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a(n) = binomial(2*n-2,n-1) - n.
1

%I #24 Apr 30 2023 18:17:05

%S 0,0,3,16,65,246,917,3424,12861,48610,184745,705420,2704143,10400586,

%T 40116585,155117504,601080373,2333606202,9075135281,35345263780,

%U 137846528799,538257874418,2104098963697,8233430727576,32247603683075,126410606437726,495918532948077

%N a(n) = binomial(2*n-2,n-1) - n.

%C a(n) is the number of ways to place n-1 indistinguishable balls into n distinguishable boxes with not all balls placed in one box.

%H Stefano Spezia, <a href="/A359176/b359176.txt">Table of n, a(n) for n = 1..1600</a>

%F a(n) = A000984(n-1) - n.

%F G.f.: x*(1/sqrt(1 - 4*x) - 1/(1 - x)^2). - _Stefano Spezia_, Dec 28 2022

%F D-finite with recurrence: (-n+1)*a(n) +(5*n-8)*a(n-1) +2*(-2*n+5)*a(n-2) +6*(n-2)=0. - _R. J. Mathar_, Jan 25 2023

%F From _Stefano Spezia_, Apr 25 2023:

%F E.g.f.: x*exp(x)*(exp(x)*(BesselI(0,2*x) - BesselI(1,2*x)) - 1).

%F a(n) ~ 2^(2*n-2)/sqrt(n*Pi). (End)

%t a[n_] := Binomial[2*n - 2, n - 1] - n; Array[a, 30] (* _Amiram Eldar_, Dec 30 2022 *)

%Y Cf. A000984.

%K nonn

%O 1,3

%A _Enrique Navarrete_, Dec 28 2022