = 10*a(n-1) - 15*a(n-2) - 50*a(n-3) - 25*a(n-4) for n >= 5.
%t A291389 z = 60; s = x + x^2; p = (1 - 5 s)^2;
%t A291389 Drop[CoefficientList[Series[s, {x, 0, z}], x], 1]  (* A019590 *)
%t A291389 u = Drop[CoefficientList[Series[1/p, {x, 0, z}], x], 1]  (* A291389 *)
%t A291389 u / 5  (* A291390 *)
%o A291389 (GAP)
%o A291389 a:=5*[2,17,130,940];; for n in [5..10^2] do a[n]:=10*a[n-1]-15*a[n-2]-50*a[n-3]-25*a[n-4]; od; a;  # _Muniru A Asiru_, Sep 07 2017
%Y A291389 Cf. A019590, A291382, A291390.
%K A291389 nonn,easy
%O A291389 0,1
%A A291389 _Clark Kimberling_, Sep 04 2017