Mixing
Find a generating function for product of Stirling numbers.
Clash Royale CLAN TAG#URR8PPP
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3
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What is a close form of the series
$$
G_i,j(z)=sum_p=0^infty S(p,i) S(p,j) z^p?
$$
Here $S(*,*)$ is the Stirling numbers of the second kind.
For $i=1$, since $S(p,1)=1$ the result is well-known
$$
G_1,j(z)=sum_p=0^infty S(p,j) z^p=fracz^j(1-z)(1-2z)cdots(1-jz)
$$
For small $i,j>1$ with computer I found that
$$
G_2,2(z)= -frac z^2(2,z+1) left( z-1 right) left( 2,z-1 right) left( 4
,z-1 right) ,\
G_2,3(z)=3,frac z^3(4,z^2+2,z-1) left( z-1 right) left( 6,z-1
right) left( 4,z-1 right) left( 3,z-1 right) left( 2,z-1
right) ,\
G_2,4(z)=-frac z^4(24,z^2+18,z-7) left( z-1 right) left( 6,z-1
right) left( 4,z-1 right) left( 3,z-1 right) left( 2,z-1
right) left( 8,z-1 right)
$$
What about the general case $G_i,j(z)?$ I have tried consider the $G_i,j(z)$ as Hadamar product $G_1,i(z) * G_1,j(z)$ but without any success.
combinatorics generating-functions stirling-numbers
asked Jul 29 at 11:01
Leox
5,0921323
add a comment |Â
up vote
3
down vote
favorite
What is a close form of the series
$$
G_i,j(z)=sum_p=0^infty S(p,i) S(p,j) z^p?
$$
Here $S(*,*)$ is the Stirling numbers of the second kind.
For $i=1$, since $S(p,1)=1$ the result is well-known
$$
G_1,j(z)=sum_p=0^infty S(p,j) z^p=fracz^j(1-z)(1-2z)cdots(1-jz)
$$
For small $i,j>1$ with computer I found that
$$
G_2,2(z)= -frac z^2(2,z+1) left( z-1 right) left( 2,z-1 right) left( 4
,z-1 right) ,\
G_2,3(z)=3,frac z^3(4,z^2+2,z-1) left( z-1 right) left( 6,z-1
right) left( 4,z-1 right) left( 3,z-1 right) left( 2,z-1
right) ,\
G_2,4(z)=-frac z^4(24,z^2+18,z-7) left( z-1 right) left( 6,z-1
right) left( 4,z-1 right) left( 3,z-1 right) left( 2,z-1
right) left( 8,z-1 right)
$$
What about the general case $G_i,j(z)?$ I have tried consider the $G_i,j(z)$ as Hadamar product $G_1,i(z) * G_1,j(z)$ but without any success.
combinatorics generating-functions stirling-numbers
asked Jul 29 at 11:01
Leox
5,0921323
1
Perhaps you're using a different convention, but AFAIK $S(p,j)=0$ for $p < j$. As a result you're missing some factors of $z$,
– Robert Israel
Jul 29 at 12:23
Yes, you are right - I have lost the factor $z^max(i,j)$. Corrected. Thank you!
– Leox
Jul 29 at 12:40
add a comment |Â
up vote
3
down vote
favorite
up vote
3
down vote
favorite
What is a close form of the series
$$
G_i,j(z)=sum_p=0^infty S(p,i) S(p,j) z^p?
$$
Here $S(*,*)$ is the Stirling numbers of the second kind.
For $i=1$, since $S(p,1)=1$ the result is well-known
$$
G_1,j(z)=sum_p=0^infty S(p,j) z^p=fracz^j(1-z)(1-2z)cdots(1-jz)
$$
For small $i,j>1$ with computer I found that
$$
G_2,2(z)= -frac z^2(2,z+1) left( z-1 right) left( 2,z-1 right) left( 4
,z-1 right) ,\
G_2,3(z)=3,frac z^3(4,z^2+2,z-1) left( z-1 right) left( 6,z-1
right) left( 4,z-1 right) left( 3,z-1 right) left( 2,z-1
right) ,\
G_2,4(z)=-frac z^4(24,z^2+18,z-7) left( z-1 right) left( 6,z-1
right) left( 4,z-1 right) left( 3,z-1 right) left( 2,z-1
right) left( 8,z-1 right)
$$
What about the general case $G_i,j(z)?$ I have tried consider the $G_i,j(z)$ as Hadamar product $G_1,i(z) * G_1,j(z)$ but without any success.
combinatorics generating-functions stirling-numbers
asked Jul 29 at 11:01
Leox
5,0921323
What is a close form of the series
$$
G_i,j(z)=sum_p=0^infty S(p,i) S(p,j) z^p?
$$
Here $S(*,*)$ is the Stirling numbers of the second kind.
For $i=1$, since $S(p,1)=1$ the result is well-known
$$
G_1,j(z)=sum_p=0^infty S(p,j) z^p=fracz^j(1-z)(1-2z)cdots(1-jz)
$$
For small $i,j>1$ with computer I found that
$$
G_2,2(z)= -frac z^2(2,z+1) left( z-1 right) left( 2,z-1 right) left( 4
,z-1 right) ,\
G_2,3(z)=3,frac z^3(4,z^2+2,z-1) left( z-1 right) left( 6,z-1
right) left( 4,z-1 right) left( 3,z-1 right) left( 2,z-1
right) ,\
G_2,4(z)=-frac z^4(24,z^2+18,z-7) left( z-1 right) left( 6,z-1
right) left( 4,z-1 right) left( 3,z-1 right) left( 2,z-1
right) left( 8,z-1 right)
$$
What about the general case $G_i,j(z)?$ I have tried consider the $G_i,j(z)$ as Hadamar product $G_1,i(z) * G_1,j(z)$ but without any success.
combinatorics generating-functions stirling-numbers
asked Jul 29 at 11:01
Leox
5,0921323
edited Jul 29 at 12:40
asked Jul 29 at 11:01
Leox
5,0921323
asked Jul 29 at 11:01
Leox
5,0921323
asked Jul 29 at 11:01
Leox
5,0921323
5,0921323
1
Perhaps you're using a different convention, but AFAIK $S(p,j)=0$ for $p < j$. As a result you're missing some factors of $z$,
– Robert Israel
Jul 29 at 12:23
Yes, you are right - I have lost the factor $z^max(i,j)$. Corrected. Thank you!
– Leox
Jul 29 at 12:40
add a comment |Â
1
Perhaps you're using a different convention, but AFAIK $S(p,j)=0$ for $p < j$. As a result you're missing some factors of $z$,
– Robert Israel
Jul 29 at 12:23
Yes, you are right - I have lost the factor $z^max(i,j)$. Corrected. Thank you!
– Leox
Jul 29 at 12:40
1
1
Perhaps you're using a different convention, but AFAIK $S(p,j)=0$ for $p < j$. As a result you're missing some factors of $z$,
– Robert Israel
Jul 29 at 12:23
Perhaps you're using a different convention, but AFAIK $S(p,j)=0$ for $p < j$. As a result you're missing some factors of $z$,
– Robert Israel
Jul 29 at 12:23
Yes, you are right - I have lost the factor $z^max(i,j)$. Corrected. Thank you!
– Leox
Jul 29 at 12:40
Yes, you are right - I have lost the factor $z^max(i,j)$. Corrected. Thank you!
– Leox
Jul 29 at 12:40
add a comment |Â
1 Answer
1
active
oldest
votes
up vote
2
down vote
accepted
Starting from the EGF
$$nbrace k = n! [z^n] frac(exp(z)-1)^kk!$$
which is the labeled combinatorial class
$$deftextsc#1dosc#1csod
defdosc#1#2csodrm #1small #2
textscSET_=k(textscSET_ge 1(mathcalZ))$$
we obtain
$$frac1k! n! [z^n]
sum_p=0^k kchoose p (-1)^k-p exp(pz)
= frac1k!
sum_p=0^k kchoose p (-1)^k-p p^n.$$
Documenting the choice of variables we also write
$$nbrace m = frac1m!
sum_q=0^m mchoose q (-1)^m-q q^n.$$
We thus have for
$$G_k,m(z) = sum_nge 0 nbrace k nbrace m z^n
\ = sum_nge 0 z^n
frac1k!
sum_p=0^k kchoose p (-1)^k-p p^n
frac1m!
sum_q=0^m mchoose q (-1)^m-q q^n
\ =
sum_nge 0 z^n frac1k!
(-1)^k [[ n = 0 ]]
frac1m!
sum_q=0^m mchoose q (-1)^m-q q^n
\ + sum_nge 0 z^n
frac1k!
sum_p=1^k kchoose p (-1)^k-p p^n
frac1m!
sum_q=0^m mchoose q (-1)^m-q q^n.$$
The first term vanishes here and we continue with
$$sum_nge 0 z^n
frac1k!
sum_p=1^k kchoose p (-1)^k-p p^n
frac1m! (-1)^m [[ n = 0 ]]
\ + sum_nge 0 z^n
frac1k!
sum_p=1^k kchoose p (-1)^k-p p^n
frac1m!
sum_q=1^m mchoose q (-1)^m-q q^n.$$
The first term again has a closed form and we obtain
$$frac(-1)^k+m+1k!times m!
+ sum_nge 0 z^n
frac1k!
sum_p=1^k kchoose p (-1)^k-p p^n
frac1m!
sum_q=1^m mchoose q (-1)^m-q q^n.$$
We continue with the triple sum:
$$frac1k!
sum_p=1^k kchoose p (-1)^k-p
frac1m!
sum_q=1^m mchoose q (-1)^m-q
sum_nge 0 z^n (pq)^n
\ = frac1k!
sum_p=1^k kchoose p (-1)^k-p
frac1m!
sum_q=1^m mchoose q (-1)^m-q
frac11-pqz.$$
Re-writing we find
$$frac1k! frac1m!
sum_l=1^km frac11-lz
sum_l wedge ple k wedge l/p le m
kchoose p (-1)^k-p
mchoose l/p (-1)^m-l/p.$$
Simplifying and collecting everything now yields
$$frac(-1)^k+m+1k!times m!
+ frac(-1)^k+mk! times m!
sum_l=1^km frac11-lz
sum_l wedge ple k wedge l/p le m
(-1)^p+l/p
kchoose p mchoose l/p.$$
The binomial coefficients control the range, being zero when $pgt k$
and / or $l/p gt m$ and we may simplify even more to get
$$bbox[5px,border:2px solid #00A000]
-frac(-1)^k+mk!times m!
+ frac(-1)^k+mk! times m!
sum_l=1^km frac11-lz
sum_l (-1)^p+l/p
kchoose p mchoose l/p.$$
We have computed the partial fraction decomposition of the desired
generating function $G_k,m(z).$ Observe that this will confirm the
three formulae provided by OP.
answered Jul 29 at 14:35
Marko Riedel
36.4k333107
So, for some $l$ the sum $sum_pmid l$ is equals to $0$ and the corresponding term $dfrac11-lz$ is absent?
– Leox
Jul 29 at 15:33
This is correct, e.g. there is no $1/(1-7z)$ in $G_2,4(z)$ and no $1/(1-5z)$ in $G_3,3(z)$ etc.
– Marko Riedel
Jul 29 at 15:37
is it possyble to describe exactly for what $l$ such sum is equals to $0?$
– Leox
Jul 29 at 15:40
1
According to the derivation if $l$ does not factor into $p$ and $l/p$ where $ple k$ and $l/ple m$, then there is no term $1/(1-lz)$ in the sum. Compare $1/(1-5z)$ in $G_3,3(z)$ with $5=1times 5 = 5times 1.$
– Marko Riedel
Jul 29 at 15:42
add a comment |Â
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
2
down vote
accepted
Starting from the EGF
$$nbrace k = n! [z^n] frac(exp(z)-1)^kk!$$
which is the labeled combinatorial class
$$deftextsc#1dosc#1csod
defdosc#1#2csodrm #1small #2
textscSET_=k(textscSET_ge 1(mathcalZ))$$
we obtain
$$frac1k! n! [z^n]
sum_p=0^k kchoose p (-1)^k-p exp(pz)
= frac1k!
sum_p=0^k kchoose p (-1)^k-p p^n.$$
Documenting the choice of variables we also write
$$nbrace m = frac1m!
sum_q=0^m mchoose q (-1)^m-q q^n.$$
We thus have for
$$G_k,m(z) = sum_nge 0 nbrace k nbrace m z^n
\ = sum_nge 0 z^n
frac1k!
sum_p=0^k kchoose p (-1)^k-p p^n
frac1m!
sum_q=0^m mchoose q (-1)^m-q q^n
\ =
sum_nge 0 z^n frac1k!
(-1)^k [[ n = 0 ]]
frac1m!
sum_q=0^m mchoose q (-1)^m-q q^n
\ + sum_nge 0 z^n
frac1k!
sum_p=1^k kchoose p (-1)^k-p p^n
frac1m!
sum_q=0^m mchoose q (-1)^m-q q^n.$$
The first term vanishes here and we continue with
$$sum_nge 0 z^n
frac1k!
sum_p=1^k kchoose p (-1)^k-p p^n
frac1m! (-1)^m [[ n = 0 ]]
\ + sum_nge 0 z^n
frac1k!
sum_p=1^k kchoose p (-1)^k-p p^n
frac1m!
sum_q=1^m mchoose q (-1)^m-q q^n.$$
The first term again has a closed form and we obtain
$$frac(-1)^k+m+1k!times m!
+ sum_nge 0 z^n
frac1k!
sum_p=1^k kchoose p (-1)^k-p p^n
frac1m!
sum_q=1^m mchoose q (-1)^m-q q^n.$$
We continue with the triple sum:
$$frac1k!
sum_p=1^k kchoose p (-1)^k-p
frac1m!
sum_q=1^m mchoose q (-1)^m-q
sum_nge 0 z^n (pq)^n
\ = frac1k!
sum_p=1^k kchoose p (-1)^k-p
frac1m!
sum_q=1^m mchoose q (-1)^m-q
frac11-pqz.$$
Re-writing we find
$$frac1k! frac1m!
sum_l=1^km frac11-lz
sum_l wedge ple k wedge l/p le m
kchoose p (-1)^k-p
mchoose l/p (-1)^m-l/p.$$
Simplifying and collecting everything now yields
$$frac(-1)^k+m+1k!times m!
+ frac(-1)^k+mk! times m!
sum_l=1^km frac11-lz
sum_l wedge ple k wedge l/p le m
(-1)^p+l/p
kchoose p mchoose l/p.$$
The binomial coefficients control the range, being zero when $pgt k$
and / or $l/p gt m$ and we may simplify even more to get
$$bbox[5px,border:2px solid #00A000]
-frac(-1)^k+mk!times m!
+ frac(-1)^k+mk! times m!
sum_l=1^km frac11-lz
sum_l (-1)^p+l/p
kchoose p mchoose l/p.$$
We have computed the partial fraction decomposition of the desired
generating function $G_k,m(z).$ Observe that this will confirm the
three formulae provided by OP.
answered Jul 29 at 14:35
Marko Riedel
36.4k333107
So, for some $l$ the sum $sum_pmid l$ is equals to $0$ and the corresponding term $dfrac11-lz$ is absent?
– Leox
Jul 29 at 15:33
This is correct, e.g. there is no $1/(1-7z)$ in $G_2,4(z)$ and no $1/(1-5z)$ in $G_3,3(z)$ etc.
– Marko Riedel
Jul 29 at 15:37
is it possyble to describe exactly for what $l$ such sum is equals to $0?$
– Leox
Jul 29 at 15:40
1
According to the derivation if $l$ does not factor into $p$ and $l/p$ where $ple k$ and $l/ple m$, then there is no term $1/(1-lz)$ in the sum. Compare $1/(1-5z)$ in $G_3,3(z)$ with $5=1times 5 = 5times 1.$
– Marko Riedel
Jul 29 at 15:42
add a comment |Â
up vote
2
down vote
accepted
Starting from the EGF
$$nbrace k = n! [z^n] frac(exp(z)-1)^kk!$$
which is the labeled combinatorial class
$$deftextsc#1dosc#1csod
defdosc#1#2csodrm #1small #2
textscSET_=k(textscSET_ge 1(mathcalZ))$$
we obtain
$$frac1k! n! [z^n]
sum_p=0^k kchoose p (-1)^k-p exp(pz)
= frac1k!
sum_p=0^k kchoose p (-1)^k-p p^n.$$
Documenting the choice of variables we also write
$$nbrace m = frac1m!
sum_q=0^m mchoose q (-1)^m-q q^n.$$
We thus have for
$$G_k,m(z) = sum_nge 0 nbrace k nbrace m z^n
\ = sum_nge 0 z^n
frac1k!
sum_p=0^k kchoose p (-1)^k-p p^n
frac1m!
sum_q=0^m mchoose q (-1)^m-q q^n
\ =
sum_nge 0 z^n frac1k!
(-1)^k [[ n = 0 ]]
frac1m!
sum_q=0^m mchoose q (-1)^m-q q^n
\ + sum_nge 0 z^n
frac1k!
sum_p=1^k kchoose p (-1)^k-p p^n
frac1m!
sum_q=0^m mchoose q (-1)^m-q q^n.$$
The first term vanishes here and we continue with
$$sum_nge 0 z^n
frac1k!
sum_p=1^k kchoose p (-1)^k-p p^n
frac1m! (-1)^m [[ n = 0 ]]
\ + sum_nge 0 z^n
frac1k!
sum_p=1^k kchoose p (-1)^k-p p^n
frac1m!
sum_q=1^m mchoose q (-1)^m-q q^n.$$
The first term again has a closed form and we obtain
$$frac(-1)^k+m+1k!times m!
+ sum_nge 0 z^n
frac1k!
sum_p=1^k kchoose p (-1)^k-p p^n
frac1m!
sum_q=1^m mchoose q (-1)^m-q q^n.$$
We continue with the triple sum:
$$frac1k!
sum_p=1^k kchoose p (-1)^k-p
frac1m!
sum_q=1^m mchoose q (-1)^m-q
sum_nge 0 z^n (pq)^n
\ = frac1k!
sum_p=1^k kchoose p (-1)^k-p
frac1m!
sum_q=1^m mchoose q (-1)^m-q
frac11-pqz.$$
Re-writing we find
$$frac1k! frac1m!
sum_l=1^km frac11-lz
sum_l wedge ple k wedge l/p le m
kchoose p (-1)^k-p
mchoose l/p (-1)^m-l/p.$$
Simplifying and collecting everything now yields
$$frac(-1)^k+m+1k!times m!
+ frac(-1)^k+mk! times m!
sum_l=1^km frac11-lz
sum_l wedge ple k wedge l/p le m
(-1)^p+l/p
kchoose p mchoose l/p.$$
The binomial coefficients control the range, being zero when $pgt k$
and / or $l/p gt m$ and we may simplify even more to get
$$bbox[5px,border:2px solid #00A000]
-frac(-1)^k+mk!times m!
+ frac(-1)^k+mk! times m!
sum_l=1^km frac11-lz
sum_l (-1)^p+l/p
kchoose p mchoose l/p.$$
We have computed the partial fraction decomposition of the desired
generating function $G_k,m(z).$ Observe that this will confirm the
three formulae provided by OP.
answered Jul 29 at 14:35
Marko Riedel
36.4k333107
So, for some $l$ the sum $sum_pmid l$ is equals to $0$ and the corresponding term $dfrac11-lz$ is absent?
– Leox
Jul 29 at 15:33
This is correct, e.g. there is no $1/(1-7z)$ in $G_2,4(z)$ and no $1/(1-5z)$ in $G_3,3(z)$ etc.
– Marko Riedel
Jul 29 at 15:37
is it possyble to describe exactly for what $l$ such sum is equals to $0?$
– Leox
Jul 29 at 15:40
1
According to the derivation if $l$ does not factor into $p$ and $l/p$ where $ple k$ and $l/ple m$, then there is no term $1/(1-lz)$ in the sum. Compare $1/(1-5z)$ in $G_3,3(z)$ with $5=1times 5 = 5times 1.$
– Marko Riedel
Jul 29 at 15:42
add a comment |Â
up vote
2
down vote
accepted
up vote
2
down vote
accepted
Starting from the EGF
$$nbrace k = n! [z^n] frac(exp(z)-1)^kk!$$
which is the labeled combinatorial class
$$deftextsc#1dosc#1csod
defdosc#1#2csodrm #1small #2
textscSET_=k(textscSET_ge 1(mathcalZ))$$
we obtain
$$frac1k! n! [z^n]
sum_p=0^k kchoose p (-1)^k-p exp(pz)
= frac1k!
sum_p=0^k kchoose p (-1)^k-p p^n.$$
Documenting the choice of variables we also write
$$nbrace m = frac1m!
sum_q=0^m mchoose q (-1)^m-q q^n.$$
We thus have for
$$G_k,m(z) = sum_nge 0 nbrace k nbrace m z^n
\ = sum_nge 0 z^n
frac1k!
sum_p=0^k kchoose p (-1)^k-p p^n
frac1m!
sum_q=0^m mchoose q (-1)^m-q q^n
\ =
sum_nge 0 z^n frac1k!
(-1)^k [[ n = 0 ]]
frac1m!
sum_q=0^m mchoose q (-1)^m-q q^n
\ + sum_nge 0 z^n
frac1k!
sum_p=1^k kchoose p (-1)^k-p p^n
frac1m!
sum_q=0^m mchoose q (-1)^m-q q^n.$$
The first term vanishes here and we continue with
$$sum_nge 0 z^n
frac1k!
sum_p=1^k kchoose p (-1)^k-p p^n
frac1m! (-1)^m [[ n = 0 ]]
\ + sum_nge 0 z^n
frac1k!
sum_p=1^k kchoose p (-1)^k-p p^n
frac1m!
sum_q=1^m mchoose q (-1)^m-q q^n.$$
The first term again has a closed form and we obtain
$$frac(-1)^k+m+1k!times m!
+ sum_nge 0 z^n
frac1k!
sum_p=1^k kchoose p (-1)^k-p p^n
frac1m!
sum_q=1^m mchoose q (-1)^m-q q^n.$$
We continue with the triple sum:
$$frac1k!
sum_p=1^k kchoose p (-1)^k-p
frac1m!
sum_q=1^m mchoose q (-1)^m-q
sum_nge 0 z^n (pq)^n
\ = frac1k!
sum_p=1^k kchoose p (-1)^k-p
frac1m!
sum_q=1^m mchoose q (-1)^m-q
frac11-pqz.$$
Re-writing we find
$$frac1k! frac1m!
sum_l=1^km frac11-lz
sum_l wedge ple k wedge l/p le m
kchoose p (-1)^k-p
mchoose l/p (-1)^m-l/p.$$
Simplifying and collecting everything now yields
$$frac(-1)^k+m+1k!times m!
+ frac(-1)^k+mk! times m!
sum_l=1^km frac11-lz
sum_l wedge ple k wedge l/p le m
(-1)^p+l/p
kchoose p mchoose l/p.$$
The binomial coefficients control the range, being zero when $pgt k$
and / or $l/p gt m$ and we may simplify even more to get
$$bbox[5px,border:2px solid #00A000]
-frac(-1)^k+mk!times m!
+ frac(-1)^k+mk! times m!
sum_l=1^km frac11-lz
sum_l (-1)^p+l/p
kchoose p mchoose l/p.$$
We have computed the partial fraction decomposition of the desired
generating function $G_k,m(z).$ Observe that this will confirm the
three formulae provided by OP.
answered Jul 29 at 14:35
Marko Riedel
36.4k333107
Starting from the EGF
$$nbrace k = n! [z^n] frac(exp(z)-1)^kk!$$
which is the labeled combinatorial class
$$deftextsc#1dosc#1csod
defdosc#1#2csodrm #1small #2
textscSET_=k(textscSET_ge 1(mathcalZ))$$
we obtain
$$frac1k! n! [z^n]
sum_p=0^k kchoose p (-1)^k-p exp(pz)
= frac1k!
sum_p=0^k kchoose p (-1)^k-p p^n.$$
Documenting the choice of variables we also write
$$nbrace m = frac1m!
sum_q=0^m mchoose q (-1)^m-q q^n.$$
We thus have for
$$G_k,m(z) = sum_nge 0 nbrace k nbrace m z^n
\ = sum_nge 0 z^n
frac1k!
sum_p=0^k kchoose p (-1)^k-p p^n
frac1m!
sum_q=0^m mchoose q (-1)^m-q q^n
\ =
sum_nge 0 z^n frac1k!
(-1)^k [[ n = 0 ]]
frac1m!
sum_q=0^m mchoose q (-1)^m-q q^n
\ + sum_nge 0 z^n
frac1k!
sum_p=1^k kchoose p (-1)^k-p p^n
frac1m!
sum_q=0^m mchoose q (-1)^m-q q^n.$$
The first term vanishes here and we continue with
$$sum_nge 0 z^n
frac1k!
sum_p=1^k kchoose p (-1)^k-p p^n
frac1m! (-1)^m [[ n = 0 ]]
\ + sum_nge 0 z^n
frac1k!
sum_p=1^k kchoose p (-1)^k-p p^n
frac1m!
sum_q=1^m mchoose q (-1)^m-q q^n.$$
The first term again has a closed form and we obtain
$$frac(-1)^k+m+1k!times m!
+ sum_nge 0 z^n
frac1k!
sum_p=1^k kchoose p (-1)^k-p p^n
frac1m!
sum_q=1^m mchoose q (-1)^m-q q^n.$$
We continue with the triple sum:
$$frac1k!
sum_p=1^k kchoose p (-1)^k-p
frac1m!
sum_q=1^m mchoose q (-1)^m-q
sum_nge 0 z^n (pq)^n
\ = frac1k!
sum_p=1^k kchoose p (-1)^k-p
frac1m!
sum_q=1^m mchoose q (-1)^m-q
frac11-pqz.$$
Re-writing we find
$$frac1k! frac1m!
sum_l=1^km frac11-lz
sum_l wedge ple k wedge l/p le m
kchoose p (-1)^k-p
mchoose l/p (-1)^m-l/p.$$
Simplifying and collecting everything now yields
$$frac(-1)^k+m+1k!times m!
+ frac(-1)^k+mk! times m!
sum_l=1^km frac11-lz
sum_l wedge ple k wedge l/p le m
(-1)^p+l/p
kchoose p mchoose l/p.$$
The binomial coefficients control the range, being zero when $pgt k$
and / or $l/p gt m$ and we may simplify even more to get
$$bbox[5px,border:2px solid #00A000]
-frac(-1)^k+mk!times m!
+ frac(-1)^k+mk! times m!
sum_l=1^km frac11-lz
sum_l (-1)^p+l/p
kchoose p mchoose l/p.$$
We have computed the partial fraction decomposition of the desired
generating function $G_k,m(z).$ Observe that this will confirm the
three formulae provided by OP.
answered Jul 29 at 14:35
Marko Riedel
36.4k333107
edited Jul 29 at 16:33
answered Jul 29 at 14:35
Marko Riedel
36.4k333107
answered Jul 29 at 14:35
Marko Riedel
36.4k333107
answered Jul 29 at 14:35
Marko Riedel
36.4k333107
36.4k333107
So, for some $l$ the sum $sum_pmid l$ is equals to $0$ and the corresponding term $dfrac11-lz$ is absent?
– Leox
Jul 29 at 15:33
This is correct, e.g. there is no $1/(1-7z)$ in $G_2,4(z)$ and no $1/(1-5z)$ in $G_3,3(z)$ etc.
– Marko Riedel
Jul 29 at 15:37
is it possyble to describe exactly for what $l$ such sum is equals to $0?$
– Leox
Jul 29 at 15:40
1
According to the derivation if $l$ does not factor into $p$ and $l/p$ where $ple k$ and $l/ple m$, then there is no term $1/(1-lz)$ in the sum. Compare $1/(1-5z)$ in $G_3,3(z)$ with $5=1times 5 = 5times 1.$
– Marko Riedel
Jul 29 at 15:42
add a comment |Â
So, for some $l$ the sum $sum_pmid l$ is equals to $0$ and the corresponding term $dfrac11-lz$ is absent?
– Leox
Jul 29 at 15:33
This is correct, e.g. there is no $1/(1-7z)$ in $G_2,4(z)$ and no $1/(1-5z)$ in $G_3,3(z)$ etc.
– Marko Riedel
Jul 29 at 15:37
is it possyble to describe exactly for what $l$ such sum is equals to $0?$
– Leox
Jul 29 at 15:40
1
According to the derivation if $l$ does not factor into $p$ and $l/p$ where $ple k$ and $l/ple m$, then there is no term $1/(1-lz)$ in the sum. Compare $1/(1-5z)$ in $G_3,3(z)$ with $5=1times 5 = 5times 1.$
– Marko Riedel
Jul 29 at 15:42
So, for some $l$ the sum $sum_pmid l$ is equals to $0$ and the corresponding term $dfrac11-lz$ is absent?
– Leox
Jul 29 at 15:33
So, for some $l$ the sum $sum_pmid l$ is equals to $0$ and the corresponding term $dfrac11-lz$ is absent?
– Leox
Jul 29 at 15:33
This is correct, e.g. there is no $1/(1-7z)$ in $G_2,4(z)$ and no $1/(1-5z)$ in $G_3,3(z)$ etc.
– Marko Riedel
Jul 29 at 15:37
This is correct, e.g. there is no $1/(1-7z)$ in $G_2,4(z)$ and no $1/(1-5z)$ in $G_3,3(z)$ etc.
– Marko Riedel
Jul 29 at 15:37
is it possyble to describe exactly for what $l$ such sum is equals to $0?$
– Leox
Jul 29 at 15:40
is it possyble to describe exactly for what $l$ such sum is equals to $0?$
– Leox
Jul 29 at 15:40
1
1
According to the derivation if $l$ does not factor into $p$ and $l/p$ where $ple k$ and $l/ple m$, then there is no term $1/(1-lz)$ in the sum. Compare $1/(1-5z)$ in $G_3,3(z)$ with $5=1times 5 = 5times 1.$
– Marko Riedel
Jul 29 at 15:42
According to the derivation if $l$ does not factor into $p$ and $l/p$ where $ple k$ and $l/ple m$, then there is no term $1/(1-lz)$ in the sum. Compare $1/(1-5z)$ in $G_3,3(z)$ with $5=1times 5 = 5times 1.$
– Marko Riedel
Jul 29 at 15:42
add a comment |Â
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1
Perhaps you're using a different convention, but AFAIK $S(p,j)=0$ for $p < j$. As a result you're missing some factors of $z$,
– Robert Israel
Jul 29 at 12:23
Yes, you are right - I have lost the factor $z^max(i,j)$. Corrected. Thank you!
– Leox
Jul 29 at 12:40