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A336630 a(n) = 2*F(2*n+1) + 4*F(n+1)*F(n-1) for n > 0, with a(0) = 0 and F(n) = A000045(n).

%I A336630 #14 Sep 11 2020 12:04:52
%S A336630 0,4,18,38,108,274,726,1892,4962,12982,33996,88994,232998,609988,
%T A336630 1596978,4180934,10945836,28656562,75023862,196415012,514221186,
%U A336630 1346248534,3524524428,9227324738,24157449798,63245024644,165577624146,433487847782
%N A336630 a(n) = 2*F(2*n+1) + 4*F(n+1)*F(n-1) for n > 0, with a(0) = 0 and F(n) = A000045(n).
%H A336630 <a href="/index/Rec#order_03">Index entries for linear recurrences with constant coefficients</a>, signature (2,2,-1).
%F A336630 a(n) = (F(4n+1) - F(n+1)^4)/F(n)^2 for n > 0 and a(0) = 0, where F(n) = A000045(n).
%F A336630 a(n) = 2*A001519(n+1) + 4*A059929(n-1) for n > 0.
%F A336630 From _Stefano Spezia_, Jul 28 2020: (Start)
%F A336630 O.g.f.: 2*x*(2 + 5*x - 3*x^2)/(1 - 2*x - 2*x^2 + x^3).
%F A336630 a(n) = 2*a(n-1) + 2*a(n-2) - a(n-2) for n > 3. (End)
%Y A336630 Cf. A000045, A001519, A059929.
%K A336630 nonn,easy
%O A336630 0,2
%A A336630 _Michael Tulskikh_, Jul 28 2020