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A291724 p-INVERT of (1,0,1,0,0,0,0,...), where p(S) = 1 - S^5.

%I A291724 #6 Sep 11 2017 12:06:42
%S A291724 0,0,0,0,1,0,5,0,10,1,10,10,5,45,2,120,15,210,105,253,455,230,1365,
%T A291724 310,3004,1185,5030,4855,6735,15506,8735,38790,17655,77955,56134,
%U A291724 130030,178500,195365,481750,327263,1088225,761775,2095350,2162550,3593394,5940325
%N A291724 p-INVERT of (1,0,1,0,0,0,0,...), where p(S) = 1 - S^5.
%C A291724 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 A291724 See A291728 for a guide to related sequences.
%H A291724 Clark Kimberling, <a href="/A291724/b291724.txt">Table of n, a(n) for n = 0..1000</a>
%H A291724 <a href="/index/Rec#order_15">Index entries for linear recurrences with constant coefficients</a>, signature (0, 0, 0, 0, 1, 0, 5, 0, 10, 0, 10, 0, 5, 0, 1)
%F A291724 G.f.: -((x^4 (1 + x^2)^5)/((-1 + x + x^3) (1 + x + x^2 + 2 x^3 + 3 x^4 + 3 x^5 + 5 x^6 + 3 x^7 + 6 x^8 + x^9 + 4 x^10 + x^12))).
%F A291724 a(n) = a(n-5) + 5*a(n-7) + 10*a(n-9) + 10*a(n-11) + 5*a(n-13) + a(n-15) for n >= 15.
%t A291724 z = 60; s = x + x^3; p = 1 - s^5;
%t A291724 Drop[CoefficientList[Series[s, {x, 0, z}], x], 1]  (* A154272 *)
%t A291724 Drop[CoefficientList[Series[1/p, {x, 0, z}], x], 1]  (* A291724 *)
%t A291724 LinearRecurrence[{0, 0, 0, 0, 1, 0, 5, 0, 10, 0, 10, 0, 5, 0, 1}, {0, 0, 0, 0, 1, 0, 5, 0, 10, 1, 10, 10, 5, 45, 2}, 50] (* _Vincenzo Librandi_, Sep 10 2017 *)
%Y A291724 Cf. A154272, A291728.
%K A291724 nonn,easy
%O A291724 0,7
%A A291724 _Clark Kimberling_, Sep 08 2017