Where can I find a proof, that $s(n+1,k+1) = sum_{i = 0}^{n} binom{i}{k}s(n,i)$?
$begingroup$
In our Combinatorics Script it is written, that
$$s_{n+1,k+1} = sum_{i = 0}^{n} binom{i}{k}s_{n,i}$$ for $n,k in mathbb{N}$.
The problem is that I can't find a combinatorial proof for that, not in the Script and not online.
I thought about looking at a set $S_{n,k}$ of the permutations of $[n]$, which are the products of exactly $k$ disjoint cycles, but that's just a thought...
The Stirling numbers of the first kind are defined recursively by:
And how can one give a bijection between a set of the cardinality $sum_{i=0}^{n} binom{i}{k} s_{n,i}$ and $S_{n+1, k+1}$?
Apparently defining such a transformation can be extracted from the following example. Does someone know how it's done?
$(1,3,8,7)(2,9,5)(4,12,10,6)(11,13) to (14,4,12,10,6,1,3,8,7)(2,9,5)(11,13)$
combinatorics transformation stirling-numbers
asked Dec 6 '18 at 7:14
NotEinsteinNotEinstein
2146
$endgroup$
add a comment |
$begingroup$
In our Combinatorics Script it is written, that
$$s_{n+1,k+1} = sum_{i = 0}^{n} binom{i}{k}s_{n,i}$$ for $n,k in mathbb{N}$.
The problem is that I can't find a combinatorial proof for that, not in the Script and not online.
I thought about looking at a set $S_{n,k}$ of the permutations of $[n]$, which are the products of exactly $k$ disjoint cycles, but that's just a thought...
The Stirling numbers of the first kind are defined recursively by:
And how can one give a bijection between a set of the cardinality $sum_{i=0}^{n} binom{i}{k} s_{n,i}$ and $S_{n+1, k+1}$?
Apparently defining such a transformation can be extracted from the following example. Does someone know how it's done?
$(1,3,8,7)(2,9,5)(4,12,10,6)(11,13) to (14,4,12,10,6,1,3,8,7)(2,9,5)(11,13)$
combinatorics transformation stirling-numbers
asked Dec 6 '18 at 7:14
NotEinsteinNotEinstein
2146
$endgroup$
$begingroup$
What is definition of $S(a,b)$? And of $s-{u,v}$ [lower case $s$ with subscripts] Or should the latter be also upper case as in title?
$endgroup$
– coffeemath
Dec 6 '18 at 7:16
$begingroup$
Are you sure these are second kind Stirling numbers? Your example contains permutations, not partitions.
$endgroup$
– Lord Shark the Unknown
Dec 6 '18 at 7:30
$begingroup$
@LordSharktheUnknown Sorry, you are right. It was the first kind. I edited it.
$endgroup$
– NotEinstein
Dec 6 '18 at 7:38
$begingroup$
Can you please give a reference to your script? Are you sure you have written all the correct correctly? Also, you're "thought" in the third line, isn't it the definition of Stirling numbers of the first kind (should've been lowercase s)?
$endgroup$
– Sri Krishna Sahoo
Dec 6 '18 at 9:38
add a comment |
$begingroup$
In our Combinatorics Script it is written, that
$$s_{n+1,k+1} = sum_{i = 0}^{n} binom{i}{k}s_{n,i}$$ for $n,k in mathbb{N}$.
The problem is that I can't find a combinatorial proof for that, not in the Script and not online.
I thought about looking at a set $S_{n,k}$ of the permutations of $[n]$, which are the products of exactly $k$ disjoint cycles, but that's just a thought...
The Stirling numbers of the first kind are defined recursively by:
And how can one give a bijection between a set of the cardinality $sum_{i=0}^{n} binom{i}{k} s_{n,i}$ and $S_{n+1, k+1}$?
Apparently defining such a transformation can be extracted from the following example. Does someone know how it's done?
$(1,3,8,7)(2,9,5)(4,12,10,6)(11,13) to (14,4,12,10,6,1,3,8,7)(2,9,5)(11,13)$
combinatorics transformation stirling-numbers
asked Dec 6 '18 at 7:14
NotEinsteinNotEinstein
2146
$endgroup$
In our Combinatorics Script it is written, that
$$s_{n+1,k+1} = sum_{i = 0}^{n} binom{i}{k}s_{n,i}$$ for $n,k in mathbb{N}$.
The problem is that I can't find a combinatorial proof for that, not in the Script and not online.
I thought about looking at a set $S_{n,k}$ of the permutations of $[n]$, which are the products of exactly $k$ disjoint cycles, but that's just a thought...
The Stirling numbers of the first kind are defined recursively by:
And how can one give a bijection between a set of the cardinality $sum_{i=0}^{n} binom{i}{k} s_{n,i}$ and $S_{n+1, k+1}$?
Apparently defining such a transformation can be extracted from the following example. Does someone know how it's done?
$(1,3,8,7)(2,9,5)(4,12,10,6)(11,13) to (14,4,12,10,6,1,3,8,7)(2,9,5)(11,13)$
combinatorics transformation stirling-numbers
combinatorics transformation stirling-numbers
asked Dec 6 '18 at 7:14
NotEinsteinNotEinstein
2146
asked Dec 6 '18 at 7:14
NotEinsteinNotEinstein
2146
edited Dec 6 '18 at 7:40
NotEinstein
asked Dec 6 '18 at 7:14
NotEinsteinNotEinstein
2146
asked Dec 6 '18 at 7:14
NotEinsteinNotEinstein
2146
asked Dec 6 '18 at 7:14
NotEinsteinNotEinstein
2146
2146
$begingroup$
What is definition of $S(a,b)$? And of $s-{u,v}$ [lower case $s$ with subscripts] Or should the latter be also upper case as in title?
$endgroup$
– coffeemath
Dec 6 '18 at 7:16
$begingroup$
Are you sure these are second kind Stirling numbers? Your example contains permutations, not partitions.
$endgroup$
– Lord Shark the Unknown
Dec 6 '18 at 7:30
$begingroup$
@LordSharktheUnknown Sorry, you are right. It was the first kind. I edited it.
$endgroup$
– NotEinstein
Dec 6 '18 at 7:38
$begingroup$
Can you please give a reference to your script? Are you sure you have written all the correct correctly? Also, you're "thought" in the third line, isn't it the definition of Stirling numbers of the first kind (should've been lowercase s)?
$endgroup$
– Sri Krishna Sahoo
Dec 6 '18 at 9:38
add a comment |
$begingroup$
What is definition of $S(a,b)$? And of $s-{u,v}$ [lower case $s$ with subscripts] Or should the latter be also upper case as in title?
$endgroup$
– coffeemath
Dec 6 '18 at 7:16
$begingroup$
Are you sure these are second kind Stirling numbers? Your example contains permutations, not partitions.
$endgroup$
– Lord Shark the Unknown
Dec 6 '18 at 7:30
$begingroup$
@LordSharktheUnknown Sorry, you are right. It was the first kind. I edited it.
$endgroup$
– NotEinstein
Dec 6 '18 at 7:38
$begingroup$
Can you please give a reference to your script? Are you sure you have written all the correct correctly? Also, you're "thought" in the third line, isn't it the definition of Stirling numbers of the first kind (should've been lowercase s)?
$endgroup$
– Sri Krishna Sahoo
Dec 6 '18 at 9:38
$begingroup$
What is definition of $S(a,b)$? And of $s-{u,v}$ [lower case $s$ with subscripts] Or should the latter be also upper case as in title?
$endgroup$
– coffeemath
Dec 6 '18 at 7:16
$begingroup$
What is definition of $S(a,b)$? And of $s-{u,v}$ [lower case $s$ with subscripts] Or should the latter be also upper case as in title?
$endgroup$
– coffeemath
Dec 6 '18 at 7:16
$begingroup$
Are you sure these are second kind Stirling numbers? Your example contains permutations, not partitions.
$endgroup$
– Lord Shark the Unknown
Dec 6 '18 at 7:30
$begingroup$
Are you sure these are second kind Stirling numbers? Your example contains permutations, not partitions.
$endgroup$
– Lord Shark the Unknown
Dec 6 '18 at 7:30
$begingroup$
@LordSharktheUnknown Sorry, you are right. It was the first kind. I edited it.
$endgroup$
– NotEinstein
Dec 6 '18 at 7:38
$begingroup$
@LordSharktheUnknown Sorry, you are right. It was the first kind. I edited it.
$endgroup$
– NotEinstein
Dec 6 '18 at 7:38
$begingroup$
Can you please give a reference to your script? Are you sure you have written all the correct correctly? Also, you're "thought" in the third line, isn't it the definition of Stirling numbers of the first kind (should've been lowercase s)?
$endgroup$
– Sri Krishna Sahoo
Dec 6 '18 at 9:38
$begingroup$
Can you please give a reference to your script? Are you sure you have written all the correct correctly? Also, you're "thought" in the third line, isn't it the definition of Stirling numbers of the first kind (should've been lowercase s)?
$endgroup$
– Sri Krishna Sahoo
Dec 6 '18 at 9:38
add a comment |
2 Answers
2
active
oldest
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$begingroup$
Let $A={a_1,cdots,a_m } $ be a finite set, that does not contain the element $n+1$. We begin by noting a one to one correspondence between permutations of this set and the conjugacy class of $A cup {n+1 }$.
Given an element $pi in S_A$ ($pi(a_i)=pi_{a_i}$) we form an element of $S_{A cup {n+1 }}$ by
begin{eqnarray*}
pi rightarrow (n+1,pi_{a_1}, cdots , pi_{a_m} ).
end{eqnarray*}
We shall now use this to establish the following formula
begin{eqnarray*}
sum_{i=k}^{n} {n brack i} binom{i}{k} = {n+1 brack k+1}.
end{eqnarray*}
Let $ sigma$ be an element of $S_n$ with $i$ cycles. Choose $k$ of these cycles and let $A$ be the set of elements in the other $i-k$ cycles. Now form an element of $S_{n+1}$ consisting of the $k$ chosen cycles and a cycle formed by the correspondence $S_A rightarrow S_{A cup {n+1 }}$ defined above. Now let $k$ vary over its admissable values and the formula is proven.
answered Dec 7 '18 at 1:36
Donald SplutterwitDonald Splutterwit
22.9k21446
$endgroup$
add a comment |
$begingroup$
You know what method works (almost) always to prove an identity about a sequence defined inductively? Induction.
Base case is trivial.
$$ s_{n+1,k+1}=s_{n,k}+ns_{n,k+1}= sum_{i = 0}^{n-1} binom{i}{k-1}s_{n-1,i}+ nsum_{i = 0}^{n-1} binom{i}{k}s_{n-1,i}$$$$= sum_{i = 0}^{n-1} (binom{i}{k-1}s_{n-1,i}+ binom{i}{k}s_{n-1,i}) +(n-1)sum_{i = 0}^{n-1} binom{i}{k}s_{n-1,i}$$$$=sum_{i = 0}^{n-1} binom{i+1}{k}s_{n-1,i} +(n-1)sum_{i = 0}^{n-1} binom{i}{k}s_{n-1,i}$$$$=sum_{i = 1}^{n} binom{i}{k}s_{n-1,i-1} +(n-1)sum_{i = 0}^{n-1} binom{i}{k}s_{n-1,i}$$$$=sum_{i = 0}^{n-1} binom{i}{k}s_{n,i}+binom{n}{k}=sum_{i = 0}^{n} binom{i}{k}s_{n,i}$$
$blacksquare$
answered Dec 6 '18 at 13:18
Anubhab GhosalAnubhab Ghosal
1,22319
$endgroup$
add a comment |
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2 Answers
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2 Answers
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$begingroup$
Let $A={a_1,cdots,a_m } $ be a finite set, that does not contain the element $n+1$. We begin by noting a one to one correspondence between permutations of this set and the conjugacy class of $A cup {n+1 }$.
Given an element $pi in S_A$ ($pi(a_i)=pi_{a_i}$) we form an element of $S_{A cup {n+1 }}$ by
begin{eqnarray*}
pi rightarrow (n+1,pi_{a_1}, cdots , pi_{a_m} ).
end{eqnarray*}
We shall now use this to establish the following formula
begin{eqnarray*}
sum_{i=k}^{n} {n brack i} binom{i}{k} = {n+1 brack k+1}.
end{eqnarray*}
Let $ sigma$ be an element of $S_n$ with $i$ cycles. Choose $k$ of these cycles and let $A$ be the set of elements in the other $i-k$ cycles. Now form an element of $S_{n+1}$ consisting of the $k$ chosen cycles and a cycle formed by the correspondence $S_A rightarrow S_{A cup {n+1 }}$ defined above. Now let $k$ vary over its admissable values and the formula is proven.
answered Dec 7 '18 at 1:36
Donald SplutterwitDonald Splutterwit
22.9k21446
$endgroup$
add a comment |
$begingroup$
Let $A={a_1,cdots,a_m } $ be a finite set, that does not contain the element $n+1$. We begin by noting a one to one correspondence between permutations of this set and the conjugacy class of $A cup {n+1 }$.
Given an element $pi in S_A$ ($pi(a_i)=pi_{a_i}$) we form an element of $S_{A cup {n+1 }}$ by
begin{eqnarray*}
pi rightarrow (n+1,pi_{a_1}, cdots , pi_{a_m} ).
end{eqnarray*}
We shall now use this to establish the following formula
begin{eqnarray*}
sum_{i=k}^{n} {n brack i} binom{i}{k} = {n+1 brack k+1}.
end{eqnarray*}
Let $ sigma$ be an element of $S_n$ with $i$ cycles. Choose $k$ of these cycles and let $A$ be the set of elements in the other $i-k$ cycles. Now form an element of $S_{n+1}$ consisting of the $k$ chosen cycles and a cycle formed by the correspondence $S_A rightarrow S_{A cup {n+1 }}$ defined above. Now let $k$ vary over its admissable values and the formula is proven.
answered Dec 7 '18 at 1:36
Donald SplutterwitDonald Splutterwit
22.9k21446
$endgroup$
add a comment |
$begingroup$
Let $A={a_1,cdots,a_m } $ be a finite set, that does not contain the element $n+1$. We begin by noting a one to one correspondence between permutations of this set and the conjugacy class of $A cup {n+1 }$.
Given an element $pi in S_A$ ($pi(a_i)=pi_{a_i}$) we form an element of $S_{A cup {n+1 }}$ by
begin{eqnarray*}
pi rightarrow (n+1,pi_{a_1}, cdots , pi_{a_m} ).
end{eqnarray*}
We shall now use this to establish the following formula
begin{eqnarray*}
sum_{i=k}^{n} {n brack i} binom{i}{k} = {n+1 brack k+1}.
end{eqnarray*}
Let $ sigma$ be an element of $S_n$ with $i$ cycles. Choose $k$ of these cycles and let $A$ be the set of elements in the other $i-k$ cycles. Now form an element of $S_{n+1}$ consisting of the $k$ chosen cycles and a cycle formed by the correspondence $S_A rightarrow S_{A cup {n+1 }}$ defined above. Now let $k$ vary over its admissable values and the formula is proven.
answered Dec 7 '18 at 1:36
Donald SplutterwitDonald Splutterwit
22.9k21446
$endgroup$
Let $A={a_1,cdots,a_m } $ be a finite set, that does not contain the element $n+1$. We begin by noting a one to one correspondence between permutations of this set and the conjugacy class of $A cup {n+1 }$.
Given an element $pi in S_A$ ($pi(a_i)=pi_{a_i}$) we form an element of $S_{A cup {n+1 }}$ by
begin{eqnarray*}
pi rightarrow (n+1,pi_{a_1}, cdots , pi_{a_m} ).
end{eqnarray*}
We shall now use this to establish the following formula
begin{eqnarray*}
sum_{i=k}^{n} {n brack i} binom{i}{k} = {n+1 brack k+1}.
end{eqnarray*}
Let $ sigma$ be an element of $S_n$ with $i$ cycles. Choose $k$ of these cycles and let $A$ be the set of elements in the other $i-k$ cycles. Now form an element of $S_{n+1}$ consisting of the $k$ chosen cycles and a cycle formed by the correspondence $S_A rightarrow S_{A cup {n+1 }}$ defined above. Now let $k$ vary over its admissable values and the formula is proven.
answered Dec 7 '18 at 1:36
Donald SplutterwitDonald Splutterwit
22.9k21446
answered Dec 7 '18 at 1:36
Donald SplutterwitDonald Splutterwit
22.9k21446
answered Dec 7 '18 at 1:36
Donald SplutterwitDonald Splutterwit
22.9k21446
answered Dec 7 '18 at 1:36
Donald SplutterwitDonald Splutterwit
22.9k21446
22.9k21446
add a comment |
add a comment |
$begingroup$
You know what method works (almost) always to prove an identity about a sequence defined inductively? Induction.
Base case is trivial.
$$ s_{n+1,k+1}=s_{n,k}+ns_{n,k+1}= sum_{i = 0}^{n-1} binom{i}{k-1}s_{n-1,i}+ nsum_{i = 0}^{n-1} binom{i}{k}s_{n-1,i}$$$$= sum_{i = 0}^{n-1} (binom{i}{k-1}s_{n-1,i}+ binom{i}{k}s_{n-1,i}) +(n-1)sum_{i = 0}^{n-1} binom{i}{k}s_{n-1,i}$$$$=sum_{i = 0}^{n-1} binom{i+1}{k}s_{n-1,i} +(n-1)sum_{i = 0}^{n-1} binom{i}{k}s_{n-1,i}$$$$=sum_{i = 1}^{n} binom{i}{k}s_{n-1,i-1} +(n-1)sum_{i = 0}^{n-1} binom{i}{k}s_{n-1,i}$$$$=sum_{i = 0}^{n-1} binom{i}{k}s_{n,i}+binom{n}{k}=sum_{i = 0}^{n} binom{i}{k}s_{n,i}$$
$blacksquare$
answered Dec 6 '18 at 13:18
Anubhab GhosalAnubhab Ghosal
1,22319
$endgroup$
add a comment |
$begingroup$
You know what method works (almost) always to prove an identity about a sequence defined inductively? Induction.
Base case is trivial.
$$ s_{n+1,k+1}=s_{n,k}+ns_{n,k+1}= sum_{i = 0}^{n-1} binom{i}{k-1}s_{n-1,i}+ nsum_{i = 0}^{n-1} binom{i}{k}s_{n-1,i}$$$$= sum_{i = 0}^{n-1} (binom{i}{k-1}s_{n-1,i}+ binom{i}{k}s_{n-1,i}) +(n-1)sum_{i = 0}^{n-1} binom{i}{k}s_{n-1,i}$$$$=sum_{i = 0}^{n-1} binom{i+1}{k}s_{n-1,i} +(n-1)sum_{i = 0}^{n-1} binom{i}{k}s_{n-1,i}$$$$=sum_{i = 1}^{n} binom{i}{k}s_{n-1,i-1} +(n-1)sum_{i = 0}^{n-1} binom{i}{k}s_{n-1,i}$$$$=sum_{i = 0}^{n-1} binom{i}{k}s_{n,i}+binom{n}{k}=sum_{i = 0}^{n} binom{i}{k}s_{n,i}$$
$blacksquare$
answered Dec 6 '18 at 13:18
Anubhab GhosalAnubhab Ghosal
1,22319
$endgroup$
add a comment |
$begingroup$
You know what method works (almost) always to prove an identity about a sequence defined inductively? Induction.
Base case is trivial.
$$ s_{n+1,k+1}=s_{n,k}+ns_{n,k+1}= sum_{i = 0}^{n-1} binom{i}{k-1}s_{n-1,i}+ nsum_{i = 0}^{n-1} binom{i}{k}s_{n-1,i}$$$$= sum_{i = 0}^{n-1} (binom{i}{k-1}s_{n-1,i}+ binom{i}{k}s_{n-1,i}) +(n-1)sum_{i = 0}^{n-1} binom{i}{k}s_{n-1,i}$$$$=sum_{i = 0}^{n-1} binom{i+1}{k}s_{n-1,i} +(n-1)sum_{i = 0}^{n-1} binom{i}{k}s_{n-1,i}$$$$=sum_{i = 1}^{n} binom{i}{k}s_{n-1,i-1} +(n-1)sum_{i = 0}^{n-1} binom{i}{k}s_{n-1,i}$$$$=sum_{i = 0}^{n-1} binom{i}{k}s_{n,i}+binom{n}{k}=sum_{i = 0}^{n} binom{i}{k}s_{n,i}$$
$blacksquare$
answered Dec 6 '18 at 13:18
Anubhab GhosalAnubhab Ghosal
1,22319
$endgroup$
You know what method works (almost) always to prove an identity about a sequence defined inductively? Induction.
Base case is trivial.
$$ s_{n+1,k+1}=s_{n,k}+ns_{n,k+1}= sum_{i = 0}^{n-1} binom{i}{k-1}s_{n-1,i}+ nsum_{i = 0}^{n-1} binom{i}{k}s_{n-1,i}$$$$= sum_{i = 0}^{n-1} (binom{i}{k-1}s_{n-1,i}+ binom{i}{k}s_{n-1,i}) +(n-1)sum_{i = 0}^{n-1} binom{i}{k}s_{n-1,i}$$$$=sum_{i = 0}^{n-1} binom{i+1}{k}s_{n-1,i} +(n-1)sum_{i = 0}^{n-1} binom{i}{k}s_{n-1,i}$$$$=sum_{i = 1}^{n} binom{i}{k}s_{n-1,i-1} +(n-1)sum_{i = 0}^{n-1} binom{i}{k}s_{n-1,i}$$$$=sum_{i = 0}^{n-1} binom{i}{k}s_{n,i}+binom{n}{k}=sum_{i = 0}^{n} binom{i}{k}s_{n,i}$$
$blacksquare$
answered Dec 6 '18 at 13:18
Anubhab GhosalAnubhab Ghosal
1,22319
answered Dec 6 '18 at 13:18
Anubhab GhosalAnubhab Ghosal
1,22319
answered Dec 6 '18 at 13:18
Anubhab GhosalAnubhab Ghosal
1,22319
answered Dec 6 '18 at 13:18
Anubhab GhosalAnubhab Ghosal
1,22319
1,22319
add a comment |
add a comment |
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What is definition of $S(a,b)$? And of $s-{u,v}$ [lower case $s$ with subscripts] Or should the latter be also upper case as in title?
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– coffeemath
Dec 6 '18 at 7:16
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Are you sure these are second kind Stirling numbers? Your example contains permutations, not partitions.
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– Lord Shark the Unknown
Dec 6 '18 at 7:30
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@LordSharktheUnknown Sorry, you are right. It was the first kind. I edited it.
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– NotEinstein
Dec 6 '18 at 7:38
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Can you please give a reference to your script? Are you sure you have written all the correct correctly? Also, you're "thought" in the third line, isn't it the definition of Stirling numbers of the first kind (should've been lowercase s)?
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– Sri Krishna Sahoo
Dec 6 '18 at 9:38