Closed form for $q_k(0,0)$ from recurrence
up vote
0
down vote
favorite
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 at 3:20
user514787
676110
add a comment |
up vote
0
down vote
favorite
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 at 3:20
user514787
676110
add a comment |
up vote
0
down vote
favorite
up vote
0
down vote
favorite
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 at 3:20
user514787
676110
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 at 3:20
user514787
676110
asked Nov 21 at 3:20
user514787
676110
edited Nov 23 at 2:01
asked Nov 21 at 3:20
user514787
676110
asked Nov 21 at 3:20
user514787
676110
asked Nov 21 at 3:20
user514787
676110
676110
add a comment |
add a comment |
active
oldest
votes
active
oldest
votes
active
oldest
votes
active
oldest
votes
active
oldest
votes
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
StackExchange.ready(
function () {
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f3007212%2fclosed-form-for-q-k0-0-from-recurrence%23new-answer', 'question_page');
}
);
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
