Mixing
Closed form for product of Stirling numbers of the second kind
$begingroup$
What does the following expression evaluate to:
begin{equation}
sumlimits_{k=1}^n dbinom{n}{k} cdot k! begin{Bmatrix} n \ k end{Bmatrix} cdot k! begin{Bmatrix} n \ k end{Bmatrix}
end{equation}
We know that $k! begin{Bmatrix} n \ k end{Bmatrix} = n![x^n]:(e^x-1)^k$, where $[x^k]:f(x)$ represents the coefficient of $x^k$ in the power series for $f(x)$. I was wondering if squaring $left(text{i.e., } k! begin{Bmatrix} n \ k end{Bmatrix} cdot k! begin{Bmatrix} n \ k end{Bmatrix}right)$ takes us to a different power series or just to a different coefficient in the same power series? I am looking for some clean closed form. A related expression:
begin{equation}
sumlimits_{k=1}^n dbinom{n}{k} cdot k! begin{Bmatrix} n \ k end{Bmatrix}
end{equation}
is proven to be equal to $n^n$ in this answer https://math.stackexchange.com/q/3076350.
Note: The series $1,6,147,6940,536405,62352066, dots$ is not on oeis.org
combinatorics power-series binomial-coefficients stirling-numbers
asked Jan 21 at 14:52
SinghSingh
454
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add a comment |
$begingroup$
What does the following expression evaluate to:
begin{equation}
sumlimits_{k=1}^n dbinom{n}{k} cdot k! begin{Bmatrix} n \ k end{Bmatrix} cdot k! begin{Bmatrix} n \ k end{Bmatrix}
end{equation}
We know that $k! begin{Bmatrix} n \ k end{Bmatrix} = n![x^n]:(e^x-1)^k$, where $[x^k]:f(x)$ represents the coefficient of $x^k$ in the power series for $f(x)$. I was wondering if squaring $left(text{i.e., } k! begin{Bmatrix} n \ k end{Bmatrix} cdot k! begin{Bmatrix} n \ k end{Bmatrix}right)$ takes us to a different power series or just to a different coefficient in the same power series? I am looking for some clean closed form. A related expression:
begin{equation}
sumlimits_{k=1}^n dbinom{n}{k} cdot k! begin{Bmatrix} n \ k end{Bmatrix}
end{equation}
is proven to be equal to $n^n$ in this answer https://math.stackexchange.com/q/3076350.
Note: The series $1,6,147,6940,536405,62352066, dots$ is not on oeis.org
combinatorics power-series binomial-coefficients stirling-numbers
asked Jan 21 at 14:52
SinghSingh
454
$endgroup$
add a comment |
$begingroup$
What does the following expression evaluate to:
begin{equation}
sumlimits_{k=1}^n dbinom{n}{k} cdot k! begin{Bmatrix} n \ k end{Bmatrix} cdot k! begin{Bmatrix} n \ k end{Bmatrix}
end{equation}
We know that $k! begin{Bmatrix} n \ k end{Bmatrix} = n![x^n]:(e^x-1)^k$, where $[x^k]:f(x)$ represents the coefficient of $x^k$ in the power series for $f(x)$. I was wondering if squaring $left(text{i.e., } k! begin{Bmatrix} n \ k end{Bmatrix} cdot k! begin{Bmatrix} n \ k end{Bmatrix}right)$ takes us to a different power series or just to a different coefficient in the same power series? I am looking for some clean closed form. A related expression:
begin{equation}
sumlimits_{k=1}^n dbinom{n}{k} cdot k! begin{Bmatrix} n \ k end{Bmatrix}
end{equation}
is proven to be equal to $n^n$ in this answer https://math.stackexchange.com/q/3076350.
Note: The series $1,6,147,6940,536405,62352066, dots$ is not on oeis.org
combinatorics power-series binomial-coefficients stirling-numbers
asked Jan 21 at 14:52
SinghSingh
454
$endgroup$
What does the following expression evaluate to:
begin{equation}
sumlimits_{k=1}^n dbinom{n}{k} cdot k! begin{Bmatrix} n \ k end{Bmatrix} cdot k! begin{Bmatrix} n \ k end{Bmatrix}
end{equation}
We know that $k! begin{Bmatrix} n \ k end{Bmatrix} = n![x^n]:(e^x-1)^k$, where $[x^k]:f(x)$ represents the coefficient of $x^k$ in the power series for $f(x)$. I was wondering if squaring $left(text{i.e., } k! begin{Bmatrix} n \ k end{Bmatrix} cdot k! begin{Bmatrix} n \ k end{Bmatrix}right)$ takes us to a different power series or just to a different coefficient in the same power series? I am looking for some clean closed form. A related expression:
begin{equation}
sumlimits_{k=1}^n dbinom{n}{k} cdot k! begin{Bmatrix} n \ k end{Bmatrix}
end{equation}
is proven to be equal to $n^n$ in this answer https://math.stackexchange.com/q/3076350.
Note: The series $1,6,147,6940,536405,62352066, dots$ is not on oeis.org
combinatorics power-series binomial-coefficients stirling-numbers
combinatorics power-series binomial-coefficients stirling-numbers
asked Jan 21 at 14:52
SinghSingh
454
asked Jan 21 at 14:52
SinghSingh
454
edited Jan 21 at 18:06
Singh
asked Jan 21 at 14:52
SinghSingh
454
asked Jan 21 at 14:52
SinghSingh
454
asked Jan 21 at 14:52
SinghSingh
454
454
add a comment |
add a comment |
1 Answer
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oldest
votes
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Through the Eulerian Number of 1st kind $ leftlangle matrix{n cr mcr} rightrangle$ we get the following identities
$$
m!left{ matrix{ n cr m cr} right}
= sumlimits_{left( {0, le } right),k,left( { le ,n} right)} {
leftlangle matrix{n cr k cr} rightrangle left( matrix{ k cr n - m cr} right)}
quad Leftrightarrow quad left( {n - m} right)!left{ matrix{ n cr n - m cr} right}
= sumlimits_{left( {0, le } right),k,left( { le ,n} right)} {
leftlangle matrix{n cr k cr} rightrangle left( matrix{ k cr m cr} right)}
$$
therefrom we can write our sum as
$$
eqalign{
& S(n) = sumlimits_{left( {0, le } right),k,left( { le ,n} right)} {
left( matrix{ n cr k cr} right)k!left{ matrix{ n cr k cr} right}k!left{ matrix{ n cr k cr} right}} = cr
& = sumlimits_{left( {0, le } right),k,left( { le ,n} right)} {
left( matrix{ n cr k cr} right)k!left{ matrix{ n cr k cr} right}
sumlimits_{left( {0, le } right),j,left( { le ,n} right)}
{leftlangle matrix{ n cr j cr} rightrangle left( matrix{ j cr n - k cr} right)} } = cr
& = n!sumlimits_{left( {0, le } right),k,left( { le ,n} right)} {
left{ matrix{ n cr k cr} right}left( {sumlimits_{left( {0, le } right),j,left( { le ,n} right)}
{leftlangle matrix{ n cr j cr} rightrangle left( matrix{ j cr n - k cr} right)} } right){1 over {left( {n - k} right)!}}} cr}
$$
The e.g.f. for $S(n)$ then is
$$
sumlimits_{0, le ,n} {S(n){{x^{,n} } over {n!}}}
= sumlimits_{0, le ,n} {sumlimits_{left( {0, le } right),k,left( { le ,n} right)} {
left{ matrix{n cr k cr} right}x^{,k} left( {sumlimits_{left( {0, le } right),j,left( { le ,n} right)} {
leftlangle matrix{ n cr j cr} rightrangle binom{j}{n-k}
} } right){{x^{,n - k} } over {left( {n - k} right)!}}} }
$$
Indicating the Touchard polynomials as
$$
T_{,n} (x) = sumlimits_{left( {0, le } right),k,left( { le ,n} right)} {
left{ matrix{n cr k cr} right}x^{,k} }
= e^{, - ,x} sumlimits_{0, le ,k} {{{k^{,n} } over {k!}}x^{,k} }
$$
and the second polynomial as
$$
P_n (x) = sumlimits_{left( {0, le } right),k,left( { le ,n} right)} {left( {sumlimits_{left( {0, le } right),j,left( { le ,n} right)} {
leftlangle matrix{ n cr j cr} rightrangle left( matrix{ j cr k cr} right)} } right){{x^{,k} } over {k!}}}
= sumlimits_{left( {0, le } right),k,left( { le ,n} right)} {{{left( {n - k} right)!} over {k!}}left{ matrix{ n cr n - k cr} right}x^{,k} }
$$
we get
$$
S(n) = n!left[ {x^{,n} } right]left( {T_{,n} (x)P_{,n} (x)} right)
$$
Concerning the polynomial $P_n(x)$ let's evidence that, since
$$
{1 over {1 - yleft( {e^{,x} - 1} right)}} = sumlimits_{0, le ,j} {left( {e^{,x} - 1} right)^{,j} y^{,j} } quad
= sumlimits_{0, le ,k} {{{e^{,x,k} y^{,k} } over {left( {1 + y} right)^{,k + 1} }};}
= sumlimits_{0, le ,k} {sumlimits_{0, le ,j} {{{j!} over {k!}}left{ matrix{ k cr j cr} right}x^{,k} y^{,j} } }
$$
then denoting $P_{, n, , m}(x)$ as
$$
P_{n,,m} (x) = sumlimits_{left( {0, le } right),k,left( { le ,n} right)} {
left( {n - k} right)!left{ matrix{m cr n - k cr} right}{{x^{,k} } over {k!}}}
$$
we easily reach to
$$
eqalign{
& sumlimits_{0, le ,m} {sumlimits_{0, le ,n} {P_{n,,m} (x)y^{,n} {{z^{,m} } over {m!}}} }
= e^{,x,y} sumlimits_{0, le ,m} {sumlimits_{0, le ,n} {{{n!} over {m!}}left{ matrix{m cr n cr} right}y^{,n} z^{,m} } } = cr
& = {{e^{,x,y} } over {1 - yleft( {e^{,z} - 1} right)}} cr}
$$
so that we can define $P_n(x)$ in another way as
$$
P_n (x) = n!left[ {left( {yz} right)^n } right]left( {{{e^{,x,y} } over {1 - yleft( {e^{,z} - 1} right)}}} right)
$$
answered Jan 25 at 17:07
G CabG Cab
19.9k31340
$endgroup$
2
$begingroup$
thanks for the appreciation and .. bounty!
$endgroup$
– G Cab
Jan 27 at 15:50
2
$begingroup$
Thanks for your help, and @NoviceGeek for offering the bounty.
$endgroup$
– Singh
Jan 27 at 16:12
add a comment |
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1 Answer
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1 Answer
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$begingroup$
Through the Eulerian Number of 1st kind $ leftlangle matrix{n cr mcr} rightrangle$ we get the following identities
$$
m!left{ matrix{ n cr m cr} right}
= sumlimits_{left( {0, le } right),k,left( { le ,n} right)} {
leftlangle matrix{n cr k cr} rightrangle left( matrix{ k cr n - m cr} right)}
quad Leftrightarrow quad left( {n - m} right)!left{ matrix{ n cr n - m cr} right}
= sumlimits_{left( {0, le } right),k,left( { le ,n} right)} {
leftlangle matrix{n cr k cr} rightrangle left( matrix{ k cr m cr} right)}
$$
therefrom we can write our sum as
$$
eqalign{
& S(n) = sumlimits_{left( {0, le } right),k,left( { le ,n} right)} {
left( matrix{ n cr k cr} right)k!left{ matrix{ n cr k cr} right}k!left{ matrix{ n cr k cr} right}} = cr
& = sumlimits_{left( {0, le } right),k,left( { le ,n} right)} {
left( matrix{ n cr k cr} right)k!left{ matrix{ n cr k cr} right}
sumlimits_{left( {0, le } right),j,left( { le ,n} right)}
{leftlangle matrix{ n cr j cr} rightrangle left( matrix{ j cr n - k cr} right)} } = cr
& = n!sumlimits_{left( {0, le } right),k,left( { le ,n} right)} {
left{ matrix{ n cr k cr} right}left( {sumlimits_{left( {0, le } right),j,left( { le ,n} right)}
{leftlangle matrix{ n cr j cr} rightrangle left( matrix{ j cr n - k cr} right)} } right){1 over {left( {n - k} right)!}}} cr}
$$
The e.g.f. for $S(n)$ then is
$$
sumlimits_{0, le ,n} {S(n){{x^{,n} } over {n!}}}
= sumlimits_{0, le ,n} {sumlimits_{left( {0, le } right),k,left( { le ,n} right)} {
left{ matrix{n cr k cr} right}x^{,k} left( {sumlimits_{left( {0, le } right),j,left( { le ,n} right)} {
leftlangle matrix{ n cr j cr} rightrangle binom{j}{n-k}
} } right){{x^{,n - k} } over {left( {n - k} right)!}}} }
$$
Indicating the Touchard polynomials as
$$
T_{,n} (x) = sumlimits_{left( {0, le } right),k,left( { le ,n} right)} {
left{ matrix{n cr k cr} right}x^{,k} }
= e^{, - ,x} sumlimits_{0, le ,k} {{{k^{,n} } over {k!}}x^{,k} }
$$
and the second polynomial as
$$
P_n (x) = sumlimits_{left( {0, le } right),k,left( { le ,n} right)} {left( {sumlimits_{left( {0, le } right),j,left( { le ,n} right)} {
leftlangle matrix{ n cr j cr} rightrangle left( matrix{ j cr k cr} right)} } right){{x^{,k} } over {k!}}}
= sumlimits_{left( {0, le } right),k,left( { le ,n} right)} {{{left( {n - k} right)!} over {k!}}left{ matrix{ n cr n - k cr} right}x^{,k} }
$$
we get
$$
S(n) = n!left[ {x^{,n} } right]left( {T_{,n} (x)P_{,n} (x)} right)
$$
Concerning the polynomial $P_n(x)$ let's evidence that, since
$$
{1 over {1 - yleft( {e^{,x} - 1} right)}} = sumlimits_{0, le ,j} {left( {e^{,x} - 1} right)^{,j} y^{,j} } quad
= sumlimits_{0, le ,k} {{{e^{,x,k} y^{,k} } over {left( {1 + y} right)^{,k + 1} }};}
= sumlimits_{0, le ,k} {sumlimits_{0, le ,j} {{{j!} over {k!}}left{ matrix{ k cr j cr} right}x^{,k} y^{,j} } }
$$
then denoting $P_{, n, , m}(x)$ as
$$
P_{n,,m} (x) = sumlimits_{left( {0, le } right),k,left( { le ,n} right)} {
left( {n - k} right)!left{ matrix{m cr n - k cr} right}{{x^{,k} } over {k!}}}
$$
we easily reach to
$$
eqalign{
& sumlimits_{0, le ,m} {sumlimits_{0, le ,n} {P_{n,,m} (x)y^{,n} {{z^{,m} } over {m!}}} }
= e^{,x,y} sumlimits_{0, le ,m} {sumlimits_{0, le ,n} {{{n!} over {m!}}left{ matrix{m cr n cr} right}y^{,n} z^{,m} } } = cr
& = {{e^{,x,y} } over {1 - yleft( {e^{,z} - 1} right)}} cr}
$$
so that we can define $P_n(x)$ in another way as
$$
P_n (x) = n!left[ {left( {yz} right)^n } right]left( {{{e^{,x,y} } over {1 - yleft( {e^{,z} - 1} right)}}} right)
$$
answered Jan 25 at 17:07
G CabG Cab
19.9k31340
$endgroup$
2
$begingroup$
thanks for the appreciation and .. bounty!
$endgroup$
– G Cab
Jan 27 at 15:50
2
$begingroup$
Thanks for your help, and @NoviceGeek for offering the bounty.
$endgroup$
– Singh
Jan 27 at 16:12
add a comment |
$begingroup$
Through the Eulerian Number of 1st kind $ leftlangle matrix{n cr mcr} rightrangle$ we get the following identities
$$
m!left{ matrix{ n cr m cr} right}
= sumlimits_{left( {0, le } right),k,left( { le ,n} right)} {
leftlangle matrix{n cr k cr} rightrangle left( matrix{ k cr n - m cr} right)}
quad Leftrightarrow quad left( {n - m} right)!left{ matrix{ n cr n - m cr} right}
= sumlimits_{left( {0, le } right),k,left( { le ,n} right)} {
leftlangle matrix{n cr k cr} rightrangle left( matrix{ k cr m cr} right)}
$$
therefrom we can write our sum as
$$
eqalign{
& S(n) = sumlimits_{left( {0, le } right),k,left( { le ,n} right)} {
left( matrix{ n cr k cr} right)k!left{ matrix{ n cr k cr} right}k!left{ matrix{ n cr k cr} right}} = cr
& = sumlimits_{left( {0, le } right),k,left( { le ,n} right)} {
left( matrix{ n cr k cr} right)k!left{ matrix{ n cr k cr} right}
sumlimits_{left( {0, le } right),j,left( { le ,n} right)}
{leftlangle matrix{ n cr j cr} rightrangle left( matrix{ j cr n - k cr} right)} } = cr
& = n!sumlimits_{left( {0, le } right),k,left( { le ,n} right)} {
left{ matrix{ n cr k cr} right}left( {sumlimits_{left( {0, le } right),j,left( { le ,n} right)}
{leftlangle matrix{ n cr j cr} rightrangle left( matrix{ j cr n - k cr} right)} } right){1 over {left( {n - k} right)!}}} cr}
$$
The e.g.f. for $S(n)$ then is
$$
sumlimits_{0, le ,n} {S(n){{x^{,n} } over {n!}}}
= sumlimits_{0, le ,n} {sumlimits_{left( {0, le } right),k,left( { le ,n} right)} {
left{ matrix{n cr k cr} right}x^{,k} left( {sumlimits_{left( {0, le } right),j,left( { le ,n} right)} {
leftlangle matrix{ n cr j cr} rightrangle binom{j}{n-k}
} } right){{x^{,n - k} } over {left( {n - k} right)!}}} }
$$
Indicating the Touchard polynomials as
$$
T_{,n} (x) = sumlimits_{left( {0, le } right),k,left( { le ,n} right)} {
left{ matrix{n cr k cr} right}x^{,k} }
= e^{, - ,x} sumlimits_{0, le ,k} {{{k^{,n} } over {k!}}x^{,k} }
$$
and the second polynomial as
$$
P_n (x) = sumlimits_{left( {0, le } right),k,left( { le ,n} right)} {left( {sumlimits_{left( {0, le } right),j,left( { le ,n} right)} {
leftlangle matrix{ n cr j cr} rightrangle left( matrix{ j cr k cr} right)} } right){{x^{,k} } over {k!}}}
= sumlimits_{left( {0, le } right),k,left( { le ,n} right)} {{{left( {n - k} right)!} over {k!}}left{ matrix{ n cr n - k cr} right}x^{,k} }
$$
we get
$$
S(n) = n!left[ {x^{,n} } right]left( {T_{,n} (x)P_{,n} (x)} right)
$$
Concerning the polynomial $P_n(x)$ let's evidence that, since
$$
{1 over {1 - yleft( {e^{,x} - 1} right)}} = sumlimits_{0, le ,j} {left( {e^{,x} - 1} right)^{,j} y^{,j} } quad
= sumlimits_{0, le ,k} {{{e^{,x,k} y^{,k} } over {left( {1 + y} right)^{,k + 1} }};}
= sumlimits_{0, le ,k} {sumlimits_{0, le ,j} {{{j!} over {k!}}left{ matrix{ k cr j cr} right}x^{,k} y^{,j} } }
$$
then denoting $P_{, n, , m}(x)$ as
$$
P_{n,,m} (x) = sumlimits_{left( {0, le } right),k,left( { le ,n} right)} {
left( {n - k} right)!left{ matrix{m cr n - k cr} right}{{x^{,k} } over {k!}}}
$$
we easily reach to
$$
eqalign{
& sumlimits_{0, le ,m} {sumlimits_{0, le ,n} {P_{n,,m} (x)y^{,n} {{z^{,m} } over {m!}}} }
= e^{,x,y} sumlimits_{0, le ,m} {sumlimits_{0, le ,n} {{{n!} over {m!}}left{ matrix{m cr n cr} right}y^{,n} z^{,m} } } = cr
& = {{e^{,x,y} } over {1 - yleft( {e^{,z} - 1} right)}} cr}
$$
so that we can define $P_n(x)$ in another way as
$$
P_n (x) = n!left[ {left( {yz} right)^n } right]left( {{{e^{,x,y} } over {1 - yleft( {e^{,z} - 1} right)}}} right)
$$
answered Jan 25 at 17:07
G CabG Cab
19.9k31340
$endgroup$
2
$begingroup$
thanks for the appreciation and .. bounty!
$endgroup$
– G Cab
Jan 27 at 15:50
2
$begingroup$
Thanks for your help, and @NoviceGeek for offering the bounty.
$endgroup$
– Singh
Jan 27 at 16:12
add a comment |
$begingroup$
Through the Eulerian Number of 1st kind $ leftlangle matrix{n cr mcr} rightrangle$ we get the following identities
$$
m!left{ matrix{ n cr m cr} right}
= sumlimits_{left( {0, le } right),k,left( { le ,n} right)} {
leftlangle matrix{n cr k cr} rightrangle left( matrix{ k cr n - m cr} right)}
quad Leftrightarrow quad left( {n - m} right)!left{ matrix{ n cr n - m cr} right}
= sumlimits_{left( {0, le } right),k,left( { le ,n} right)} {
leftlangle matrix{n cr k cr} rightrangle left( matrix{ k cr m cr} right)}
$$
therefrom we can write our sum as
$$
eqalign{
& S(n) = sumlimits_{left( {0, le } right),k,left( { le ,n} right)} {
left( matrix{ n cr k cr} right)k!left{ matrix{ n cr k cr} right}k!left{ matrix{ n cr k cr} right}} = cr
& = sumlimits_{left( {0, le } right),k,left( { le ,n} right)} {
left( matrix{ n cr k cr} right)k!left{ matrix{ n cr k cr} right}
sumlimits_{left( {0, le } right),j,left( { le ,n} right)}
{leftlangle matrix{ n cr j cr} rightrangle left( matrix{ j cr n - k cr} right)} } = cr
& = n!sumlimits_{left( {0, le } right),k,left( { le ,n} right)} {
left{ matrix{ n cr k cr} right}left( {sumlimits_{left( {0, le } right),j,left( { le ,n} right)}
{leftlangle matrix{ n cr j cr} rightrangle left( matrix{ j cr n - k cr} right)} } right){1 over {left( {n - k} right)!}}} cr}
$$
The e.g.f. for $S(n)$ then is
$$
sumlimits_{0, le ,n} {S(n){{x^{,n} } over {n!}}}
= sumlimits_{0, le ,n} {sumlimits_{left( {0, le } right),k,left( { le ,n} right)} {
left{ matrix{n cr k cr} right}x^{,k} left( {sumlimits_{left( {0, le } right),j,left( { le ,n} right)} {
leftlangle matrix{ n cr j cr} rightrangle binom{j}{n-k}
} } right){{x^{,n - k} } over {left( {n - k} right)!}}} }
$$
Indicating the Touchard polynomials as
$$
T_{,n} (x) = sumlimits_{left( {0, le } right),k,left( { le ,n} right)} {
left{ matrix{n cr k cr} right}x^{,k} }
= e^{, - ,x} sumlimits_{0, le ,k} {{{k^{,n} } over {k!}}x^{,k} }
$$
and the second polynomial as
$$
P_n (x) = sumlimits_{left( {0, le } right),k,left( { le ,n} right)} {left( {sumlimits_{left( {0, le } right),j,left( { le ,n} right)} {
leftlangle matrix{ n cr j cr} rightrangle left( matrix{ j cr k cr} right)} } right){{x^{,k} } over {k!}}}
= sumlimits_{left( {0, le } right),k,left( { le ,n} right)} {{{left( {n - k} right)!} over {k!}}left{ matrix{ n cr n - k cr} right}x^{,k} }
$$
we get
$$
S(n) = n!left[ {x^{,n} } right]left( {T_{,n} (x)P_{,n} (x)} right)
$$
Concerning the polynomial $P_n(x)$ let's evidence that, since
$$
{1 over {1 - yleft( {e^{,x} - 1} right)}} = sumlimits_{0, le ,j} {left( {e^{,x} - 1} right)^{,j} y^{,j} } quad
= sumlimits_{0, le ,k} {{{e^{,x,k} y^{,k} } over {left( {1 + y} right)^{,k + 1} }};}
= sumlimits_{0, le ,k} {sumlimits_{0, le ,j} {{{j!} over {k!}}left{ matrix{ k cr j cr} right}x^{,k} y^{,j} } }
$$
then denoting $P_{, n, , m}(x)$ as
$$
P_{n,,m} (x) = sumlimits_{left( {0, le } right),k,left( { le ,n} right)} {
left( {n - k} right)!left{ matrix{m cr n - k cr} right}{{x^{,k} } over {k!}}}
$$
we easily reach to
$$
eqalign{
& sumlimits_{0, le ,m} {sumlimits_{0, le ,n} {P_{n,,m} (x)y^{,n} {{z^{,m} } over {m!}}} }
= e^{,x,y} sumlimits_{0, le ,m} {sumlimits_{0, le ,n} {{{n!} over {m!}}left{ matrix{m cr n cr} right}y^{,n} z^{,m} } } = cr
& = {{e^{,x,y} } over {1 - yleft( {e^{,z} - 1} right)}} cr}
$$
so that we can define $P_n(x)$ in another way as
$$
P_n (x) = n!left[ {left( {yz} right)^n } right]left( {{{e^{,x,y} } over {1 - yleft( {e^{,z} - 1} right)}}} right)
$$
answered Jan 25 at 17:07
G CabG Cab
19.9k31340
$endgroup$
Through the Eulerian Number of 1st kind $ leftlangle matrix{n cr mcr} rightrangle$ we get the following identities
$$
m!left{ matrix{ n cr m cr} right}
= sumlimits_{left( {0, le } right),k,left( { le ,n} right)} {
leftlangle matrix{n cr k cr} rightrangle left( matrix{ k cr n - m cr} right)}
quad Leftrightarrow quad left( {n - m} right)!left{ matrix{ n cr n - m cr} right}
= sumlimits_{left( {0, le } right),k,left( { le ,n} right)} {
leftlangle matrix{n cr k cr} rightrangle left( matrix{ k cr m cr} right)}
$$
therefrom we can write our sum as
$$
eqalign{
& S(n) = sumlimits_{left( {0, le } right),k,left( { le ,n} right)} {
left( matrix{ n cr k cr} right)k!left{ matrix{ n cr k cr} right}k!left{ matrix{ n cr k cr} right}} = cr
& = sumlimits_{left( {0, le } right),k,left( { le ,n} right)} {
left( matrix{ n cr k cr} right)k!left{ matrix{ n cr k cr} right}
sumlimits_{left( {0, le } right),j,left( { le ,n} right)}
{leftlangle matrix{ n cr j cr} rightrangle left( matrix{ j cr n - k cr} right)} } = cr
& = n!sumlimits_{left( {0, le } right),k,left( { le ,n} right)} {
left{ matrix{ n cr k cr} right}left( {sumlimits_{left( {0, le } right),j,left( { le ,n} right)}
{leftlangle matrix{ n cr j cr} rightrangle left( matrix{ j cr n - k cr} right)} } right){1 over {left( {n - k} right)!}}} cr}
$$
The e.g.f. for $S(n)$ then is
$$
sumlimits_{0, le ,n} {S(n){{x^{,n} } over {n!}}}
= sumlimits_{0, le ,n} {sumlimits_{left( {0, le } right),k,left( { le ,n} right)} {
left{ matrix{n cr k cr} right}x^{,k} left( {sumlimits_{left( {0, le } right),j,left( { le ,n} right)} {
leftlangle matrix{ n cr j cr} rightrangle binom{j}{n-k}
} } right){{x^{,n - k} } over {left( {n - k} right)!}}} }
$$
Indicating the Touchard polynomials as
$$
T_{,n} (x) = sumlimits_{left( {0, le } right),k,left( { le ,n} right)} {
left{ matrix{n cr k cr} right}x^{,k} }
= e^{, - ,x} sumlimits_{0, le ,k} {{{k^{,n} } over {k!}}x^{,k} }
$$
and the second polynomial as
$$
P_n (x) = sumlimits_{left( {0, le } right),k,left( { le ,n} right)} {left( {sumlimits_{left( {0, le } right),j,left( { le ,n} right)} {
leftlangle matrix{ n cr j cr} rightrangle left( matrix{ j cr k cr} right)} } right){{x^{,k} } over {k!}}}
= sumlimits_{left( {0, le } right),k,left( { le ,n} right)} {{{left( {n - k} right)!} over {k!}}left{ matrix{ n cr n - k cr} right}x^{,k} }
$$
we get
$$
S(n) = n!left[ {x^{,n} } right]left( {T_{,n} (x)P_{,n} (x)} right)
$$
Concerning the polynomial $P_n(x)$ let's evidence that, since
$$
{1 over {1 - yleft( {e^{,x} - 1} right)}} = sumlimits_{0, le ,j} {left( {e^{,x} - 1} right)^{,j} y^{,j} } quad
= sumlimits_{0, le ,k} {{{e^{,x,k} y^{,k} } over {left( {1 + y} right)^{,k + 1} }};}
= sumlimits_{0, le ,k} {sumlimits_{0, le ,j} {{{j!} over {k!}}left{ matrix{ k cr j cr} right}x^{,k} y^{,j} } }
$$
then denoting $P_{, n, , m}(x)$ as
$$
P_{n,,m} (x) = sumlimits_{left( {0, le } right),k,left( { le ,n} right)} {
left( {n - k} right)!left{ matrix{m cr n - k cr} right}{{x^{,k} } over {k!}}}
$$
we easily reach to
$$
eqalign{
& sumlimits_{0, le ,m} {sumlimits_{0, le ,n} {P_{n,,m} (x)y^{,n} {{z^{,m} } over {m!}}} }
= e^{,x,y} sumlimits_{0, le ,m} {sumlimits_{0, le ,n} {{{n!} over {m!}}left{ matrix{m cr n cr} right}y^{,n} z^{,m} } } = cr
& = {{e^{,x,y} } over {1 - yleft( {e^{,z} - 1} right)}} cr}
$$
so that we can define $P_n(x)$ in another way as
$$
P_n (x) = n!left[ {left( {yz} right)^n } right]left( {{{e^{,x,y} } over {1 - yleft( {e^{,z} - 1} right)}}} right)
$$
answered Jan 25 at 17:07
G CabG Cab
19.9k31340
edited Jan 25 at 21:23
answered Jan 25 at 17:07
G CabG Cab
19.9k31340
answered Jan 25 at 17:07
G CabG Cab
19.9k31340
answered Jan 25 at 17:07
G CabG Cab
19.9k31340
19.9k31340
2
$begingroup$
thanks for the appreciation and .. bounty!
$endgroup$
– G Cab
Jan 27 at 15:50
2
$begingroup$
Thanks for your help, and @NoviceGeek for offering the bounty.
$endgroup$
– Singh
Jan 27 at 16:12
add a comment |
2
$begingroup$
thanks for the appreciation and .. bounty!
$endgroup$
– G Cab
Jan 27 at 15:50
2
$begingroup$
Thanks for your help, and @NoviceGeek for offering the bounty.
$endgroup$
– Singh
Jan 27 at 16:12
2
2
$begingroup$
thanks for the appreciation and .. bounty!
$endgroup$
– G Cab
Jan 27 at 15:50
$begingroup$
thanks for the appreciation and .. bounty!
$endgroup$
– G Cab
Jan 27 at 15:50
2
2
$begingroup$
Thanks for your help, and @NoviceGeek for offering the bounty.
$endgroup$
– Singh
Jan 27 at 16:12
$begingroup$
Thanks for your help, and @NoviceGeek for offering the bounty.
$endgroup$
– Singh
Jan 27 at 16:12
add a comment |
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