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A100335 An inverse Catalan transform of J(2n).

%I A100335 #22 Jan 25 2023 20:36:46
%S A100335 0,1,4,11,27,64,149,341,768,1707,3755,8192,17749,38229,81920,174763,
%T A100335 371371,786432,1660245,3495253,7340032,15379115,32156331,67108864,
%U A100335 139810133,290805077,603979776,1252698795,2594876075,5368709120
%N A100335 An inverse Catalan transform of J(2n).
%C A100335 The g.f. is obtained from that of A002450 through the mapping g(x) -> g(x*(1-x)). A002450 may be retrieved through the mapping g(x) -> g(x*c(x)), where c(x) is the g.f. of A000108.
%H A100335 G. C. Greubel, <a href="/A100335/b100335.txt">Table of n, a(n) for n = 0..1000</a>
%H A100335 <a href="/index/Rec#order_04">Index entries for linear recurrences with constant coefficients</a>, signature (5,-9,8,-4).
%F A100335 G.f.: x*(1-x)/(1 - 5*x + 9*x^2 - 8*x^3 + 4*x^4).
%F A100335 a(n) = 5*a(n-1) - 9*a(n-2) + 8*a(n-3) - 4*a(n-4).
%F A100335 a(n) = Sum_{k=0..floor(n/2)} binomial(n-k, k)*(-1)^k*(4^(n-k) - 1)/3.
%F A100335 a(n) = (1/3)*((n+1)*2^n - A010892(n)). - _Ralf Stephan_, May 15 2007
%F A100335 Binomial transform of A042965: (1, 3, 4, 5, 7, 8, 9, 11, 12, 13, ...), also row sums of triangle A133110. - _Gary W. Adamson_, Sep 12 2007
%F A100335 a(n) = Sum_{k=0..n} A109466(n,k)*A002450(k). - _Philippe Deléham_, Oct 30 2008
%t A100335 LinearRecurrence[{5,-9,8,-4}, {0,1,4,11}, 41] (* _G. C. Greubel_, Jan 24 2023 *)
%o A100335 (Magma) I:=[0,1,4,11]; [n le 4 select I[n] else 5*Self(n-1) -9*Self(n-2) +8*Self(n-3) -4*Self(n-4): n in [1..41]]; // _G. C. Greubel_, Jan 24 2023
%o A100335 (SageMath)
%o A100335 def A100335(n): return (1/3)*((n+1)*2^n - chebyshev_U(n,1/2))
%o A100335 [A100335(n) for n in range(41)] # _G. C. Greubel_, Jan 24 2023
%Y A100335 Cf. A000108, A001045, A002450, A010892.
%Y A100335 Cf. A042965, A100334, A109466, A133110.
%K A100335 easy,nonn
%O A100335 0,3
%A A100335 _Paul Barry_, Nov 17 2004