C++ :

#include<bits/stdc++.h>
#define int long long
#define rep(i,j,k) for(int i=j;i<=k;i++)
#define per(i,j,k) for(int i=j;i>=k;i--)
using namespace std;
inline int read(){
	char ch=getchar();
	int x=0,t=1;
	while((ch<'0'||ch>'9')&&ch!='-'){ch=getchar();}
	if(ch=='-'){t=-1,ch=getchar();}
	while(ch>='0'&&ch<='9'){x=(x<<3)+(x<<1)+(ch^48),ch=getchar();}
	return x*t;
}
const int N=55,mod=1e9+7;
int f[N][N][N*N];//最后一维表示把现在未配对的都假设配在目前的第i位的花费
int n,kk;
signed main(){
	n=read(),kk=read();
	f[0][0][0]=1;
	rep(i,1,n){
		rep(j,0,i){
			rep(k,0,kk){
				if(j-1>=0&&k-2*(j-1)>=0){
					f[i][j][k]+=f[i-1][j-1][k-2*(j-1)];
				}
				if(k-2*j>=0){
					f[i][j][k]+=f[i-1][j][k-2*j]*(2*j+1);
				}
				if(k-2*(j+1)>=0){
					f[i][j][k]+=f[i-1][j+1][k-2*(j+1)]*(j+1)*(j+1);
				}
				f[i][j][k]%=mod;
			}
		}
	}
	cout<<f[n][0][kk]<<endl;
	return 0;
}

Python :

# coding=utf-8
mod = 10 ** 9 + 7

n, K = map(int, input().split())
dp = [[0] * (K + 1) for j in range(n)]
dp[0][0] = 1
for i in range(n):
    ndp = [[0] * (K + 1) for j in range(n)]
    for j in range(i + 1):
        for k in range(K + 1):
            dpi = dp[j][k]
            if dpi == 0:
                continue
            for nj, co in [(j - 1, j * j), (j, 1 + j * 2), (j + 1, 1)]:
                nk = k + nj * 2
                if 0 <= nj < n and nk <= K:
                    ndp[nj][nk] += dpi * co
                    ndp[nj][nk] %= mod
    dp = ndp
ans = dp[0][K]
print(ans)