• Let $a(n)$ be A204262 i.e. permanent of the matrix $n\times n$ with elements $\min(i,j)$.
• Let $$ f_{n,\ell}(x)=\int (n-\ell)^2 f_{n-1,\ell}(x)+f_{n,\ell-1}(\ell)-\int (n-\ell)^2 f_{n-1,\ell}(\ell), \\ f_{n,0}(x)=n!x^n $$
• Let $$ R(n,q)=\sum\limits_{j=0}^{q+1}\binom{q+1}{j}(j+1)R(n-1,j), \\ R(0,q)=1 $$

I conjecture that $$R(n,0)=f_{n+1,n+1}(0)=a(n+1).$$

Here is the PARI/GP prog to check it numerically:

a(n)=matpermanent(matrix(n, n, i, j, min(i, j)))
f_upto(n)=my(v1); v1=vector(n+1, i, i--; i!*x^i); for(i=1, n, for(j=i, n, my(A=intformal((j-i)^2*v1[j])); v1[j+1] = A + subst(v1[j+1] - A, x, i))); v1
R_upto(n)=my(v1, v2, v3, v4); v1=vector(n+1, i, 1); v2=v1; v3=vector(n+1, i, 0); v3[1]=1; v4=vector(n, i, vector(i+1, j, binomial(i, j-1)*j)); for(i=1, n, for(q=0, n-i, v2[q+1]=sum(j=0, q+1, v4[q+1][j+1]*v1[j+1])); v1=v2; v3[i+1]=v1[1];); v3
test1(n)=f_upto(n)==concat(1, R_upto(n-1))
test2(n)=f_upto(n)==vector(n+1, i, a(i-1))

Is there a way to prove it?

• Let $a(n)$ be A204262 i.e. permanent of the matrix $n\times n$ with elements $\min(i,j)$.
• Let $$ f_{n,\ell}(x)=g_{n,\ell}(x)+f_{n,\ell-1}(\ell)-g_{n,\ell}(\ell), \\ g_{n,\ell}(x)=\int (n-\ell)^2 f_{n-1,\ell}(x)\,dx, \\ f_{n,0}(x)=n!x^n $$
• Let $$ R(n,q)=\sum\limits_{j=0}^{q+1}\binom{q+1}{j}(j+1)R(n-1,j), \\ R(0,q)=1 $$

I conjecture that $$R(n,0)=f_{n+1,n+1}(0)=a(n+1).$$

Here is the PARI/GP prog to check it numerically:

a(n)=matpermanent(matrix(n, n, i, j, min(i, j)))
f_upto(n)=my(v1); v1=vector(n+1, i, i--; i!*x^i); for(i=1, n, for(j=i, n, my(A=intformal((j-i)^2*v1[j])); v1[j+1] = A + subst(v1[j+1] - A, x, i))); v1
R_upto(n)=my(v1, v2, v3, v4); v1=vector(n+1, i, 1); v2=v1; v3=vector(n+1, i, 0); v3[1]=1; v4=vector(n, i, vector(i+1, j, binomial(i, j-1)*j)); for(i=1, n, for(q=0, n-i, v2[q+1]=sum(j=0, q+1, v4[q+1][j+1]*v1[j+1])); v1=v2; v3[i+1]=v1[1];); v3
test1(n)=f_upto(n)==concat(1, R_upto(n-1))
test2(n)=f_upto(n)==vector(n+1, i, a(i-1))

Is there a way to prove it?

• Let $a(n)$ be A204262 i.e. permanent of the matrix $n\times n$ with elements $\min(i,j)$.
• Let $$ f_{n,\ell}(x)=g_{n,\ell}(x)+f_{n,\ell-1}(\ell)-g_{n,\ell}(\ell), \\ g_{n,\ell}(x)=\int (n-\ell)^2 f_{n-1,\ell}(x), \\ f_{n,0}(x)=n!x^n $$
• Let $$ R(n,q)=\sum\limits_{j=0}^{q+1}\binom{q+1}{j}(j+1)R(n-1,j), \\ R(0,q)=1 $$

I conjecture that $$R(n,0)=f_{n+1,n+1}(0)=a(n+1).$$

Here is the PARI/GP prog to check it numerically:

a(n)=matpermanent(matrix(n, n, i, j, min(i, j)))
f_upto(n)=my(v1); v1=vector(n+1, i, i--; i!*x^i); for(i=1, n, for(j=i, n, my(A=intformal((j-i)^2*v1[j])); v1[j+1] = A + subst(v1[j+1] - A, x, i))); v1
R_upto(n)=my(v1, v2, v3, v4); v1=vector(n+1, i, 1); v2=v1; v3=vector(n+1, i, 0); v3[1]=1; v4=vector(n, i, vector(i+1, j, binomial(i, j-1)*j)); for(i=1, n, for(q=0, n-i, v2[q+1]=sum(j=0, q+1, v4[q+1][j+1]*v1[j+1])); v1=v2; v3[i+1]=v1[1];); v3
test1(n)=f_upto(n)==concat(1, R_upto(n-1))
test2(n)=f_upto(n)==vector(n+1, i, a(i-1))

Is there a way to prove it?

• Let $a(n)$ be A204262 i.e. permanent of the matrix $n\times n$ with elements $\min(i,j)$.
• Let $$ f_{n,\ell}(x)=\int (n-\ell)^2 f_{n-1,\ell}(x)+f_{n,\ell-1}(\ell)-\int (n-\ell)^2 f_{n-1,\ell}(\ell), \\ f_{n,0}(x)=n!x^n $$
• Let $$ R(n,q)=\sum\limits_{j=0}^{q+1}\binom{q+1}{j}(j+1)R(n-1,j), \\ R(0,q)=1 $$

I conjecture that $$R(n,0)=f_{n+1,n+1}(0)=a(n+1).$$

Here is the PARI/GP prog to check it numerically:

a(n)=matpermanent(matrix(n, n, i, j, min(i, j)))
f_upto(n)=my(v1); v1=vector(n+1, i, i--; i!*x^i); for(i=1, n, for(j=i, n, my(A=intformal((j-i)^2*v1[j])); v1[j+1] = A + subst(v1[j+1] - A, x, i))); v1
R_upto(n)=my(v1, v2, v3, v4); v1=vector(n+1, i, 1); v2=v1; v3=vector(n+1, i, 0); v3[1]=1; v4=vector(n, i, vector(i+1, j, binomial(i, j-1)*j)); for(i=1, n, for(q=0, n-i, v2[q+1]=sum(j=0, q+1, v4[q+1][j+1]*v1[j+1])); v1=v2; v3[i+1]=v1[1];); v3
test1(n)=f_upto(n)==concat(1, R_upto(n-1))
test2(n)=f_upto(n)==vector(n+1, i, a(i-1))

Is there a way to prove it?