stirling numbers of the 1st kindmyweb.facstaff.wwu.edu/trenees/math_566/stirling.pdf · stirling...
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Stirling Numbers of the 1st Kind
Daniel Reiss, Colebrook Jackson, Brad Dallas
Western Washington University
November 28, 2012
Combinatorics Stirling Numbers of the 1st Kind
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Introduction
The set of all permutations of a set N is denoted S(N), while theset of all permutations of {1, 2, . . . , n} is denoted S(n).
A permutation σ ∈ S(n) is a bijective mapping whose canonicalnotation is
σ =
(1 2 . . . n
σ(1)σ(2) . . . σ(n)
).
We call σ = σ(1)σ(2) . . . σ(n) the word representation of σ.
Combinatorics Stirling Numbers of the 1st Kind
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With composition S(n) forms the symmetric group of order n. Weread a product always from right to left, thus for
σ =
(123456
234165
), τ =
(123456
134526
),
we have
τσ =
(123456
345162
), and στ =
(123456
241635
).
For example, 3→ τ(σ(3)) = τ(4) = 5.
Combinatorics Stirling Numbers of the 1st Kind
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Another way to describe σ is by its cycle decomposition.
For every i , the sequence i , σ(i), σ2(i), . . . must eventuallyterminate with, say, σk(i) = i .
We denote the cycle containing i by (i , σ(i), σ2(i), . . . , σk−1(i)).
Repeating this for all elements, we arrive at the cycledecomposition σ = σ1σ2 · · ·σt .
Combinatorics Stirling Numbers of the 1st Kind
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Example. (i). σ =
(12345678
35146827
)has word representation
σ = 35146827 and cycle form σ = (13)(25678)(4).
(ii). σ =
(12345678
12354786
)has word representation
σ = 12354786 and cycle form σ = (1)(2)(3)(45)(678).
Combinatorics Stirling Numbers of the 1st Kind
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Cycle Decomposition
I We may start a cycle with any element in the cycle.Thus, (2568) and (8256) are the same cycle.
I Order of cycles is irrelevant.
I Cycles of length 1 are fixed points in σ.
I Cycles of length 2 are transpositions i ↔ j .
I A permutation with only 1 cycle is called cyclic.There are (n − 1)! cyclic permutations of n.
Combinatorics Stirling Numbers of the 1st Kind
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Stirling Numbers of the First Kind
The Stirling number s(n, k) of the first kind is the number ofpermutations of an n-set with precisely k cycles.
Define s(0, 0) = 1 and s(0, k) = 0 for k > 0.
Several different notations for the Stirling numbers are in use.Stirling numbers of the first kind are written with a small s, andthose of the second kind with a large S . The Stirling numbers ofthe second kind are never negative, but those of the first kind canbe negative; hence, there is a separate notation for the unsignedStirling numbers of the first kind.
Combinatorics Stirling Numbers of the 1st Kind
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Common notations for the Stirling numbers include:
I s(n, k) = sn,k =[nk
](−1)n−k
for the ordinary signed Stirling numbers of the first kind
I c(n, k) =[nk
]= |s(n, k)|
for the unsigned Stirling numbers of first kind
I S(n, k) ={nk
}= S
(k)n = Sn,k
for the Stirling numbers of the second kind
Note: Aigner uses sn,k for the signless Stirling numbers and for thispresentation we will use s(n, k) for the signless Stirling numbers.
Combinatorics Stirling Numbers of the 1st Kind
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The table lists the first values of the Stirling matrix s(n, k).
n k 0 1 2 3 4 5 6 7 8
0 1
1 0 1
2 0 1 1
3 0 2 3 1
4 0 6 11 6 1
5 0 24 50 35 10 1
6 0 120 274 225 85 15 1
7 0 720 1764 1624 735 175 21 1
8 0 5040 13068 13132 6769 1960 322 28 1
Stirling numbers of the first kind s(n, k)
Combinatorics Stirling Numbers of the 1st Kind
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A useful graphical representation is to interpret σ ∈ S(n) as adirected graph with i → j if j = σ(i). The following graphsillustrate the Stirling numbers s(5, k) for k = 1, . . . , 5.
s(5, 1) = 24 (5 objects with 1 cycle)
Combinatorics Stirling Numbers of the 1st Kind
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s(5, 2) = 50 (5 objects with 2 cycles)
Combinatorics Stirling Numbers of the 1st Kind
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s(5, 3) = 35 (5 objects with 3 cycles)
Combinatorics Stirling Numbers of the 1st Kind
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s(5, 4) = 10 (5 objects with 4 cycles)
s(5, 5) = 1 (5 objects with 5 cycles)
Combinatorics Stirling Numbers of the 1st Kind
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The Stirling numbers, s(n, k), satisfy the recurrence relation:
s(n, k) = s(n − 1, k − 1) + (n − 1)s(n − 1, k) (n ≥ 1)
with initial conditions s(0, 0) = 1 and s(n, 0) = s(0, n) = 0, n > 0.
Combinatorics Stirling Numbers of the 1st Kind
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Proof.
Consider forming a new permutation with n objects from apermutation of n − 1 objects by adding a distinguished object.There are exactly two ways in which this can be accomplished.
First, we could form a singleton cycle, leaving the extra objectfixed. This increases the number of cycles by 1 and so accounts forthe s(n − 1, k − 1) term in the recurrence.
Second, we could insert the object into one of the existing cycles.Consider an arbitrary permutation of n − 1 objects with k cycles.To form the new permutation, we insert the new object before anyof the n − 1 objects already present. This explains the(n − 1)s(n − 1, k) term of the recurrence.
These two cases include all of the possibilities, so the recurrencerelation follows with the given initial conditions.
Combinatorics Stirling Numbers of the 1st Kind
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The Stirling numbers, s(n, k), satisfy the following identities:
I s(n, k) = s(n − 1, k − 1) + (n − 1)s(n − 1, k)
I
n∑k=0
s(n, k) = n!
I s(n, 1) = (n − 1)!
I s(n, n) = 1
I s(n, n − 1) =(n2
)I s(n, n − 2) = 1
4(3n − 1)(n3
)I s(n, n − 3) =
(n2
)(n4
)
Combinatorics Stirling Numbers of the 1st Kind
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The Stirling numbers, s(n, k), satisfy the following identities:
I s(n, k) = s(n − 1, k − 1) + (n − 1)s(n − 1, k)
I
n∑k=0
s(n, k) = n!
I s(n, 1) = (n − 1)!
I s(n, n) = 1
I s(n, n − 1) =(n2
)I s(n, n − 2) = 1
4(3n − 1)(n3
)I s(n, n − 3) =
(n2
)(n4
)
Combinatorics Stirling Numbers of the 1st Kind
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The Stirling numbers, s(n, k), satisfy the following identities:
I s(n, k) = s(n − 1, k − 1) + (n − 1)s(n − 1, k)
I
n∑k=0
s(n, k) = n!
I s(n, 1) = (n − 1)!
I s(n, n) = 1
I s(n, n − 1) =(n2
)I s(n, n − 2) = 1
4(3n − 1)(n3
)I s(n, n − 3) =
(n2
)(n4
)
Combinatorics Stirling Numbers of the 1st Kind
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The Stirling numbers, s(n, k), satisfy the following identities:
I s(n, k) = s(n − 1, k − 1) + (n − 1)s(n − 1, k)
I
n∑k=0
s(n, k) = n!
I s(n, 1) = (n − 1)!
I s(n, n) = 1
I s(n, n − 1) =(n2
)I s(n, n − 2) = 1
4(3n − 1)(n3
)I s(n, n − 3) =
(n2
)(n4
)
Combinatorics Stirling Numbers of the 1st Kind
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The Stirling numbers, s(n, k), satisfy the following identities:
I s(n, k) = s(n − 1, k − 1) + (n − 1)s(n − 1, k)
I
n∑k=0
s(n, k) = n!
I s(n, 1) = (n − 1)!
I s(n, n) = 1
I s(n, n − 1) =(n2
)I s(n, n − 2) = 1
4(3n − 1)(n3
)I s(n, n − 3) =
(n2
)(n4
)
Combinatorics Stirling Numbers of the 1st Kind
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The Stirling numbers, s(n, k), satisfy the following identities:
I s(n, k) = s(n − 1, k − 1) + (n − 1)s(n − 1, k)
I
n∑k=0
s(n, k) = n!
I s(n, 1) = (n − 1)!
I s(n, n) = 1
I s(n, n − 1) =(n2
)
I s(n, n − 2) = 14(3n − 1)
(n3
)I s(n, n − 3) =
(n2
)(n4
)
Combinatorics Stirling Numbers of the 1st Kind
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The Stirling numbers, s(n, k), satisfy the following identities:
I s(n, k) = s(n − 1, k − 1) + (n − 1)s(n − 1, k)
I
n∑k=0
s(n, k) = n!
I s(n, 1) = (n − 1)!
I s(n, n) = 1
I s(n, n − 1) =(n2
)I s(n, n − 2) = 1
4(3n − 1)(n3
)
I s(n, n − 3) =(n2
)(n4
)
Combinatorics Stirling Numbers of the 1st Kind
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The Stirling numbers, s(n, k), satisfy the following identities:
I s(n, k) = s(n − 1, k − 1) + (n − 1)s(n − 1, k)
I
n∑k=0
s(n, k) = n!
I s(n, 1) = (n − 1)!
I s(n, n) = 1
I s(n, n − 1) =(n2
)I s(n, n − 2) = 1
4(3n − 1)(n3
)I s(n, n − 3) =
(n2
)(n4
)
Combinatorics Stirling Numbers of the 1st Kind
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Proof.
The recursion has already been established so we prove first that
n∑k=0
s(n, k) = n!.
We proceed by induction on n. The case when n = 0 is triviallytrue since s(0, 0) = 1 = 0!. Now we assume that the summation istrue for n = `. That is,
∑̀k=0
s(`, k) = `!.
For n = `+ 1 we have,
Combinatorics Stirling Numbers of the 1st Kind
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Proof.
The recursion has already been established so we prove first that
n∑k=0
s(n, k) = n!.
We proceed by induction on n.
The case when n = 0 is triviallytrue since s(0, 0) = 1 = 0!. Now we assume that the summation istrue for n = `. That is,
∑̀k=0
s(`, k) = `!.
For n = `+ 1 we have,
Combinatorics Stirling Numbers of the 1st Kind
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Proof.
The recursion has already been established so we prove first that
n∑k=0
s(n, k) = n!.
We proceed by induction on n. The case when n = 0 is triviallytrue since s(0, 0) = 1 = 0!.
Now we assume that the summation istrue for n = `. That is,
∑̀k=0
s(`, k) = `!.
For n = `+ 1 we have,
Combinatorics Stirling Numbers of the 1st Kind
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Proof.
The recursion has already been established so we prove first that
n∑k=0
s(n, k) = n!.
We proceed by induction on n. The case when n = 0 is triviallytrue since s(0, 0) = 1 = 0!. Now we assume that the summation istrue for n = `. That is,
∑̀k=0
s(`, k) = `!.
For n = `+ 1 we have,
Combinatorics Stirling Numbers of the 1st Kind
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Proof.
The recursion has already been established so we prove first that
n∑k=0
s(n, k) = n!.
We proceed by induction on n. The case when n = 0 is triviallytrue since s(0, 0) = 1 = 0!. Now we assume that the summation istrue for n = `. That is,
∑̀k=0
s(`, k) = `!.
For n = `+ 1 we have,
Combinatorics Stirling Numbers of the 1st Kind
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Proof.
`+1∑k=0
s(`+ 1, k) =`+1∑k=0
(s(`, k − 1) + `s(`, k)
)=
`+1∑k=0
s(`, k − 1) + `
`+1∑k=0
s(`, k)
=`+1∑k=1
s(`, k − 1) + `
`+1∑k=0
s(`, k)
=∑̀k=0
s(`, k) + `∑̀k=0
s(`, k)
= `! + `(`!) = (`+ 1)!
This completes the inductive step and the proof of the claim.
Combinatorics Stirling Numbers of the 1st Kind
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Proof.
`+1∑k=0
s(`+ 1, k) =`+1∑k=0
(s(`, k − 1) + `s(`, k)
)=
`+1∑k=0
s(`, k − 1) + `
`+1∑k=0
s(`, k)
=`+1∑k=1
s(`, k − 1) + `
`+1∑k=0
s(`, k)
=∑̀k=0
s(`, k) + `∑̀k=0
s(`, k)
= `! + `(`!) = (`+ 1)!
This completes the inductive step and the proof of the claim.
Combinatorics Stirling Numbers of the 1st Kind
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Proof.
We now show the remaining claims starting with the second whichsays that s(n, 1) = (n − 1)!.
Permute {1, . . . , n} in one cycle.There are n! ways to do this. Since we can start at any one of thevalues in a given cycle we have overcounted the total by a factorof n. Thus,
s(n, 1) = (n − 1)!.
The proof that s(n, n) = 1 is trivial. Consider now s(n, n − 1). Tocount these permutations we need only choose which 2 of{1, . . . , n} are going to share a cycle while the others arerepresented by a singleton cycle. Thus,
s(n, n − 1) =(n2
).
Combinatorics Stirling Numbers of the 1st Kind
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Proof.
We now show the remaining claims starting with the second whichsays that s(n, 1) = (n − 1)!. Permute {1, . . . , n} in one cycle.There are n! ways to do this. Since we can start at any one of thevalues in a given cycle we have overcounted the total by a factorof n. Thus,
s(n, 1) = (n − 1)!.
The proof that s(n, n) = 1 is trivial. Consider now s(n, n − 1). Tocount these permutations we need only choose which 2 of{1, . . . , n} are going to share a cycle while the others arerepresented by a singleton cycle. Thus,
s(n, n − 1) =(n2
).
Combinatorics Stirling Numbers of the 1st Kind
![Page 33: Stirling Numbers of the 1st Kindmyweb.facstaff.wwu.edu/trenees/Math_566/stirling.pdf · Stirling Numbers of the 1st Kind Daniel Reiss, Colebrook Jackson, Brad Dallas Western Washington](https://reader031.vdocuments.pub/reader031/viewer/2022021504/5a9d86837f8b9a28388b523b/html5/thumbnails/33.jpg)
Proof.
We now show the remaining claims starting with the second whichsays that s(n, 1) = (n − 1)!. Permute {1, . . . , n} in one cycle.There are n! ways to do this. Since we can start at any one of thevalues in a given cycle we have overcounted the total by a factorof n. Thus,
s(n, 1) = (n − 1)!.
The proof that s(n, n) = 1 is trivial.
Consider now s(n, n − 1). Tocount these permutations we need only choose which 2 of{1, . . . , n} are going to share a cycle while the others arerepresented by a singleton cycle. Thus,
s(n, n − 1) =(n2
).
Combinatorics Stirling Numbers of the 1st Kind
![Page 34: Stirling Numbers of the 1st Kindmyweb.facstaff.wwu.edu/trenees/Math_566/stirling.pdf · Stirling Numbers of the 1st Kind Daniel Reiss, Colebrook Jackson, Brad Dallas Western Washington](https://reader031.vdocuments.pub/reader031/viewer/2022021504/5a9d86837f8b9a28388b523b/html5/thumbnails/34.jpg)
Proof.
We now show the remaining claims starting with the second whichsays that s(n, 1) = (n − 1)!. Permute {1, . . . , n} in one cycle.There are n! ways to do this. Since we can start at any one of thevalues in a given cycle we have overcounted the total by a factorof n. Thus,
s(n, 1) = (n − 1)!.
The proof that s(n, n) = 1 is trivial. Consider now s(n, n − 1). Tocount these permutations we need only choose which 2 of{1, . . . , n} are going to share a cycle while the others arerepresented by a singleton cycle. Thus,
s(n, n − 1) =(n2
).
Combinatorics Stirling Numbers of the 1st Kind
![Page 35: Stirling Numbers of the 1st Kindmyweb.facstaff.wwu.edu/trenees/Math_566/stirling.pdf · Stirling Numbers of the 1st Kind Daniel Reiss, Colebrook Jackson, Brad Dallas Western Washington](https://reader031.vdocuments.pub/reader031/viewer/2022021504/5a9d86837f8b9a28388b523b/html5/thumbnails/35.jpg)
The remaining two have similar combinatorial arguments butrequire a little algebra to get the desired form.
s(n, n − 2) = 14(3n − 1)
(n3
)
s(n, n − 3) =(n2
)(n4
)
Combinatorics Stirling Numbers of the 1st Kind
![Page 36: Stirling Numbers of the 1st Kindmyweb.facstaff.wwu.edu/trenees/Math_566/stirling.pdf · Stirling Numbers of the 1st Kind Daniel Reiss, Colebrook Jackson, Brad Dallas Western Washington](https://reader031.vdocuments.pub/reader031/viewer/2022021504/5a9d86837f8b9a28388b523b/html5/thumbnails/36.jpg)
The remaining two have similar combinatorial arguments butrequire a little algebra to get the desired form.
s(n, n − 2) = 14(3n − 1)
(n3
)
s(n, n − 3) =(n2
)(n4
)
Combinatorics Stirling Numbers of the 1st Kind
![Page 37: Stirling Numbers of the 1st Kindmyweb.facstaff.wwu.edu/trenees/Math_566/stirling.pdf · Stirling Numbers of the 1st Kind Daniel Reiss, Colebrook Jackson, Brad Dallas Western Washington](https://reader031.vdocuments.pub/reader031/viewer/2022021504/5a9d86837f8b9a28388b523b/html5/thumbnails/37.jpg)
The Stirling numbers also appear in the following relationships:
I s(n, 2) = (n − 1)!Hn−1
where Hn−1 is the (n − 1)st Harmonic number
I
n∑k=0
(−1)n−ks(n, k)xk = xn
I∑n≥0
s(j , n)S(n, k) = δjk
Combinatorics Stirling Numbers of the 1st Kind
![Page 38: Stirling Numbers of the 1st Kindmyweb.facstaff.wwu.edu/trenees/Math_566/stirling.pdf · Stirling Numbers of the 1st Kind Daniel Reiss, Colebrook Jackson, Brad Dallas Western Washington](https://reader031.vdocuments.pub/reader031/viewer/2022021504/5a9d86837f8b9a28388b523b/html5/thumbnails/38.jpg)
The Stirling numbers also appear in the following relationships:
I s(n, 2) = (n − 1)!Hn−1
where Hn−1 is the (n − 1)st Harmonic number
I
n∑k=0
(−1)n−ks(n, k)xk = xn
I∑n≥0
s(j , n)S(n, k) = δjk
Combinatorics Stirling Numbers of the 1st Kind
![Page 39: Stirling Numbers of the 1st Kindmyweb.facstaff.wwu.edu/trenees/Math_566/stirling.pdf · Stirling Numbers of the 1st Kind Daniel Reiss, Colebrook Jackson, Brad Dallas Western Washington](https://reader031.vdocuments.pub/reader031/viewer/2022021504/5a9d86837f8b9a28388b523b/html5/thumbnails/39.jpg)
The Stirling numbers also appear in the following relationships:
I s(n, 2) = (n − 1)!Hn−1
where Hn−1 is the (n − 1)st Harmonic number
I
n∑k=0
(−1)n−ks(n, k)xk = xn
I∑n≥0
s(j , n)S(n, k) = δjk
Combinatorics Stirling Numbers of the 1st Kind
![Page 40: Stirling Numbers of the 1st Kindmyweb.facstaff.wwu.edu/trenees/Math_566/stirling.pdf · Stirling Numbers of the 1st Kind Daniel Reiss, Colebrook Jackson, Brad Dallas Western Washington](https://reader031.vdocuments.pub/reader031/viewer/2022021504/5a9d86837f8b9a28388b523b/html5/thumbnails/40.jpg)
The Stirling numbers also appear in the following relationships:
I s(n, 2) = (n − 1)!Hn−1
where Hn−1 is the (n − 1)st Harmonic number
I
n∑k=0
(−1)n−ks(n, k)xk = xn
I∑n≥0
s(j , n)S(n, k) = δjk
Combinatorics Stirling Numbers of the 1st Kind
![Page 41: Stirling Numbers of the 1st Kindmyweb.facstaff.wwu.edu/trenees/Math_566/stirling.pdf · Stirling Numbers of the 1st Kind Daniel Reiss, Colebrook Jackson, Brad Dallas Western Washington](https://reader031.vdocuments.pub/reader031/viewer/2022021504/5a9d86837f8b9a28388b523b/html5/thumbnails/41.jpg)
Proof.
We start with the first claim.
To compute s(n, 2), we make use ofthe recurrence. Dividing by (n − 1)! we obtain,
s(n, 2)
(n − 1)!=
(n − 2)!
(n − 1)!+
(n − 1)s(n − 1, 2)
(n − 1)!=
s(n − 1, 2)
(n − 2)!+
1
n − 1.
By repeated iteration we arrive at the following,
s(n, 2)
(n − 1)!=
1
n − 1+
1
n − 2+ · · ·+ 1
2+
1
1= Hn−1.
Multiplying by (n − 1)! completes the proof.
Combinatorics Stirling Numbers of the 1st Kind
![Page 42: Stirling Numbers of the 1st Kindmyweb.facstaff.wwu.edu/trenees/Math_566/stirling.pdf · Stirling Numbers of the 1st Kind Daniel Reiss, Colebrook Jackson, Brad Dallas Western Washington](https://reader031.vdocuments.pub/reader031/viewer/2022021504/5a9d86837f8b9a28388b523b/html5/thumbnails/42.jpg)
Proof.
We start with the first claim. To compute s(n, 2), we make use ofthe recurrence. Dividing by (n − 1)! we obtain,
s(n, 2)
(n − 1)!=
(n − 2)!
(n − 1)!+
(n − 1)s(n − 1, 2)
(n − 1)!=
s(n − 1, 2)
(n − 2)!+
1
n − 1.
By repeated iteration we arrive at the following,
s(n, 2)
(n − 1)!=
1
n − 1+
1
n − 2+ · · ·+ 1
2+
1
1= Hn−1.
Multiplying by (n − 1)! completes the proof.
Combinatorics Stirling Numbers of the 1st Kind
![Page 43: Stirling Numbers of the 1st Kindmyweb.facstaff.wwu.edu/trenees/Math_566/stirling.pdf · Stirling Numbers of the 1st Kind Daniel Reiss, Colebrook Jackson, Brad Dallas Western Washington](https://reader031.vdocuments.pub/reader031/viewer/2022021504/5a9d86837f8b9a28388b523b/html5/thumbnails/43.jpg)
Proof.
We start with the first claim. To compute s(n, 2), we make use ofthe recurrence. Dividing by (n − 1)! we obtain,
s(n, 2)
(n − 1)!=
(n − 2)!
(n − 1)!+
(n − 1)s(n − 1, 2)
(n − 1)!=
s(n − 1, 2)
(n − 2)!+
1
n − 1.
By repeated iteration we arrive at the following,
s(n, 2)
(n − 1)!=
1
n − 1+
1
n − 2+ · · ·+ 1
2+
1
1= Hn−1.
Multiplying by (n − 1)! completes the proof.
Combinatorics Stirling Numbers of the 1st Kind
![Page 44: Stirling Numbers of the 1st Kindmyweb.facstaff.wwu.edu/trenees/Math_566/stirling.pdf · Stirling Numbers of the 1st Kind Daniel Reiss, Colebrook Jackson, Brad Dallas Western Washington](https://reader031.vdocuments.pub/reader031/viewer/2022021504/5a9d86837f8b9a28388b523b/html5/thumbnails/44.jpg)
Proof.
We start with the first claim. To compute s(n, 2), we make use ofthe recurrence. Dividing by (n − 1)! we obtain,
s(n, 2)
(n − 1)!=
(n − 2)!
(n − 1)!+
(n − 1)s(n − 1, 2)
(n − 1)!=
s(n − 1, 2)
(n − 2)!+
1
n − 1.
By repeated iteration we arrive at the following,
s(n, 2)
(n − 1)!=
1
n − 1+
1
n − 2+ · · ·+ 1
2+
1
1= Hn−1.
Multiplying by (n − 1)! completes the proof.
Combinatorics Stirling Numbers of the 1st Kind
![Page 45: Stirling Numbers of the 1st Kindmyweb.facstaff.wwu.edu/trenees/Math_566/stirling.pdf · Stirling Numbers of the 1st Kind Daniel Reiss, Colebrook Jackson, Brad Dallas Western Washington](https://reader031.vdocuments.pub/reader031/viewer/2022021504/5a9d86837f8b9a28388b523b/html5/thumbnails/45.jpg)
Proof.
We start with the first claim. To compute s(n, 2), we make use ofthe recurrence. Dividing by (n − 1)! we obtain,
s(n, 2)
(n − 1)!=
(n − 2)!
(n − 1)!+
(n − 1)s(n − 1, 2)
(n − 1)!=
s(n − 1, 2)
(n − 2)!+
1
n − 1.
By repeated iteration we arrive at the following,
s(n, 2)
(n − 1)!=
1
n − 1+
1
n − 2+ · · ·+ 1
2+
1
1= Hn−1.
Multiplying by (n − 1)! completes the proof.
Combinatorics Stirling Numbers of the 1st Kind
![Page 46: Stirling Numbers of the 1st Kindmyweb.facstaff.wwu.edu/trenees/Math_566/stirling.pdf · Stirling Numbers of the 1st Kind Daniel Reiss, Colebrook Jackson, Brad Dallas Western Washington](https://reader031.vdocuments.pub/reader031/viewer/2022021504/5a9d86837f8b9a28388b523b/html5/thumbnails/46.jpg)
Proof.
We start with the first claim. To compute s(n, 2), we make use ofthe recurrence. Dividing by (n − 1)! we obtain,
s(n, 2)
(n − 1)!=
(n − 2)!
(n − 1)!+
(n − 1)s(n − 1, 2)
(n − 1)!=
s(n − 1, 2)
(n − 2)!+
1
n − 1.
By repeated iteration we arrive at the following,
s(n, 2)
(n − 1)!=
1
n − 1+
1
n − 2+ · · ·+ 1
2+
1
1= Hn−1.
Multiplying by (n − 1)! completes the proof.
Combinatorics Stirling Numbers of the 1st Kind
![Page 47: Stirling Numbers of the 1st Kindmyweb.facstaff.wwu.edu/trenees/Math_566/stirling.pdf · Stirling Numbers of the 1st Kind Daniel Reiss, Colebrook Jackson, Brad Dallas Western Washington](https://reader031.vdocuments.pub/reader031/viewer/2022021504/5a9d86837f8b9a28388b523b/html5/thumbnails/47.jpg)
Next we would like to show that
xn =n∑
k=0
(−1)n−ks(n, k)xk
Proof.
To prove this claim we first show that xn =n∑
k=0
s(n, k)xk .
We proceed by induction on n. For n = 0 and n = 1 we have that,
x0 = s(0, 0) = 1 and x1 = s(1, 0) + x s(1, 1) = x .
Now assume the claim is true for n = `− 1. That is,
x`−1 =`−1∑k=0
s(`− 1, k)xk .
Combinatorics Stirling Numbers of the 1st Kind
![Page 48: Stirling Numbers of the 1st Kindmyweb.facstaff.wwu.edu/trenees/Math_566/stirling.pdf · Stirling Numbers of the 1st Kind Daniel Reiss, Colebrook Jackson, Brad Dallas Western Washington](https://reader031.vdocuments.pub/reader031/viewer/2022021504/5a9d86837f8b9a28388b523b/html5/thumbnails/48.jpg)
Next we would like to show that
xn =n∑
k=0
(−1)n−ks(n, k)xk
Proof.
To prove this claim we first show that xn =n∑
k=0
s(n, k)xk .
We proceed by induction on n.
For n = 0 and n = 1 we have that,
x0 = s(0, 0) = 1 and x1 = s(1, 0) + x s(1, 1) = x .
Now assume the claim is true for n = `− 1. That is,
x`−1 =`−1∑k=0
s(`− 1, k)xk .
Combinatorics Stirling Numbers of the 1st Kind
![Page 49: Stirling Numbers of the 1st Kindmyweb.facstaff.wwu.edu/trenees/Math_566/stirling.pdf · Stirling Numbers of the 1st Kind Daniel Reiss, Colebrook Jackson, Brad Dallas Western Washington](https://reader031.vdocuments.pub/reader031/viewer/2022021504/5a9d86837f8b9a28388b523b/html5/thumbnails/49.jpg)
Next we would like to show that
xn =n∑
k=0
(−1)n−ks(n, k)xk
Proof.
To prove this claim we first show that xn =n∑
k=0
s(n, k)xk .
We proceed by induction on n. For n = 0 and n = 1 we have that,
x0 = s(0, 0) = 1 and x1 = s(1, 0) + x s(1, 1) = x .
Now assume the claim is true for n = `− 1. That is,
x`−1 =`−1∑k=0
s(`− 1, k)xk .
Combinatorics Stirling Numbers of the 1st Kind
![Page 50: Stirling Numbers of the 1st Kindmyweb.facstaff.wwu.edu/trenees/Math_566/stirling.pdf · Stirling Numbers of the 1st Kind Daniel Reiss, Colebrook Jackson, Brad Dallas Western Washington](https://reader031.vdocuments.pub/reader031/viewer/2022021504/5a9d86837f8b9a28388b523b/html5/thumbnails/50.jpg)
Next we would like to show that
xn =n∑
k=0
(−1)n−ks(n, k)xk
Proof.
To prove this claim we first show that xn =n∑
k=0
s(n, k)xk .
We proceed by induction on n. For n = 0 and n = 1 we have that,
x0 = s(0, 0) = 1 and x1 = s(1, 0) + x s(1, 1) = x .
Now assume the claim is true for n = `− 1. That is,
x`−1 =`−1∑k=0
s(`− 1, k)xk .
Combinatorics Stirling Numbers of the 1st Kind
![Page 51: Stirling Numbers of the 1st Kindmyweb.facstaff.wwu.edu/trenees/Math_566/stirling.pdf · Stirling Numbers of the 1st Kind Daniel Reiss, Colebrook Jackson, Brad Dallas Western Washington](https://reader031.vdocuments.pub/reader031/viewer/2022021504/5a9d86837f8b9a28388b523b/html5/thumbnails/51.jpg)
Proof.
For n = `, we have
x` = x`−1 (x + `− 1)
= x · x`−1 + (`− 1) x`−1
= x∑k≥0
s(`− 1, k)xk + (`− 1)∑k≥0
s(`− 1, k)xk
=∑k≥0
s(`− 1, k)xk+1 + (`− 1)∑k≥0
s(`− 1, k)xk
=∑k≥1
s(`− 1, k − 1)xk + (`− 1)∑k≥0
s(`− 1, k)xk
=∑k≥0
[s(`− 1, k − 1) + (`− 1)s(`− 1, k)
]xk
=∑k≥0
s(`, k)xk .
Combinatorics Stirling Numbers of the 1st Kind
![Page 52: Stirling Numbers of the 1st Kindmyweb.facstaff.wwu.edu/trenees/Math_566/stirling.pdf · Stirling Numbers of the 1st Kind Daniel Reiss, Colebrook Jackson, Brad Dallas Western Washington](https://reader031.vdocuments.pub/reader031/viewer/2022021504/5a9d86837f8b9a28388b523b/html5/thumbnails/52.jpg)
Proof.
By the reciprocity law, we have xn = (−1)n(−x)n. Thus,
xn = (−1)nn∑
k=0
s(n, k)(−x)k
=n∑
k=0
(−1)n−ks(n, k)xk
Combinatorics Stirling Numbers of the 1st Kind
![Page 53: Stirling Numbers of the 1st Kindmyweb.facstaff.wwu.edu/trenees/Math_566/stirling.pdf · Stirling Numbers of the 1st Kind Daniel Reiss, Colebrook Jackson, Brad Dallas Western Washington](https://reader031.vdocuments.pub/reader031/viewer/2022021504/5a9d86837f8b9a28388b523b/html5/thumbnails/53.jpg)
Proof.
Next we prove the so called inversion formula.
Let A be an infinitelower triangular matrix. Then A has a unique inverse A−1 which isalso lower triangular if and only if A has non-zero entries along themain diagonal. Let A = (a(i , j)) and A−1 = (a−1(i , j)). Then astraightforward calculation shows that,∑
n≥0
a(j , n)a−1(n, k) = δjk .
Note now that we have the Stirling connection,
xn =∑k≥0
S(n, k)xk ⇔ xn =∑k≥0
(−1)n−ks(n, k)xk .
This statement implies that the matrices for the signed Stirlingnumbers of the first kind and the Stirling numbers of the secondkind are in fact inverses of one another. This together with theprevious establishes the claim.
Combinatorics Stirling Numbers of the 1st Kind
![Page 54: Stirling Numbers of the 1st Kindmyweb.facstaff.wwu.edu/trenees/Math_566/stirling.pdf · Stirling Numbers of the 1st Kind Daniel Reiss, Colebrook Jackson, Brad Dallas Western Washington](https://reader031.vdocuments.pub/reader031/viewer/2022021504/5a9d86837f8b9a28388b523b/html5/thumbnails/54.jpg)
Proof.
Next we prove the so called inversion formula. Let A be an infinitelower triangular matrix. Then A has a unique inverse A−1 which isalso lower triangular if and only if A has non-zero entries along themain diagonal.
Let A = (a(i , j)) and A−1 = (a−1(i , j)). Then astraightforward calculation shows that,∑
n≥0
a(j , n)a−1(n, k) = δjk .
Note now that we have the Stirling connection,
xn =∑k≥0
S(n, k)xk ⇔ xn =∑k≥0
(−1)n−ks(n, k)xk .
This statement implies that the matrices for the signed Stirlingnumbers of the first kind and the Stirling numbers of the secondkind are in fact inverses of one another. This together with theprevious establishes the claim.
Combinatorics Stirling Numbers of the 1st Kind
![Page 55: Stirling Numbers of the 1st Kindmyweb.facstaff.wwu.edu/trenees/Math_566/stirling.pdf · Stirling Numbers of the 1st Kind Daniel Reiss, Colebrook Jackson, Brad Dallas Western Washington](https://reader031.vdocuments.pub/reader031/viewer/2022021504/5a9d86837f8b9a28388b523b/html5/thumbnails/55.jpg)
Proof.
Next we prove the so called inversion formula. Let A be an infinitelower triangular matrix. Then A has a unique inverse A−1 which isalso lower triangular if and only if A has non-zero entries along themain diagonal. Let A = (a(i , j)) and A−1 = (a−1(i , j)).
Then astraightforward calculation shows that,∑
n≥0
a(j , n)a−1(n, k) = δjk .
Note now that we have the Stirling connection,
xn =∑k≥0
S(n, k)xk ⇔ xn =∑k≥0
(−1)n−ks(n, k)xk .
This statement implies that the matrices for the signed Stirlingnumbers of the first kind and the Stirling numbers of the secondkind are in fact inverses of one another. This together with theprevious establishes the claim.
Combinatorics Stirling Numbers of the 1st Kind
![Page 56: Stirling Numbers of the 1st Kindmyweb.facstaff.wwu.edu/trenees/Math_566/stirling.pdf · Stirling Numbers of the 1st Kind Daniel Reiss, Colebrook Jackson, Brad Dallas Western Washington](https://reader031.vdocuments.pub/reader031/viewer/2022021504/5a9d86837f8b9a28388b523b/html5/thumbnails/56.jpg)
Proof.
Next we prove the so called inversion formula. Let A be an infinitelower triangular matrix. Then A has a unique inverse A−1 which isalso lower triangular if and only if A has non-zero entries along themain diagonal. Let A = (a(i , j)) and A−1 = (a−1(i , j)). Then astraightforward calculation shows that,
∑n≥0
a(j , n)a−1(n, k) = δjk .
Note now that we have the Stirling connection,
xn =∑k≥0
S(n, k)xk ⇔ xn =∑k≥0
(−1)n−ks(n, k)xk .
This statement implies that the matrices for the signed Stirlingnumbers of the first kind and the Stirling numbers of the secondkind are in fact inverses of one another. This together with theprevious establishes the claim.
Combinatorics Stirling Numbers of the 1st Kind
![Page 57: Stirling Numbers of the 1st Kindmyweb.facstaff.wwu.edu/trenees/Math_566/stirling.pdf · Stirling Numbers of the 1st Kind Daniel Reiss, Colebrook Jackson, Brad Dallas Western Washington](https://reader031.vdocuments.pub/reader031/viewer/2022021504/5a9d86837f8b9a28388b523b/html5/thumbnails/57.jpg)
Proof.
Next we prove the so called inversion formula. Let A be an infinitelower triangular matrix. Then A has a unique inverse A−1 which isalso lower triangular if and only if A has non-zero entries along themain diagonal. Let A = (a(i , j)) and A−1 = (a−1(i , j)). Then astraightforward calculation shows that,∑
n≥0
a(j , n)a−1(n, k) = δjk .
Note now that we have the Stirling connection,
xn =∑k≥0
S(n, k)xk ⇔ xn =∑k≥0
(−1)n−ks(n, k)xk .
This statement implies that the matrices for the signed Stirlingnumbers of the first kind and the Stirling numbers of the secondkind are in fact inverses of one another. This together with theprevious establishes the claim.
Combinatorics Stirling Numbers of the 1st Kind
![Page 58: Stirling Numbers of the 1st Kindmyweb.facstaff.wwu.edu/trenees/Math_566/stirling.pdf · Stirling Numbers of the 1st Kind Daniel Reiss, Colebrook Jackson, Brad Dallas Western Washington](https://reader031.vdocuments.pub/reader031/viewer/2022021504/5a9d86837f8b9a28388b523b/html5/thumbnails/58.jpg)
Proof.
Next we prove the so called inversion formula. Let A be an infinitelower triangular matrix. Then A has a unique inverse A−1 which isalso lower triangular if and only if A has non-zero entries along themain diagonal. Let A = (a(i , j)) and A−1 = (a−1(i , j)). Then astraightforward calculation shows that,∑
n≥0
a(j , n)a−1(n, k) = δjk .
Note now that we have the Stirling connection,
xn =∑k≥0
S(n, k)xk ⇔ xn =∑k≥0
(−1)n−ks(n, k)xk .
This statement implies that the matrices for the signed Stirlingnumbers of the first kind and the Stirling numbers of the secondkind are in fact inverses of one another. This together with theprevious establishes the claim.
Combinatorics Stirling Numbers of the 1st Kind
![Page 59: Stirling Numbers of the 1st Kindmyweb.facstaff.wwu.edu/trenees/Math_566/stirling.pdf · Stirling Numbers of the 1st Kind Daniel Reiss, Colebrook Jackson, Brad Dallas Western Washington](https://reader031.vdocuments.pub/reader031/viewer/2022021504/5a9d86837f8b9a28388b523b/html5/thumbnails/59.jpg)
Proof.
Next we prove the so called inversion formula. Let A be an infinitelower triangular matrix. Then A has a unique inverse A−1 which isalso lower triangular if and only if A has non-zero entries along themain diagonal. Let A = (a(i , j)) and A−1 = (a−1(i , j)). Then astraightforward calculation shows that,∑
n≥0
a(j , n)a−1(n, k) = δjk .
Note now that we have the Stirling connection,
xn =∑k≥0
S(n, k)xk ⇔ xn =∑k≥0
(−1)n−ks(n, k)xk .
This statement implies that the matrices for the signed Stirlingnumbers of the first kind and the Stirling numbers of the secondkind are in fact inverses of one another. This together with theprevious establishes the claim.
Combinatorics Stirling Numbers of the 1st Kind
![Page 60: Stirling Numbers of the 1st Kindmyweb.facstaff.wwu.edu/trenees/Math_566/stirling.pdf · Stirling Numbers of the 1st Kind Daniel Reiss, Colebrook Jackson, Brad Dallas Western Washington](https://reader031.vdocuments.pub/reader031/viewer/2022021504/5a9d86837f8b9a28388b523b/html5/thumbnails/60.jpg)
A variety of identities may be derived by manipulating thegenerating function. The following result gives us the exponentialgenerating function for the signed Stirling numbers:
I∑n≥0
(−1)n−ks(n, k)zn
n!=
(log(1 + z))k
k!
Combinatorics Stirling Numbers of the 1st Kind
![Page 61: Stirling Numbers of the 1st Kindmyweb.facstaff.wwu.edu/trenees/Math_566/stirling.pdf · Stirling Numbers of the 1st Kind Daniel Reiss, Colebrook Jackson, Brad Dallas Western Washington](https://reader031.vdocuments.pub/reader031/viewer/2022021504/5a9d86837f8b9a28388b523b/html5/thumbnails/61.jpg)
A variety of identities may be derived by manipulating thegenerating function. The following result gives us the exponentialgenerating function for the signed Stirling numbers:
I∑n≥0
(−1)n−ks(n, k)zn
n!=
(log(1 + z))k
k!
Combinatorics Stirling Numbers of the 1st Kind
![Page 62: Stirling Numbers of the 1st Kindmyweb.facstaff.wwu.edu/trenees/Math_566/stirling.pdf · Stirling Numbers of the 1st Kind Daniel Reiss, Colebrook Jackson, Brad Dallas Western Washington](https://reader031.vdocuments.pub/reader031/viewer/2022021504/5a9d86837f8b9a28388b523b/html5/thumbnails/62.jpg)
Proof.
We utilize one of our many familiar generating functions as astarting point.
We have that
(1 + z)m =∑n≥0
(m
n
)zn
=∑n≥0
mn
n!zn
=∑n≥0
zn
n!
∑k≥0
(−1)n−ks(n, k)mk
=∑k≥0
mk∑n≥0
zn
n!(−1)n−ks(n, k)
Combinatorics Stirling Numbers of the 1st Kind
![Page 63: Stirling Numbers of the 1st Kindmyweb.facstaff.wwu.edu/trenees/Math_566/stirling.pdf · Stirling Numbers of the 1st Kind Daniel Reiss, Colebrook Jackson, Brad Dallas Western Washington](https://reader031.vdocuments.pub/reader031/viewer/2022021504/5a9d86837f8b9a28388b523b/html5/thumbnails/63.jpg)
Proof.
We utilize one of our many familiar generating functions as astarting point. We have that
(1 + z)m =∑n≥0
(m
n
)zn
=∑n≥0
mn
n!zn
=∑n≥0
zn
n!
∑k≥0
(−1)n−ks(n, k)mk
=∑k≥0
mk∑n≥0
zn
n!(−1)n−ks(n, k)
Combinatorics Stirling Numbers of the 1st Kind
![Page 64: Stirling Numbers of the 1st Kindmyweb.facstaff.wwu.edu/trenees/Math_566/stirling.pdf · Stirling Numbers of the 1st Kind Daniel Reiss, Colebrook Jackson, Brad Dallas Western Washington](https://reader031.vdocuments.pub/reader031/viewer/2022021504/5a9d86837f8b9a28388b523b/html5/thumbnails/64.jpg)
Proof.
Now we use the following identity,
(1 + z)m = emlog(1+z)
=∑k≥0
(log(1 + z))kmk
k!
Comparing the two we get the desired result.
Combinatorics Stirling Numbers of the 1st Kind
![Page 65: Stirling Numbers of the 1st Kindmyweb.facstaff.wwu.edu/trenees/Math_566/stirling.pdf · Stirling Numbers of the 1st Kind Daniel Reiss, Colebrook Jackson, Brad Dallas Western Washington](https://reader031.vdocuments.pub/reader031/viewer/2022021504/5a9d86837f8b9a28388b523b/html5/thumbnails/65.jpg)
Proof.
Now we use the following identity,
(1 + z)m = emlog(1+z)
=∑k≥0
(log(1 + z))kmk
k!
Comparing the two we get the desired result.
Combinatorics Stirling Numbers of the 1st Kind
![Page 66: Stirling Numbers of the 1st Kindmyweb.facstaff.wwu.edu/trenees/Math_566/stirling.pdf · Stirling Numbers of the 1st Kind Daniel Reiss, Colebrook Jackson, Brad Dallas Western Washington](https://reader031.vdocuments.pub/reader031/viewer/2022021504/5a9d86837f8b9a28388b523b/html5/thumbnails/66.jpg)
Proof.
Now we use the following identity,
(1 + z)m = emlog(1+z)
=∑k≥0
(log(1 + z))kmk
k!
Comparing the two we get the desired result.
Combinatorics Stirling Numbers of the 1st Kind
![Page 67: Stirling Numbers of the 1st Kindmyweb.facstaff.wwu.edu/trenees/Math_566/stirling.pdf · Stirling Numbers of the 1st Kind Daniel Reiss, Colebrook Jackson, Brad Dallas Western Washington](https://reader031.vdocuments.pub/reader031/viewer/2022021504/5a9d86837f8b9a28388b523b/html5/thumbnails/67.jpg)
References
I Aigner, Martin (2007). ”Section 1.4 - Permutations and StirlingNumbers s(n, k)”. A Course In Enumeration. Springer. pp. 24-29.ISBN 978-3-540-39032-4.
I Stirling numbers of the First Kind(http://planetmath.org/encyclopedia/StirlingNumbers.html),planetmath.org.
I Stirling Numbers of the First Kind(http://www.robertdickau.com/stirling1.html), robertdickau.com.
I Stirling Numbers of the First Kind(http://en.wikipedia.org/wiki/Stirling numbers of the first kind),wikipedia.org.
I Stirling Number of the First Kind(http://mathworld.wolfram.com/StirlingNumberoftheFirstKind.html),wolfram.com.
Combinatorics Stirling Numbers of the 1st Kind