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A291699 a(n) = n^n*(2*n)!/(n!*(n + 1)!).

%I A291699 #9 Oct 23 2023 12:46:33
%S A291699 1,1,8,135,3584,131250,6158592,353299947,23991418880,1883638417518,
%T A291699 167960000000000,16772331868538246,1854655886442627072,
%U A291699 225005916687384753700,29718395534545380311040,4245313393689422607421875,652233889532678001886494720,107247390031799133661006687830
%N A291699 a(n) = n^n*(2*n)!/(n!*(n + 1)!).
%H A291699 Robert Israel, <a href="/A291699/b291699.txt">Table of n, a(n) for n = 0..322</a>
%F A291699 a(n) = [x^n] 2/(1 + sqrt(1 - 4*n*x)).
%F A291699 a(n) = [x^n] 1/(1 - n*x/(1 - n*x/(1 - n*x/(1 - n*x/(1 - n*x/(1 - ...)))))), a continued fraction.
%F A291699 a(n) = n! * [x^n] (BesselI(0,2*n*x) - BesselI(1,2*n*x))*exp(2*n*x).
%F A291699 a(n) = n^n*binomial(2*n,n)/(n + 1).
%F A291699 a(n) = A000312(n)*A000108(n).
%F A291699 a(n) = A290605(n,n).
%F A291699 a(n) ~ 4^n*n^(n-3/2)/sqrt(Pi).
%p A291699 seq(n^n*(2*n)!/n!/(n+1)!, n=0..50); # _Robert Israel_, Aug 30 2017
%t A291699 Join[{1}, Table[n^n (2 n)!/(n! (n + 1)!), {n, 1, 17}]]
%t A291699 Table[SeriesCoefficient[2/(1 + Sqrt[1 - 4 n x]), {x, 0, n}], {n, 0, 17}]
%t A291699 Table[SeriesCoefficient[1/(1 + ContinuedFractionK[-n x, 1, {i, 1, n}]), {x, 0, n}], {n, 0, 17}]
%o A291699 (PARI) a(n)=binomial(2*n,n)/(n+1)*n^n \\ _Charles R Greathouse IV_, Oct 23 2023
%Y A291699 Main diagonal of A290605.
%Y A291699 Cf. A000108, A000312, A001761, A061711.
%K A291699 nonn
%O A291699 0,3
%A A291699 _Ilya Gutkovskiy_, Aug 30 2017