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%I #10 Apr 14 2023 03:10:14
%S 1,2,4,9,22,55,136,330,789,1872,4433,10510,24968,59409,141470,336935,
%T 802340,1910166,4546845,10822176,25758097,61308650,145928764,
%U 347350473,826795942,1968018151,4684451824,11150316882,26540849109,63174538224,150372815489
%N p-INVERT of (1,1,1,1,1,...), where p(S) = 1 - S - 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 A291000 for a guide to related sequences.
%H Clark Kimberling, <a href="/A290996/b290996.txt">Table of n, a(n) for n = 0..1000</a>
%H <a href="/index/Rec#order_04">Index entries for linear recurrences with constant coefficients</a>, signature (5,-9,7,-1).
%F a(n) = 5*a(n-1) - 9*a(n-2) + 7*a(n-3) - a(n-4) for n >= 4.
%F G.f.: (1 - 3*x + 3*x^2) / (1 - 5*x + 9*x^2 - 7*x^3 + x^4). - _Colin Barker_, Aug 22 2017
%t z = 60; s = x/(1-x); p = 1 -s -s^4;
%t Drop[CoefficientList[Series[s, {x,0,z}], x], 1] (* A000012 *)
%t Drop[CoefficientList[Series[1/p, {x,0,z}], x], 1] (* A290996 *)
%t LinearRecurrence[{5,-9,7,-1}, {1,2,4,9}, 60] (* _G. C. Greubel_, Apr 13 2023 *)
%o (PARI) Vec((1 - 3*x + 3*x^2) / (1 - 5*x + 9*x^2 - 7*x^3 + x^4) + O(x^50)) \\ _Colin Barker_, Aug 22 2017
%o (Magma) I:=[1,2,4,9]; [n le 4 select I[n] else 5*Self(n-1) -9*Self(n-2) +7*Self(n-3) -Self(n-4): n in [1..51]]; // _G. C. Greubel_, Apr 13 2023
%o (SageMath)
%o @CachedFunction
%o def a(n): # a = A290996
%o if(n<4): return (1,2,4,9)[n]
%o else: return 5*a(n-1) - 9*a(n-2) + 7*a(n-3) - a(n-4)
%o [a(n) for n in range(61)] # _G. C. Greubel_, Apr 13 2023
%Y Cf. A000012, A033453, A289780, A291000.
%K nonn,easy
%O 0,2
%A _Clark Kimberling_, Aug 22 2017
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