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Number of binary trees of height n such that the number of nodes at depth d equals d+1 for every d = 0..n.
3

%I #23 Oct 25 2024 14:26:01

%S 1,1,4,60,3360,705600,558835200,1678182105600,19198403288064000,

%T 840083731079104512000,141100463472046393835520000,

%U 91242050302344912388163665920000,227753296409896438988240405704212480000,2199573010737856838816729366169572868096000000,82356764599728553816070191604819734458909327360000000

%N Number of binary trees of height n such that the number of nodes at depth d equals d+1 for every d = 0..n.

%C To satisfy the constraint, there must be n+1 nodes at depth n, and there are 2n allowed slots for a new node.

%C A binary tree with such a level profile contains A000217(n+1) nodes.

%F a(n) = binomial(2*n,n+1)*a(n-1), a(0)=1.

%F a(n) = Product_{k=1..n} binomial(2*k,k+1).

%F a(n) = 2^(n^2+n-1/24)*A^(3/2)*Pi^(-1/4-n/2)*G(3/2 + n)*Gamma(1 + n)/(exp(1/8)*G(3 + n)) where A is the Glaisher-Kinkelin constant and G is the Barnes G function. - _Stefano Spezia_, Nov 02 2020

%F a(n) ~ A^(3/2) * 2^(-7/24 + n + n^2) * exp(-1/8 + n/2) / (n^(11/8 + n/2) * Pi^((n+1)/2)), where A = A074962 is the Glaisher-Kinkelin constant. - _Vaclav Kotesovec_, Aug 29 2023

%F a(n) = Product_{1 <= j <= i <= n-1} (i + j + 2)/(i - j + 1). - _Peter Bala_, Oct 25 2024

%t Table[Product[Binomial[2k,k+1],{k,n}],{n,0,14}] (* or *)

%t Table[2^(n^2+n-1/24)Glaisher^(3/2)Pi^(-1/4-n/2)BarnesG[3/2+n]Gamma[1+n]/(Exp[1/8]BarnesG[3+n]),{n,0,14}] (* _Stefano Spezia_, Nov 02 2020 *)

%Y Cf. A000217, A007685, A203471, A296589.

%K nonn,easy

%O 0,3

%A _Marcel K. Goh_, Nov 02 2020