Discussiones Mathematicae
General Algebra and Applications 34 (2014) 155–166
doi:10.7151/dmgaa.1225
JORDAN NUMBERS, STIRLING NUMBERS
AND SUMS OF POWERS
Roman Witula, Konrad Kaczmarek, Piotr Lorenc,
Edyta Hetmaniok and Mariusz Pleszczyński
Institute of Mathematics
Silesian University of Technology
Kaszubska 23, 44–100 Gliwice, Poland
e-mail: roman.witula,konrad.kaczmarek,[email protected]
Abstract
In the paper a new combinatorical interpretation of the Jordan numbers
is presented. Binomial type formulae connecting both kinds of numbers
mentioned in the title are given. The decomposition of the product of polynomial of variable n into the sums of kth powers of consecutive integers from
1 to n is also studied.
Keywords: Bernoulli numbers, binomial coefficients, Jordan numbers, Stirling numbers, Živković numbers.
2010 Mathematics Subject Classification: 11B68, 11B73, 11B83.
1.
Introduction
This paper was inspired by the wish to generalize the elementary formulae (formulae (2)–(7) given below) generated by the authors in the course of discussion
on D. Knuth’s excellent publication [9] (devoted to the analysis of relations between sums of the powers of consecutive positive integers, inspired by Faulhaber’s
”old” and ”deeper” results). It led us first to find the general connection of type
(8) (known form) and (9) (probably new form). In the course of discussing
these relations a very interesting and original problem has appeared, which consisted
of the product
P in deriving a formula of type
P (11), meaning
Pn a decomposition
s . Section 3 is devoted to this
nk nl=1 lr into the sum of type k+r
a
(k,
r)
l
s=1 s
l=1
problem. Section 4 presents the general formula (26) (known formula) giving the
decomposition of the Stirling numbers of second kind in the linear combination
of binomial coefficients by using the Jordan numbers. Finally, in Section 5 the
new combinatoric interpretation of the Jordan numbers is presented.
156
R. Witula, K. Kaczmarek, P. Lorenc, E. Hetmaniok and ...
2.
Stirling numbers
It should be reminded that Stirling numbers of the second kind or the partition
numbers S(n, k), k, n ∈ N, k ≤ n, satisfy the triangular recurrence relation [12,
13, 14]:
S(n, k) = S(n − 1, k − 1) + k S(n − 1, k),
(1)
S(k, k) = 1,
S(k, 0) = 0.
Moreover, we adopt here Donald Knuth’s notation [6, 9]:
X
f (n) :=
n
X
f (l)
l=1
for every function f : N → C. For example we have
X
k
n :=
n
X
lk ,
k, n ∈ N.
l=1
We shall start our deliberations by presenting the above-mentioned basic identities which can be easily verified by direct calculations. So, the following identities
hold:
X
(2)
n = S(n + 1, n),
X
3
n +
X
(3)
(4) 3
(5)
X
n5 + 10
X
1
n+2
n =
n (n + 1) = (3 n + 1)
2
3
n+2
1
= 3(n − 1) + 4
2
3
n+2
n+2
=6
+2
= 2 S(n + 2, n),
4
3
2
X
n4 + 9
X
2
X
n2 =
n2 (n + 1)(n + 2)(3n + 1)
n+3
n+1
= 24
= 24 S(n + 3, n),
4
2
n3 + 2
X
n+3
n+3
n+3
S(n + 3, n) = 15
+ 10
+
,
6
5
4
Jordan numbers, Stirling numbers and sums of powers
X
n7 + 7
(6)
(7)
X
n6 + 17
X
n5 + 17
X
n4 + 6
X
n3 =
157
X
n3 (n + 1)2 (n + 2)(n + 3)
n+4
3
2
= 48 S(n + 4, n) = 15 n + 30 n + 5 n − 2
,
5
n+4
n+4
n+4
n+4
S(n + 4, n) = 105
+ 105
+ 25
+
.
8
7
6
5
Stirling numbers of the second kind are presented in equations (3), (5) and (7)
which is not surprising because the following inversion formulae hold.
Proposition 1. We have
X
(8)
(9)
r
n =
=
r
X
l=0
r+1
X
l=1
n+1
S(r, l) l!
l+1
n
S(r + 1, l) (l − 1)!
.
l
Proof. Formula (8) is drawn from [9], whereas the second formula seems to be
new, yet it may be easily derived from the first one:
X
r
r
X
n+1
n
n
S(r, l) l!
=
S(r, l) l!
+
l+1
l+1
l
l=0
l=0
r−1 X
n
n
= S(r, 0) +
S(r, l) l! + S(r, l + 1) (l + 1)! + S(r, r) r!
l+1
r+1
l=0
r−1 X
n
n
=
l! S(r, l) + S(r, l + 1) (l + 1) + S(r + 1, r + 1) r!
l+1
r+1
l=0
=
r+1 X
l=1
n
(l − 1)! S(r + 1, l).
l
3.
Decomposition of products nk
We note that from (9) we obtain
(n + 1)
(10)
X
r+1
X
nm
n+1
n =
S(r + 1, l) (l − 1)! (l + 1)
l+1
l=1
r+1
X
n+1
by (8) X r+1
=
n
+
S(r + 1, l) (l − 1)!
,
l+1
r
P
l=1
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R. Witula, K. Kaczmarek, P. Lorenc, E. Hetmaniok and ...
which, again by (8), suggests a formula of following form
(11)
n
k
X
r
n =
k+r
X
as (k, r)
s=1
X
nk+r+1−s .
Since we have (see [6]):
(r + 1)
X
(12)
n
r
n =
r
X
l
(−1)
l=0
r+1
= (r + 1)
r+1
Bl nr+1−l
l
X
r
⇐⇒
r
n + (r + 1) B1 n −
⌊r/2⌋ X
k=1
r+1
B2k nr+1−2k ,
2k
where Bl are the Bernoulli’s numbers, we obtain indeed formulae of the type (11).
From (12) the following formulae can be generated.
Theorem 2. We have
(13)
(r + 1) n
X
r
n = (r + 2)
X
n
r+1
s=1
(14)
(r + 1) n
2
= (r + 3)
X
X
r
n = (r + 2) n
nr+2 +
X
n
r+1 X
r+2
p
r+1
+
+
s
r X
r+1
s=1
Bp
r X
r+1
X
s
Bs n
Bs
X
X
nr−s+1
nr−s+1 ,
nr−p+2
p=1
X
Bs
+
(r − s + 3)
nr−s+2
s
r−s+2
s=1
r−s+1
X r − s + 2 X
r−s−q+2
+
Bq
n
q
q=1
r
X
X
X
2r − p + 5 r + 2
r+2
= (r + 3)
n
+
Bp
nr−p+2
r+2
p
p=1
r r−s+1
X
X
X r+1r−s+2 Bs Bq X
nr−s−q+2 .
+ (r + 2) Br+1
n+
s
q
r
−
s
+
2
s=1 q=1
r X
r+1
Jordan numbers, Stirling numbers and sums of powers
159
Hence, after the appropriate regrouping, we get
(15)
X
X
X
X
1
(r + 1) n2
nr = (r + 3)
nr+2 − (r + 2)
nr+1 + (r + 1)(r + 3)
nr
6
1 r + 1 X r−1
1 r + 1 X r−2
1 r + 1 X r−3
−
n
−
n
+
n
6
2
15
4
30
4
1 r + 1 X r−4
+
n
+ ...
21
6
"
#
r−1 X
X
1
r+1
+
(r + 5)(r + 1)Br +
(r − s + 1)Bs Br−s
n2
2
s
s=1
!
r X
X
r+1
+ (r + 2)Br+1 +
Bs Br−s+1
n.
s
s=1
The last two coefficients can be reduced to the following ones (the classical splitting formulae are applied here [1, 5, 7, 10, 11]):
jrk
i
1h
(r + 5)(r + 1)Br − 2
+ 1 (r + 1)B2⌊ r2 ⌋
2
2
since we have the equality
r−1 X
r+1
s
s=1
jrk
(r − s + 1)Bs Br−s = − 2
+ 1 (r + 1)B2⌊ r2 ⌋
2
for every r = 3, 4, . . . (we have 2
(r + 2)Br+1 +
r
2
+ 1 (r + 1)B2⌊ r2 ⌋ for r = 2), and
r X
r+1
s=1
s
Bs Br−s+1 = −(r + 1)Br .
From these relations, we obtain
r−1 X
r+1
s=1
s
(r − s + 1)Bs Br−s = (r + 1)
r−1 X
r
s=1
s
Bs Br−s
if r is odd and r ≥ 3.
Proof. We prove only (13). To this aim let us fix r ∈ N and put
(16)
Tk,r := (r + 1) kr − (k + 1)r+1 + kr+1 .
160
R. Witula, K. Kaczmarek, P. Lorenc, E. Hetmaniok and ...
We have
r
Tk,r = (r + 1) k + k
n
X
(17)
Tk,r = (r + 1)
k=1
= (r + 1)
On the other hand
n
X
(18)
X
r+1
nr +
−
n
X
k=1
X
r
r+1 X
r+1
i
i=0
i
k =−
r−1 X
r+1
i=0
i
ki ,
kr+1 − (k + 1)r+1
n + 1 − (n + 1)r+1 .
Tk,r = −
r−1 X
r+1 X
i
i=0
k=1
ni .
We note that (13) holds for n = 1 since it is equivalent to the following known
relation
r+1 X
r+1
Br+1 =
Bs .
s
s=0
Let us assume that (13) holds for some n ∈ N. Then we get
(r + 1)n
X
nr − (r + 2)
=
X
r X
r+1
s
s=1
(19)
=
−
nr+1 −
r X
r+1
s=1
r X
s=1
s
r−1 X
r+1 X
i=0
i
r−1 X
r+1 X
Bs
X
n
Bs
X
(n + 1)r−s+1
r−s+1
−
ni
i=0
i
ni
r−1 X
r+1
r+1 X i
Bs (n + 1)r−s+1 −
n,
s
i
i=0
and
−
r X
r+1
s=1
s
Bs (n + 1)r−s+1
r
X
r+1
(−1)s
Bs (n + 1)r−s+1
s
s=0
X
(12)
r+1
r
= (n + 1)
+ (r + 1)(n + 1) − (r + 1)
(n + 1)r
= (n + 1)r+1 − 2(r + 1)B1 (n + 1)r −
= (n + 1)r+1 − (r + 1)
X
nr ,
Jordan numbers, Stirling numbers and sums of powers
161
which implies
−
r X
r+1
s
s=1
(20)
r−s+1
Bs (n + 1)
= (n + 1)r+1 −
−
r−1 X
r+1 X
i
i=0
r X
i=0
ni
r+1 X i
n = 1.
i
We note that the last identity can be easily deduced after summing the following
equalities
r+1
(n + 1)
=n
r+1
r X
r+1
+
i
i=0
n
r+1
r+1
= (n − 1)
+
ni ,
r X
r+1
i
i=0
...
r+1
2
r+1
=1
+
r X
r+1
i
i=0
(n − 1)i ,
1i .
Moreover, by (17) and (18), we obtain
(r + 1)n
(21)
X
= (r + 1)n
r
n − (r + 2)
X
r
X
n − (r + 2)
= 1 + (r + 1)(n + 1)
X
n
r+1
X
−
r−1 X
r+1 X
i=0
n
r+1
i
+ (r + 1)
(n + 1)r − (r + 2)
X
X
ni
nr + 1 − (n + 1)r+1
(n + 1)r+1 .
By comparing (19)–(21) we conclude that relation (16) holds also for n replaced
by n + 1 which, by the principle of mathematical induction, ends the proof.
Corollary 3. From (15) the following special formulae can be deduced
X
X
X
X
2n2
n=4
n3 − 3
n2 +
n,
X
X
X
X
5
1X
3n2
n2 = 5
n4 − 4
n2 +
n2 −
n,
2X
2
X
X
X
X
4n2
n3 = 6
n5 − 5
n4 + 4
n3 −
n2 .
Some more general formula, than the one in (15), can be also obtained.
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R. Witula, K. Kaczmarek, P. Lorenc, E. Hetmaniok and ...
Proposition 4. We have
X
k
nm+k − (m + k)
nm+k−1
2
X
k
(m + 2 k − 1) (m + k − 1)
nm+k−2
+
12
X
k
(k − 1) (m + k − 1) (m + k − 2)
nm+k−3
−
24
X m+k−4
1
+
(m + k − 3) k (m + 2 k − 3) 3 k2 + 3 k (m − 3) − m (m + 4) + 6
n
720 X m+k−5
1 m+k−3 k
2 k2 + 2 k (m − 4) − m (m + 5) + 8
n
+ ...
−
360
2
2
X
+ coeff(m, k)
n.
(22)
(m + 1) nk
X
nm = (m + k + 1)
X
We have found the general formula only for the first six coefficients from above.
The last absent coefficient coeff(m, k) can be found by subtracting from the left
side of (22) the expression staying on the right side of this formula with sums
P
np where p ≥ 2.
For example, we find
X
X
X
2n
n=3
n2 −
n,
X
X
X
9X 2 7X
2n3
n=5
n4 − 6
n3 +
n −
n,
2
2
X
X
X
X
4n
n3 = 5
n4 − 2
n3 +
n2 ,
X
1
3n4
n2 = n5 (n + 1)(2n + 1)
2
X
X
X
X
7X 2 1X
=7
n6 − 12
n5 + 15
n4 − 10
n3 +
n −
n,
2
2
and the following special one
(r + 1)n3
X
X
X
X
3
1
nr = (r + 4)
nr+3 − (r + 3)
nr+2 + (r + 2)(r + 5)
nr+1
2
4
X
X
1
1
nr −
(r − 6)r(r + 1)(r + 3)
nr−1
− (r + 1)(r + 2)
4
240
X
1
+
(r − 2)(r − 1)r(r + 1)
nr−2 + ...
240
4.
Jordan numbers
In this section, we present the generalization of relations (3), (5) and (7).
Jordan numbers, Stirling numbers and sums of powers
163
Proposition 5. We have
(23)
S(n + k, n) =
k−1
X
l=0
al,k
n+k
,
2k − l
where
a0,k+1 = (2k + 1)a0,k ,
(24)
ak,k+1 = ak−1,k ,
al,k+1 = (k − l + 1) al−1,k + (2 k − l + 1) al,k ,
for l = 1, 2, . . . , k − 1. The numbers al,k are called the Jordan numbers [2, 3, 4, 8,
15]. Authors of the present paper obtained the above relations independently. For
the sake of selfcontainedness, the proof of (23) will be given now. We proceed by
induction.
Proof. (23) We shall employ the following basic formula
n
n
X
X
k+l
k+l
k
=
(k + l − m) + (m − l)
m
m
k=m−l
k=m−l
n
n
X
X
k+l
k+l
(25)
= (m + 1)
+ (m − l)
m+1
m
k=m−l+1
k=m−l
n+l+1
n+l+1
= (m + 1)
+ (m − l)
.
m+2
m+1
Thus, assuming that formula (23) holds for certain k, n ∈ N, from (25) and (1)
the following relation can be derived
S(n + k + 1, n) = S(n + k + 1, n) − S(k + 2, 0)
=
n−1
X
l=0
n−1
X
X
n−1
S(l + k + 2, l + 1) − S(l + k + 1, l) =
(l + 1) S(l + k + 1, l + 1)
l=0
k−1
X
X
k−1
n−1
X
l+k+1
l+k+1
=
(l + 1)
aτ,k
=
aτ,k
(l + 1)
2k − τ
2k − τ
τ =0
τ =0
l=0
l=k−1−τ
k−1
X
n+k+1
n+k+1
=
aτ,k (2 k − τ + 1)
+ (k − τ )
2k − τ + 2
2k − τ + 1
τ =0
X
k−2
n+k+1
n+k+1
= a0,k (2 k + 1)
+
aτ,k (k − τ ) + aτ +1,k (2 k − τ )
2k+2
2k−τ +1
τ =0
X
k
n+k+1
n+k+1
+ ak−1,k
=
al,k+1
.
k+2
2k − l + 2
l=0
164
R. Witula, K. Kaczmarek, P. Lorenc, E. Hetmaniok and ...
Remark 6. It is possible to prove the following formulae:
a0,k = (2 k − 1)!!,
1 2k − 2
a2,k =
(2 k − 3)!!,
12
3
ak−2,k = 2k+1 − k − 3
a1,k =
k−1
a0,k ,
3
a3,k = S2(k, 2 k + 3),
ak−1,k = 1,
where S2(n, k) denotes the 2-associated Stirling number of the second kind (see [12]
and sequences A000478 and A000247 in [14]).
In Table 1 the triangle of coefficients al,k , l = 0, 1, . . . , k − 1 is presented.
Table 1. Triangle of coefficients al,k
l,k
0
1
2
3
4
5
6
7
1
1
2
3
1
3
15
10
1
4
105
105
25
1
5
945
1260
490
56
1
6
10395
17325
9450
1918
119
1
7
135135
270270
190575
56980
6825
246
1
8
2027025
4729725
4099095
1636635
302995
22935
501
1
Remark 7. Another formula connecting Stirling numbers of the second kind
with binomial coefficients is also known
X k n+k−1−r
S(n, n − k) =
,
r
2k
r
k
where
are the Eulerian numbers of the second kind (see [6]).
r
5.
Combinatoric interpretation of the Jordan and
Živković numbers
As it can be inferred from [12, pp. 76–77], the numbers an−2k−1,n−k , for k =
0, 1, . . ., ⌊(n − 1)/2⌋, enumerate the permutations of n elements with k cycles,
Jordan numbers, Stirling numbers and sums of powers
165
none of which is a unit cycle. We note that in [12] the numbers
b(n, k) := an−2k−1,n−k
are called the associated Stirling number of the second kind. Moreover, we have
the following recurrence relation
b(n + 1, k) = k b(n, k) + n b(n − 1, k − 1).
Conversely, in [15] the Živković numbers G(k, i) are defined by the following
equality
k
X
n+i−1
,
s(n, n − k) =
(−1)i G(k, i)
k+i
i=0
where G(k, i) = 0 if k ≥ 1 and i > k or i < 1 (G(0, i) = δ0,i ), and s(n, k) denotes
the Stirling number of the first kind.
Numbers G(k, i) are determined by the initial condition G(1, i) = δ1,i and by
the recurrence relation
(26)
G(k + 1, i) = i G(k, i) + (k + i) G(k, i − 1).
It is easy to check that
(27)
G(r − l, r) = al,r
⇐⇒
G(k, k + l) = al,k+l
(r − l := k).
Hence, as it is verifiable on the grounds of relation (26), al,k+l denotes the number
of ways of placing 2k + l labeled balls into k indistinguishable boxes with at least
two balls in each box.
The values of numbers G(k, i) are compiled in Table 2 (we note that G(k, i),
k = i, i + 1, . . . make up i-th diagonal – Table 1 – for each i ∈ N; the diagonal is
from the main line upwards).
Table 2. Values of numbers G(k, i)
G(1, k)
1
1
1
1
1
1
G(2, k)
G(3, k)
G(4, k)
G(5, k)
G(6, k)
3
10
25
56
119
15
105
490
1918
105
1260
9450
945
17325
10395
166
R. Witula, K. Kaczmarek, P. Lorenc, E. Hetmaniok and ...
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[14] N.J.A. Sloane, The On-Line Encyclopedia of Integer Sequences (http://oeis.org/,).
[15] M. Živković, On a representation of Stirling’s numbers of first kind , Univ. Beograd.
Publ. Elektrotehn. Fak. Ser. Mat. Fiz. 498–541 (1975) 217–221.
Received 6 December 2013
Revised 6 January 2015