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A291003 p-INVERT of (1,1,1,1,1,...), where p(S) = (1 - S)(1 - 2*S)(1 - 3*S)(1 - 4*S).

%I A291003 #10 May 24 2023 03:57:12
%S A291003 10,75,490,2956,16944,93800,506600,2687256,14064904,72873880,
%T A291003 374671560,1914880856,9741440264,49378177560,249583291720,
%U A291003 1258711575256,6336814854024,31857331730840,159980377179080,802678826106456,4024508089842184,20167014882109720
%N A291003 p-INVERT of (1,1,1,1,1,...), where p(S) = (1 - S)(1 - 2*S)(1 - 3*S)(1 - 4*S).
%C A291003 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 A291003 See A291000 for a guide to related sequences.
%H A291003 Clark Kimberling, <a href="/A291003/b291003.txt">Table of n, a(n) for n = 0..999</a>
%H A291003 <a href="/index/Rec#order_04">Index entries for linear recurrences with constant coefficients</a>, signature (14,-71,154,-120).
%F A291003 G.f.: (10 - 65*x + 150*x^2 - 119*x^3)/(1 - 14*x + 71*x^2 - 154*x^3 + 120*x^4).
%F A291003 a(n) = 14*a(n-1) - 71*a(n-2) + 154*a(n-3) - 120*a(n-4) for n >= 5.
%F A291003 a(n) = (-2^n + 16*3^(1+n) - 243*4^n + 256*5^n) / 6. - _Colin Barker_, Aug 23 2017
%t A291003 z = 60; s = x/(1-x); p = (1 - s)(1 - 2 s)(1 - 3 s)(1 - 4 s);
%t A291003 Drop[CoefficientList[Series[s, {x, 0, z}], x], 1]  (* A000012 *)
%t A291003 Drop[CoefficientList[Series[1/p, {x, 0, z}], x], 1]  (* A291003 *)
%t A291003 LinearRecurrence[{14,-71,154,-120}, {10,75,490,2956}, 41] (* _G. C. Greubel_, May 23 2023 *)
%o A291003 (Magma) [(-2^n +48*3^n -243*4^n +256*5^n)/6: n in [0..40]]; // _G. C. Greubel_, May 23 2023
%o A291003 (SageMath) [(-2^n +48*3^n -243*4^n +256*5^n)//6 for n in range(41)] # _G. C. Greubel_, May 23 2023
%Y A291003 Cf. A000012, A289780, A291000.
%K A291003 nonn,easy
%O A291003 0,1
%A A291003 _Clark Kimberling_, Aug 22 2017