Mixing
Closed form for $q_k(0,0)$ from recurrence
We have
$$p_0(n,m)=begin{cases}
0,&text{$n=m=0$}\
(n-1)!,&text{$n>0, m=0$}\
0,&text{$ngeqslant0, m>0$}
end{cases}$$
$$q_0(n,m)=0, ngeqslant0, mgeqslant0$$
and
$$p_k(n,m)=begin{cases}
0,&text{$n=0, mgeqslant0, k>0$}\
sumlimits_{d=0}^{n-1}frac{(n-1)!}{d!}q_k(d,m),&text{$n>0, mgeqslant0, k>0$}
end{cases}$$
$$q_k(n,m)=p_{k-1}(n+1,m+1)+p_{k-1}(m+1,n), ngeqslant0, mgeqslant0, k>0$$
so when $nne0$ for $p_k(n,m)$ and $q_1(n,m)$, and when $ngeqslant0$ for $q_k(n,m)$ with $k>0$, first three cases are
$$p_1(n,m)=(n-1)!m!, q_1(n,m)=0$$
$$p_2(n,m)=2n!m!(m+2), q_2(n,m)=n!m!(m+2)$$
$$p_3(n,m)=3n!(m+1)!((n+1)(m+4)+2), q_3(n,m)=2n!(m+1)!((n+1)(m+4)+1)$$
also
$$q_1(0,m)=m!, mgeqslant0$$
General interest is to find $q_k(0,0)$ for which we need
$$q_k(0,0)=p_{k-1}(1,1)+p_{k-1}(1,0)=(k-1)(q_{k-1}(0,1)+q_{k-1}(0,0))$$
Is there closed form for it?
combinatorics permutations recurrence-relations factorial closed-form
asked Nov 21 '18 at 3:20
user514787
695210
add a comment |
We have
$$p_0(n,m)=begin{cases}
0,&text{$n=m=0$}\
(n-1)!,&text{$n>0, m=0$}\
0,&text{$ngeqslant0, m>0$}
end{cases}$$
$$q_0(n,m)=0, ngeqslant0, mgeqslant0$$
and
$$p_k(n,m)=begin{cases}
0,&text{$n=0, mgeqslant0, k>0$}\
sumlimits_{d=0}^{n-1}frac{(n-1)!}{d!}q_k(d,m),&text{$n>0, mgeqslant0, k>0$}
end{cases}$$
$$q_k(n,m)=p_{k-1}(n+1,m+1)+p_{k-1}(m+1,n), ngeqslant0, mgeqslant0, k>0$$
so when $nne0$ for $p_k(n,m)$ and $q_1(n,m)$, and when $ngeqslant0$ for $q_k(n,m)$ with $k>0$, first three cases are
$$p_1(n,m)=(n-1)!m!, q_1(n,m)=0$$
$$p_2(n,m)=2n!m!(m+2), q_2(n,m)=n!m!(m+2)$$
$$p_3(n,m)=3n!(m+1)!((n+1)(m+4)+2), q_3(n,m)=2n!(m+1)!((n+1)(m+4)+1)$$
also
$$q_1(0,m)=m!, mgeqslant0$$
General interest is to find $q_k(0,0)$ for which we need
$$q_k(0,0)=p_{k-1}(1,1)+p_{k-1}(1,0)=(k-1)(q_{k-1}(0,1)+q_{k-1}(0,0))$$
Is there closed form for it?
combinatorics permutations recurrence-relations factorial closed-form
asked Nov 21 '18 at 3:20
user514787
695210
add a comment |
We have
$$p_0(n,m)=begin{cases}
0,&text{$n=m=0$}\
(n-1)!,&text{$n>0, m=0$}\
0,&text{$ngeqslant0, m>0$}
end{cases}$$
$$q_0(n,m)=0, ngeqslant0, mgeqslant0$$
and
$$p_k(n,m)=begin{cases}
0,&text{$n=0, mgeqslant0, k>0$}\
sumlimits_{d=0}^{n-1}frac{(n-1)!}{d!}q_k(d,m),&text{$n>0, mgeqslant0, k>0$}
end{cases}$$
$$q_k(n,m)=p_{k-1}(n+1,m+1)+p_{k-1}(m+1,n), ngeqslant0, mgeqslant0, k>0$$
so when $nne0$ for $p_k(n,m)$ and $q_1(n,m)$, and when $ngeqslant0$ for $q_k(n,m)$ with $k>0$, first three cases are
$$p_1(n,m)=(n-1)!m!, q_1(n,m)=0$$
$$p_2(n,m)=2n!m!(m+2), q_2(n,m)=n!m!(m+2)$$
$$p_3(n,m)=3n!(m+1)!((n+1)(m+4)+2), q_3(n,m)=2n!(m+1)!((n+1)(m+4)+1)$$
also
$$q_1(0,m)=m!, mgeqslant0$$
General interest is to find $q_k(0,0)$ for which we need
$$q_k(0,0)=p_{k-1}(1,1)+p_{k-1}(1,0)=(k-1)(q_{k-1}(0,1)+q_{k-1}(0,0))$$
Is there closed form for it?
combinatorics permutations recurrence-relations factorial closed-form
asked Nov 21 '18 at 3:20
user514787
695210
We have
$$p_0(n,m)=begin{cases}
0,&text{$n=m=0$}\
(n-1)!,&text{$n>0, m=0$}\
0,&text{$ngeqslant0, m>0$}
end{cases}$$
$$q_0(n,m)=0, ngeqslant0, mgeqslant0$$
and
$$p_k(n,m)=begin{cases}
0,&text{$n=0, mgeqslant0, k>0$}\
sumlimits_{d=0}^{n-1}frac{(n-1)!}{d!}q_k(d,m),&text{$n>0, mgeqslant0, k>0$}
end{cases}$$
$$q_k(n,m)=p_{k-1}(n+1,m+1)+p_{k-1}(m+1,n), ngeqslant0, mgeqslant0, k>0$$
so when $nne0$ for $p_k(n,m)$ and $q_1(n,m)$, and when $ngeqslant0$ for $q_k(n,m)$ with $k>0$, first three cases are
$$p_1(n,m)=(n-1)!m!, q_1(n,m)=0$$
$$p_2(n,m)=2n!m!(m+2), q_2(n,m)=n!m!(m+2)$$
$$p_3(n,m)=3n!(m+1)!((n+1)(m+4)+2), q_3(n,m)=2n!(m+1)!((n+1)(m+4)+1)$$
also
$$q_1(0,m)=m!, mgeqslant0$$
General interest is to find $q_k(0,0)$ for which we need
$$q_k(0,0)=p_{k-1}(1,1)+p_{k-1}(1,0)=(k-1)(q_{k-1}(0,1)+q_{k-1}(0,0))$$
Is there closed form for it?
combinatorics permutations recurrence-relations factorial closed-form
combinatorics permutations recurrence-relations factorial closed-form
asked Nov 21 '18 at 3:20
user514787
695210
asked Nov 21 '18 at 3:20
user514787
695210
edited Nov 23 '18 at 2:01
asked Nov 21 '18 at 3:20
user514787
695210
asked Nov 21 '18 at 3:20
user514787
695210
asked Nov 21 '18 at 3:20
user514787
695210
695210
add a comment |
add a comment |
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