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a(n) = binomial(2*n-1,n) - binomial(n,2)*(binomial(n-1,2) + 2) - 1.
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%I #11 Mar 29 2024 10:22:48

%S 0,0,0,4,45,281,1358,5790,23229,90667,350130,1348315,5194995,20051019,

%T 77548994,300527354,1166786517,4537546535,17672605394,68923231539,

%U 269128896899,1052049432887,4116715304850,16123801771169,63205303135475,247959266375901,973469712709278

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

%C a(n) is the number of ways to place n indistinguishable balls into n distinguishable boxes with at least 3 boxes remaining empty.

%C a(n) is also the number of weak compositions of n into n parts in which at least three parts are zero.

%e a(5)=45 since 5 can be written as 5+0+0+0+0, 0+5+0+0+0, etc. (5 such compositions); 4+1+0+0+0 (20 such compositions); 3+2+0+0+0 (20 such compositions).

%t Table[Binomial[2n-1,n]-Binomial[n,2]*(Binomial[n-1,2]+2)-1,{n,27}] (* _James C. McMahon_, Mar 08 2024 *)

%o (Python)

%o from math import comb

%o def A371003(n): return comb((n<<1)-1,n)-n-((m:=(n-1)**2)*(m+3)>>2) # _Chai Wah Wu_, Mar 29 2024

%Y Cf. A010763, A352027, A359435.

%K nonn

%O 1,4

%A _Enrique Navarrete_, Mar 07 2024