// Slower C++ solution:
///*****************************************************///

///*.......................Faiyaj Bin Amin............................*///

///*...................................................................*///

///*.........................KUET CSE 2K17......................*///

///*...................................................*///

///*****************************************************///



#include<bits/stdc++.h>

//#include <ext/pb_ds/assoc_container.hpp>

//#include <ext/pb_ds/tree_policy.hpp>



#define fastio      ios_base::sync_with_stdio(0);cin.tie(0),cout.tie(0)

#define ret return

#define rep(q, i, n) for(int q=i; q<n; q++)

#define forn(i, n) for(int i=0; i<n; i++)

#define mems(a,x) memset(a,x,sizeof(a))

#define ll long long

#define pb push_back

#define em emplace_back

#define mp make_pair

#define all(v) (v).begin(), (v).end()

#define ff first

#define ss second

#define pii pair<int,int>

#define pdd pair<double, double>



#define see(x)        cerr<< __LINE__ << ": " << (#x) << " is "<< (x) <<'\n';

#define seecontent(v) for(auto x : v) { cout << x << " ";} cout << "\n\n";



#define PI 2*acos(0.0)

#define lcm(a, b)  ((a / __gcd(a, b) ) * b)



#define filein freopen ("input.txt", "r", stdin)

#define fileout freopen ("output.txt", "w", stdout)



#define mx 20005

#define INF 1e18

#define MOD 1000000007

#define MIN 0



//template <typename T>  using orderedSet =

//    tree<T, null_type, less_equal<T>,

//    rb_tree_tag, tree_order_statistics_node_update>; //find_by_order, order_of_key





/*------------------------------Graph Moves----------------------------*/

//const int fx[]={+1,-1,+0,+0}; // Directional

//const int fy[]={+0,+0,+1,-1};

//const int fx[]={+0,+0,+1,-1,-1,+1,-1,+1};   // Kings Move

//const int fy[]={-1,+1,+0,+0,+1,+1,-1,-1};  // Kings Move

//const int fx[]={-2, -2, -1, -1,  1,  1,  2,  2};  // Knights Move

//const int fy[]={-1,  1, -2,  2, -2,  2, -1,  1}; // Knights Move

/*---------------------------------------------------------------------*/



/**------------------------------------BITMASK------------------------------------------**/

//int Change(int N, int pos){return N= N^(1<<pos);}

//bool Check(int N, int pos){return (bool)(N &(1<<pos));}  /// check bit 0 or 1

/**-----------------------------------------------------------------------------------------------**/



//using namespace __gnu_pbds;

using namespace std;



bool vis[mx+5];

int dis[mx+5];

int ks[mx+1]= {0};

vector<int> adj[mx+3];



void BFS(int src, int n)

{



////    see(src);

    memset(vis,0, sizeof vis);



    for(int i=0; i<=n; i++)

        dis[i]=0;



    queue <int> q;

    vis[src]=1;

    q.push(src);



    while(!q.empty())

    {

        int u=q.front();

        q.pop();

        for(int v:adj[u])

        {

            if(!vis[v])

            {

                vis[v]=1;

                dis[v]=dis[u]+1;



////                see(u);

////                see(v);

////                see(dis[v]);

                ks[dis[v]]++;

                q.push(v);

            }

        }

    }

////    cout << endl;





}



int main()

{

    ///FASTIO;



    int n,x,y;

    cin>>n>>x>>y;



    for(int i=1; i<n; i++)

    {

        adj[i].push_back(i+1);

        adj[i+1].push_back(i);

    }

    adj[x].push_back(y);

    adj[y].push_back(x);





    for(int i=1; i<=n; i++)

        BFS(i,n);



////    for(int i=1; i<=n; i++)

////    {

////        cout << i << " " << dist[i]) << endl;

////    }

    for(int i=1; i<n; i++)

    {

        cout << ks[i]/2 << "\n";

    }



    return 0;

}








// Faster C++ solution for the same problem (measured speedup: 1.5743260845518452x):
#include<stdio.h>

#include<cstdio>

#include<iostream>

#include<algorithm>

#include<queue>

#include<cstring>

using namespace std;

const int maxn=20100;

int n,x,y;

int ans[maxn];

int main()

{

	

	scanf("%d%d%d",&n,&x,&y);

	for(int i=1;i<=n;++i)

	{

		for(int j=i+1;j<=n;++j)

		{

			int k=min(j-i,abs(x-i)+1+abs(j-y));

			k=min(k,abs(x-j)+1+abs(y-i));

			ans[k]++;

		}

	}

	for(int i=1;i<n;++i)

	{

		printf("%d\n",ans[i]);

	}

	return 0;

}