mihoubi10. 112KB Jun 04 2011 12:09:13 AM
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Journal of Integer Sequences, Vol. 14 (2011),
Article 11.3.1
23 11
Touchard Polynomials, Partial Bell
Polynomials and Polynomials of Binomial
Type
Miloud Mihoubi and Mohammed Said Maamra1
University of Science and Technology Houari Boumediene (USTHB)
Faculty of Mathematics
PB 32
El Alia, 16111, Algiers
Algeria
[email protected]
[email protected]
[email protected]
Abstract
Touchard generalized the Bell polynomials in order to give some combinatorial
interpretation on permutations. Chrysaphinou introduced and studied a class of polynomials related to Touchard’s generalization. In the present paper, we establish some
relations between Touchard polynomials, Bell polynomials and the polynomials of binomial type. Several identities and relations with Stirling numbers are obtained.
1
Introduction
Among the partition polynomials, the partial Bell polynomials, introduced by Bell [3], play
an important role in different application frameworks. Several properties and identities
are given, see [5, 7, 9, 10]. Another partition polynomials, called Touchard polynomials,
introduced by Touchard [11], present an extension of the partial Bell polynomials. Some
algebraic, combinatorial and probabilistic properties of these polynomials are studied by
Touchard [11], Chrysaphinou [6], Charalambides [5], Kuzmin and Leonova [8]. In this paper,
1
Research supported by LAID3 Laboratory of USTHB University.
1
we give some relations between Touchard polynomials and partial Bell polynomials. We
exploit these relations and the polynomials of binomial type to derive some identities for
these polynomials.
In this context, let (xi ; i ≥ 1) and (yi ; i ≥ 1) be two sequences of real numbers. The
Touchard polynomials
Tn,k (xj ; yj ) := Tn,k (x1 , . . . , xn ; y1 , . . . , yn ), n ≥ k ≥ 0,
are defined by their bivariate generating function
!
X ti X ti
Tn,k (xj ; yj )u
yi
= exp u xi +
.
n!
i! i≥1 i!
n≥0 k=0
i≥1
n
XX
n
kt
The vertical generating function of Touchard polynomials, for fixed k, is given by
!k
!
X ti
X
1 X ti
tn
yi
xi
exp
, k = 0, 1, . . . .
Tn,k (xj ; yj ) =
n!
k!
i!
i!
i≥1
i≥1
n≥k
The partial Bell polynomials are given by
Bn,k (xj ) := Bn,k (x1 , . . . , xn ) = Tn,k (x1 , . . . , xn ; 0, . . . , 0), n ≥ k ≥ 0.
The polynomials of binomial type (fn (x)) are defined by
!x
X
X
un
un
, with f0 (x) = 1 and f1 (x) 6= 0 for x 6= 0.
fn (x)
=
fn (1)
n!
n!
n≥0
n≥0
For a real number a we consider in the following the sequence (fn (x; a)) defined by
fn (x; a) :=
x
fn (an + x) .
an + x
The sequence (fn (x; a)) is also of binomial type, see [9], and for more details on sequences
of binomial type see [1].
We use the following notation and hypothesis:
d
dk
dk
k
D ≡ dx
, Dk ≡ dx
k , Dx=x0 ≡ dxk |x=x0 .
For n < 0, we set fn (x) = 0, Tn,k (xj ; yj ) = 0 and Bn,k (xj ) = 0.
For x ∈ R, where R is the set of real numbers, we set
x
x
x(x − 1) · · · (x − k + 1)
x
for k = 1, 2, . . . ,
= 1 and
= 0 otherwise.
:=
k!
0
k
k
Also, for all nnnegative integers n, m we put
(
1, if m divides n;
and
1(m|n) =
0, otherwise;
2
1(n≥m) =
(
1, if n ≥ m;
0, otherwise.
2
The main results
In this section, we establish some relations between the Touchard polynomials, partial Bell
polynomials and polynomials of binomial type. Furthermore, we use these relations to develop several identities for Touchard polynomials. We start with the following theorem:
Theorem 1. Let n, k, m be integers such that n ≥ k ≥ 1, m ≥ 1; a be a real number and
(xn ) be a sequence of real numbers. Then
!
n
j/m
Bn−m,k (xj ).
Tn,k xj ; −m(j − 1)!a 1(m|j) = Bn,k (xj ) − am!
m
Proof. Let yn = −m(n − 1)!an/m 1(m|n) .
For a = 0 the theorem is trivial, otherwise, for |t| < |a|−1/m we have
!
!
X ti
X tmj
exp
yi
= exp − aj
= exp(ln(1 − atm )) = 1 − atm .
i!
j
i≥1
j≥1
Then
tn
1 X ti
Tn,k (xj ; yj ) =
xi
n!
k!
i!
i≥1
n≥k
X
!k
X ti
yi
exp
i!
i≥1
!
!k
1 X ti
=
xi
(1 − atm )
k! i≥1 i!
n
X
t
n!
Bn−m,k (xj )
.
=
Bn,k (xj ) − a
(n − m)!
n!
n≥k
Hence the theorem is proved.
If we set xn = nfn−1 (x; b) in Theorem 1 and use Proposition 1 of [9], we obtain:
Corollary 2. Let (fn (x)) be a sequence of binomial type of polynomials. We have
!
!
f
(kx;
b)
f
(kx;
b)
n!
n−k
n−m−k
−a
.
Tn,k jfj−1 (x; b); −m(j − 1)!aj/m 1(m|j) =
k!
(n − k)!
(n − m − k)!
Example 3. For fn (x) = xn in Corollary 2 we get
Tn,k jx(b(j − 1) + x)
=x
j−2
; −m(j − 1)!a
(b(n − k) + kx)
n!
(k − 1)!
(n − k)!
j/m
n−k−1
1(m|j)
!
n−m−k−1
−a
(b(n − m − k) + kx)
(n − m − k)!
3
!
1(n≥m+k) ,
and for fn (x) = n!
x
n
in Corollary 2 we get
b(j−1)+x
Tn,k xj!
=x
j−1
b(j − 1) + x
; −m(j − 1)!aj/m 1(m|j)
!
b(n−m−k)+kx
n−m−k
b(n−k)+kx
n−k
!
n!
−a
1(n≥m+k) .
(k − 1)! b(n − k) + kx
b(n − m − k) + kx
As above, for particular cases of Touchard polynomials, the following proposition gives
another expression in term of polynomials of binomial type.
Proposition 4. Let b, α be two real numbers and (fn (x)) be a sequence of binomial type of
polynomials. We have
!
n
Tn,k jfj−1 (x); αDfj (0) =
fn−k (kx + α),
k
or more generally
!
fj−1 (b(j − 1) + x)
fj (bj)
Tn,k jx
; α
b(j − 1) + x
bj
fn−k (b(n − k) + kx + α)
n
(kx + α)
=
, b 6= 0.
k
b(n − k) + kx + α
Proof. Let
X i
X
tn
t
1+
fn (x) = exp x yi
n!
i!
n≥1
i≥1
be the exponential generating function of the sequence (fn (x)) and xn = nfn−1 (x).
Necessarily yn = Dfn (0). We have
k
n
X i
X
t
1 X ti
t
exp α yi
Tn,k xj ; αyj
=
xi
n!
k! i≥1 i!
i!
i≥1
n≥k
X ti
tk
= exp (kx + α) yi
k!
i!
i≥1
tk X
tn
fn (kx + α)
k! n≥0
n!
Xn
tn+k
.
fn−k (kx + α)
=
n!
k
n≥k
=
Then, we obtain
Tn,k
n
jfj−1 (x); αDfj (0) =
fn−k (kx + α).
k
To finish the proof, replace fn (x) by fn (x; b) in the last identity.
4
Example 5. For fn (x) = xn in Proposition 4 we get
n
j−2
j−2
=
(kx + α)(b(n − k) + kx + α)n−k−1 ,
Tn,k xj(b(j − 1) + x) ; α(j − 1)(bj)
k
and for fn (x) = n! nx in Proposition 4 we get
b(j − 1) + x
j!x
j−1
Tn,k
; α(−1) (j − 1)!
b(j − 1) + x
j−1
n!
kx + α
n(n − k) + kx + α
=
.
k! n(n − k) + kx + α
n−k
Hence we may state the following:
Corollary 6. Let r, s, p be a nonnegative integers, r ≥ 1, and (xn ) be a sequence of real
numbers with x1 = 1. We have
B(r+1)(j−1)+s, r(j−1)+s (xi ) p B(r+1)j, rj (xi )
js
;
Tn,k
(r+1)(j−1)+s
(r+1)j
(r(j − 1) + s)
rj
r(j−1)+s
rj
B(r+1)(n−k)+ks+p, r(n−k)+ks+p (xi )
ks + p
n
.
=
(r+1)(n−k)+ks+p
k r(n − k) + ks + p
r(n−k)+ks+p
Proof. For x2 6= 0, let {fn (x)} be a sequence of binomial type such that fn (1) =
n
fn−k (k) we get
From the known identity Bn,k (jfj−1 (1)) =
k
−1
n+k
fn (k) =
Bn+k,k (xi ), n ≥ 0, k ≥ 1.
k
xn+1
.
n+1
Take b = r, x = s and α = p in Proposition 4 to get
fn−k (r(n − k) + ks + p)
n
fj−1 (r(j − 1) + s) fj (rj)
; p
.
=
(ks + p)
Tn,k js
r(j − 1) + s
rj
r(n − k) + ks + p
k
−1
Bn+k,k (xi ) to express in the last
Therefore, it suffices to use the identity fn (k) = n+k
k
identity fi−1 (r(i − 1) + s) and fn−k (r(n − k) + ks + p) by the partial Bell polynomials.
For the case x2 = 0 the corollary remains true by continuity.
k
Example 7. By Corollary 6 and the identity Bn,k (1!, 2!, . . . , (q + 1)!, 0, . . .) = n!
, see
k! n−k q
[2], we get
(j − 1)!p rj
r(j − 1) + s
j!s
;
Tn,k
j−1
j q
r(j − 1) + s
r
q
ks + p
r(n − k) + ks + p
n!
,
=
k! r(n − k) + ks + p
n−k
q
P k n
k
where
is the coefficients defined by (1 + x + x2 + · · · + xq )k =
x .
n q
n q
n≥0
Also, for xn = 1, xn = (−1)n−1 (n − 1)!, xn = (n − 1)! or xn = n! we get identities related
Touchard polynomials to Stirling numbers of the first and second kind.
5
The following two propositions give relations between Touchard polynomials and the
successive derivatives of polynomials of binomial type.
Proposition 8. Let b be a real number, b 6= 0, and (fn (x)) be a sequence of binomial type
of polynomials. We have
1
Tn,k Dfj (0); xDfj (0) = Dk fn (x),
k!
or more generally
Tn,k
fj (bj)
fj (bj)
; x
bj
bj
1 k
x
= D
fn (bn + x) .
k!
bn + x
Proof. Let (fn (x)) be a sequence of binomial type defined as in the proof of Proposition 4
and xn = nfn−1 (x). We have yn = Dfn (0) and
X i
X i k
X
X
tn
t
tn
t
k
= k! Tn,k (yj ; xyj ) .
exp x yi
D fn (x) =
yi
n!
i!
i!
n!
i≥1
i≥1
n≥k
n≥k
Then
Tn,k yj ; xyj
= Tn,k
1
Dfj (0); xDfj (0) = Dk fn (x).
k!
After that, replace fn (x) by fn (x; b) in the last identity.
Example 9. For fn (x) = xn in Proposition 8 we get
n!
j−1
j−1
(bn + x)n−k−1 (b(n − 1) + x),
Tn,k (bj) ; x(bj)
=
2
(k!)
and for fn (x) = n! nx in Proposition 8 we get
j! bj
n! k
bn + x
j! bj
x
Tn,k
; x
= D
.
n
bj j
bj j
k!
bn + x
Proposition 10. Let b, α, β be real numbers, r be a positive integer and (fn (x)) be a sequence
of binomial type of polynomials. We have
r
βz
Tn,k jDz=0 (e fj−1 (x + z)); αDfj (0)
(kr)!
Tn,kr β1(j=1) + jDfj−1 (0)1(j≥2) ; (kx + α)Dfj (0) ,
=
k!
or more generally,
(x + z)fj−1 (b(j − 1) + x + z) βz
fj (bj)
r
Tn,k jDz=0 (
e ); α
b(j − 1) + x + z
bj
(kr)!
fj (bj)
fj−1 (b(j − 1))
=
.
Tn,kr β1(j=1) + j
1(j≥2) ; (kx + α)
k!
b(j − 1)
bj
6
Proof. Let (fn (x)) be a sequence of binomial type defined as in the proof of Proposition 4
r
and xn = nDz=0
(eβz fn−1 (x + z)). We have yn = Dfn (x)(0) and
i k
k X
1 X ti
t
βz
r
xi
=
iDz=0 e fi−1 (x + z)
k! i≥1 i!
i!
i≥1
tk
r
F (t)kx (Dz=0
(eβz F (t)z ))k
k!
tk
= F (t)kx (β + ln F (t))kr
k!
tk kr βz
= Dz=0
(e F (t)kx+z )
k!
kr
X i
X
ti
1
t
iyi−1
=
βt +
.
exp kx yi
k!
i!
i!
i≥2
i≥1
=
Then
k
X i
n
X
1 X ti
t
t
=
xi
exp α yi
Tn,k xj ; αyj
n!
k! i≥1 i!
i!
i≥1
n≥k
kr
X
X ti
1
ti
=
βt +
iyi−1
exp (kx + α) yi
k!
i!
i!
i≥2
i≥1
n
t
(kr)! X
.
Tn,kr β1(j=1) + jyj−1 1(j≥2) ; (kx + α)yj
=
k! n≥k
n!
Then, we obtain
Tn,k
r
jDz=0
(eβz fj−1 (x
+ z)); αDfj (0)
(kr)!
=
Tn,kr β1(j=1) + jyj−1 1(j≥2) ; (kx + α)Dfj (0) .
k!
To finish the proof, replace fn (x) by fn (x; b) in the last identity.
Example 11. For fn (x) = xn in Proposition 10 we get
r
j−2 βz
j−1
Tn,k jDz=0 ((x + z)(b(j − 1) + x + z) e ); α(bj)
(kr)!
j−2
j−1
.
Tn,kr β1(j=1) + j(b(j − 1)) 1(j≥2) ; (kx + α)(bj)
=
k!
Proposition 12. Let (xn ) and (yn ) be two sequences of real numbers and r be a positive
integer. We have
kr
r
k
Bn,k j+r−1 Tj+r−1,r (xi ; yi ) = n+(r−1)k Tn+(r−1)k,kr (xi ; kyi )
r−1
(r−1)k
7
Proof. Starting with the vertical generating function of Touchard polynomials to get
X
k
X i kr
X i
tn
1
t
t
Tn,r (xi ; yi )
=
exp k yi
xi
k
n!
(r!) i≥1 i!
i!
n≥r
i≥1
(kr)! X
tn
=
T
(x
;
ky
)
.
n,kr
i
i
(r!)k n≥kr
n!
Then
(r!)k k!
Bn,k 0, . . . , 0, Tr,r (xi ; yi ), Tr+1,r (xi ; yi ), . . . ,
Tn,kr (xi ; kyi ) =
| {z }
(kr)!
r−1
and from [5, p. 450] we have
Bn,k 0, . . . , 0, ar , ar+1 , . . . =
i!ai+r−1
n!
.
Bn−(r−1)k,k
(n − (r − 1)k)!
(i + r − 1)!
Then
(r!)k n!k!
j!
Tn,kr (xj ; kyj ) =
Bn−(r−1)k,k
Tr−1+j,r (xi ; yi ) .
(kr)!(n − (r − 1)k)!
(r − 1 + j)!
Change n by n + (r − 1)k to finish the proof.
Example 13. Let r, s, p, q be nonnegative integers with q ≥ 1.
Use Application 7 in Proposition 12 to obtain
r(j − 1) + rs + p
j!(rs + p)
Bn,k
r(j − 1) + rs + p
j−1
q
n!
r(n − k) + k(rs + p)
k(rs + p)
=
.
n−k
k! r(n − k) + k(rs + p)
q
n+s−1
From the identity Tn,r (i!; (i − 1)!s) = n!
given in [5, p. 453] we obtain
r! r+s−1
n! n + k(r + s − 1) − 1
j+r+s−2
=
Bn,k j!
.
r+s−1
k(r + s) − 1
k!
From Theorems 15 and 16 given in [4] we have
n+s
n+s
Tn,r ((i − 1)!; (i − 1)!s) =
and Tn,r (1; s1(i=1) ) =
.
r+s s
r+s s
These identities and Proposition 12 give
r
j+r+s−1
Bn,k j+r−1
=
r+s
s
r−1
r
j+r+s−1
=
Bn,k j+r−1
r+s
s
r−1
8
kr
k
n+(r−1)k
(r−1)k
kr
k
n+(r−1)k
(r−1)k
n + (r + s − 1)k
(r + s)k
n + (r + s − 1)k
(r + s)k
,
ks
ks
.
References
[1] M. Aigner, Combinatorial Theory. Springer, 1979.
[2] H. Belbachir, S. Bouroubi, and A. Khelladi, Connection between ordinary multinomials,
generalized Fibonacci numbers, partial Bell partition polynomials and convolution powers
of discrete uniform distribution. Ann. Math. Inform. 35 (2008), 21–30.
[3] E. T. Bell, Exponential polynomials. Ann. Math. 35 (1934), 258–277.
[4] A. Z. Broder, The r-Stirling numbers. Discrete Math., 49 (1984), 241–259.
[5] C. A. Charalambides, Enumerative Combinatorics. Chapman & Hall/CRC, 2001.
[6] O. Chrysaphinou, On Touchard polynomials. Discrete Math. 54 (1985), 143–152.
[7] L. Comtet, Advanced Combinatorics. D. Reidel Publishing Company, Dordrecht-Holland,
1974.
[8] O. V. Kuzmin and O. V. Leonova, Touchard polynomials and their applications. Disc.
Math. Appl. 10 (2000), 391–402.
[9] M. Mihoubi, Bell polynomials and binomial type sequences. Discrete Math. 308 (2008),
2450–2459.
[10] M. Mihoubi, The role of binomial type sequences in determination identities for Bell polynomials. To appear, Ars Combin. Preprint available at
http://arxiv.org/abs/0806.3468v1.
[11] J. Touchard, Sur les cycles des substitutions, Acta Math. 70 (1939), 243–279.
2010 Mathematics Subject Classification: Primary 05A10; Secondary 05A99, 11B73, 11B75.
Keywords: Touchard polynomials; partial Bell polynomials; polynomials of binomial type.
Received October 21 2010; revised version received February 13 2011. Published in Journal
of Integer Sequences, March 25 2011.
Return to Journal of Integer Sequences home page.
9
2
3
47
6
Journal of Integer Sequences, Vol. 14 (2011),
Article 11.3.1
23 11
Touchard Polynomials, Partial Bell
Polynomials and Polynomials of Binomial
Type
Miloud Mihoubi and Mohammed Said Maamra1
University of Science and Technology Houari Boumediene (USTHB)
Faculty of Mathematics
PB 32
El Alia, 16111, Algiers
Algeria
[email protected]
[email protected]
[email protected]
Abstract
Touchard generalized the Bell polynomials in order to give some combinatorial
interpretation on permutations. Chrysaphinou introduced and studied a class of polynomials related to Touchard’s generalization. In the present paper, we establish some
relations between Touchard polynomials, Bell polynomials and the polynomials of binomial type. Several identities and relations with Stirling numbers are obtained.
1
Introduction
Among the partition polynomials, the partial Bell polynomials, introduced by Bell [3], play
an important role in different application frameworks. Several properties and identities
are given, see [5, 7, 9, 10]. Another partition polynomials, called Touchard polynomials,
introduced by Touchard [11], present an extension of the partial Bell polynomials. Some
algebraic, combinatorial and probabilistic properties of these polynomials are studied by
Touchard [11], Chrysaphinou [6], Charalambides [5], Kuzmin and Leonova [8]. In this paper,
1
Research supported by LAID3 Laboratory of USTHB University.
1
we give some relations between Touchard polynomials and partial Bell polynomials. We
exploit these relations and the polynomials of binomial type to derive some identities for
these polynomials.
In this context, let (xi ; i ≥ 1) and (yi ; i ≥ 1) be two sequences of real numbers. The
Touchard polynomials
Tn,k (xj ; yj ) := Tn,k (x1 , . . . , xn ; y1 , . . . , yn ), n ≥ k ≥ 0,
are defined by their bivariate generating function
!
X ti X ti
Tn,k (xj ; yj )u
yi
= exp u xi +
.
n!
i! i≥1 i!
n≥0 k=0
i≥1
n
XX
n
kt
The vertical generating function of Touchard polynomials, for fixed k, is given by
!k
!
X ti
X
1 X ti
tn
yi
xi
exp
, k = 0, 1, . . . .
Tn,k (xj ; yj ) =
n!
k!
i!
i!
i≥1
i≥1
n≥k
The partial Bell polynomials are given by
Bn,k (xj ) := Bn,k (x1 , . . . , xn ) = Tn,k (x1 , . . . , xn ; 0, . . . , 0), n ≥ k ≥ 0.
The polynomials of binomial type (fn (x)) are defined by
!x
X
X
un
un
, with f0 (x) = 1 and f1 (x) 6= 0 for x 6= 0.
fn (x)
=
fn (1)
n!
n!
n≥0
n≥0
For a real number a we consider in the following the sequence (fn (x; a)) defined by
fn (x; a) :=
x
fn (an + x) .
an + x
The sequence (fn (x; a)) is also of binomial type, see [9], and for more details on sequences
of binomial type see [1].
We use the following notation and hypothesis:
d
dk
dk
k
D ≡ dx
, Dk ≡ dx
k , Dx=x0 ≡ dxk |x=x0 .
For n < 0, we set fn (x) = 0, Tn,k (xj ; yj ) = 0 and Bn,k (xj ) = 0.
For x ∈ R, where R is the set of real numbers, we set
x
x
x(x − 1) · · · (x − k + 1)
x
for k = 1, 2, . . . ,
= 1 and
= 0 otherwise.
:=
k!
0
k
k
Also, for all nnnegative integers n, m we put
(
1, if m divides n;
and
1(m|n) =
0, otherwise;
2
1(n≥m) =
(
1, if n ≥ m;
0, otherwise.
2
The main results
In this section, we establish some relations between the Touchard polynomials, partial Bell
polynomials and polynomials of binomial type. Furthermore, we use these relations to develop several identities for Touchard polynomials. We start with the following theorem:
Theorem 1. Let n, k, m be integers such that n ≥ k ≥ 1, m ≥ 1; a be a real number and
(xn ) be a sequence of real numbers. Then
!
n
j/m
Bn−m,k (xj ).
Tn,k xj ; −m(j − 1)!a 1(m|j) = Bn,k (xj ) − am!
m
Proof. Let yn = −m(n − 1)!an/m 1(m|n) .
For a = 0 the theorem is trivial, otherwise, for |t| < |a|−1/m we have
!
!
X ti
X tmj
exp
yi
= exp − aj
= exp(ln(1 − atm )) = 1 − atm .
i!
j
i≥1
j≥1
Then
tn
1 X ti
Tn,k (xj ; yj ) =
xi
n!
k!
i!
i≥1
n≥k
X
!k
X ti
yi
exp
i!
i≥1
!
!k
1 X ti
=
xi
(1 − atm )
k! i≥1 i!
n
X
t
n!
Bn−m,k (xj )
.
=
Bn,k (xj ) − a
(n − m)!
n!
n≥k
Hence the theorem is proved.
If we set xn = nfn−1 (x; b) in Theorem 1 and use Proposition 1 of [9], we obtain:
Corollary 2. Let (fn (x)) be a sequence of binomial type of polynomials. We have
!
!
f
(kx;
b)
f
(kx;
b)
n!
n−k
n−m−k
−a
.
Tn,k jfj−1 (x; b); −m(j − 1)!aj/m 1(m|j) =
k!
(n − k)!
(n − m − k)!
Example 3. For fn (x) = xn in Corollary 2 we get
Tn,k jx(b(j − 1) + x)
=x
j−2
; −m(j − 1)!a
(b(n − k) + kx)
n!
(k − 1)!
(n − k)!
j/m
n−k−1
1(m|j)
!
n−m−k−1
−a
(b(n − m − k) + kx)
(n − m − k)!
3
!
1(n≥m+k) ,
and for fn (x) = n!
x
n
in Corollary 2 we get
b(j−1)+x
Tn,k xj!
=x
j−1
b(j − 1) + x
; −m(j − 1)!aj/m 1(m|j)
!
b(n−m−k)+kx
n−m−k
b(n−k)+kx
n−k
!
n!
−a
1(n≥m+k) .
(k − 1)! b(n − k) + kx
b(n − m − k) + kx
As above, for particular cases of Touchard polynomials, the following proposition gives
another expression in term of polynomials of binomial type.
Proposition 4. Let b, α be two real numbers and (fn (x)) be a sequence of binomial type of
polynomials. We have
!
n
Tn,k jfj−1 (x); αDfj (0) =
fn−k (kx + α),
k
or more generally
!
fj−1 (b(j − 1) + x)
fj (bj)
Tn,k jx
; α
b(j − 1) + x
bj
fn−k (b(n − k) + kx + α)
n
(kx + α)
=
, b 6= 0.
k
b(n − k) + kx + α
Proof. Let
X i
X
tn
t
1+
fn (x) = exp x yi
n!
i!
n≥1
i≥1
be the exponential generating function of the sequence (fn (x)) and xn = nfn−1 (x).
Necessarily yn = Dfn (0). We have
k
n
X i
X
t
1 X ti
t
exp α yi
Tn,k xj ; αyj
=
xi
n!
k! i≥1 i!
i!
i≥1
n≥k
X ti
tk
= exp (kx + α) yi
k!
i!
i≥1
tk X
tn
fn (kx + α)
k! n≥0
n!
Xn
tn+k
.
fn−k (kx + α)
=
n!
k
n≥k
=
Then, we obtain
Tn,k
n
jfj−1 (x); αDfj (0) =
fn−k (kx + α).
k
To finish the proof, replace fn (x) by fn (x; b) in the last identity.
4
Example 5. For fn (x) = xn in Proposition 4 we get
n
j−2
j−2
=
(kx + α)(b(n − k) + kx + α)n−k−1 ,
Tn,k xj(b(j − 1) + x) ; α(j − 1)(bj)
k
and for fn (x) = n! nx in Proposition 4 we get
b(j − 1) + x
j!x
j−1
Tn,k
; α(−1) (j − 1)!
b(j − 1) + x
j−1
n!
kx + α
n(n − k) + kx + α
=
.
k! n(n − k) + kx + α
n−k
Hence we may state the following:
Corollary 6. Let r, s, p be a nonnegative integers, r ≥ 1, and (xn ) be a sequence of real
numbers with x1 = 1. We have
B(r+1)(j−1)+s, r(j−1)+s (xi ) p B(r+1)j, rj (xi )
js
;
Tn,k
(r+1)(j−1)+s
(r+1)j
(r(j − 1) + s)
rj
r(j−1)+s
rj
B(r+1)(n−k)+ks+p, r(n−k)+ks+p (xi )
ks + p
n
.
=
(r+1)(n−k)+ks+p
k r(n − k) + ks + p
r(n−k)+ks+p
Proof. For x2 6= 0, let {fn (x)} be a sequence of binomial type such that fn (1) =
n
fn−k (k) we get
From the known identity Bn,k (jfj−1 (1)) =
k
−1
n+k
fn (k) =
Bn+k,k (xi ), n ≥ 0, k ≥ 1.
k
xn+1
.
n+1
Take b = r, x = s and α = p in Proposition 4 to get
fn−k (r(n − k) + ks + p)
n
fj−1 (r(j − 1) + s) fj (rj)
; p
.
=
(ks + p)
Tn,k js
r(j − 1) + s
rj
r(n − k) + ks + p
k
−1
Bn+k,k (xi ) to express in the last
Therefore, it suffices to use the identity fn (k) = n+k
k
identity fi−1 (r(i − 1) + s) and fn−k (r(n − k) + ks + p) by the partial Bell polynomials.
For the case x2 = 0 the corollary remains true by continuity.
k
Example 7. By Corollary 6 and the identity Bn,k (1!, 2!, . . . , (q + 1)!, 0, . . .) = n!
, see
k! n−k q
[2], we get
(j − 1)!p rj
r(j − 1) + s
j!s
;
Tn,k
j−1
j q
r(j − 1) + s
r
q
ks + p
r(n − k) + ks + p
n!
,
=
k! r(n − k) + ks + p
n−k
q
P k n
k
where
is the coefficients defined by (1 + x + x2 + · · · + xq )k =
x .
n q
n q
n≥0
Also, for xn = 1, xn = (−1)n−1 (n − 1)!, xn = (n − 1)! or xn = n! we get identities related
Touchard polynomials to Stirling numbers of the first and second kind.
5
The following two propositions give relations between Touchard polynomials and the
successive derivatives of polynomials of binomial type.
Proposition 8. Let b be a real number, b 6= 0, and (fn (x)) be a sequence of binomial type
of polynomials. We have
1
Tn,k Dfj (0); xDfj (0) = Dk fn (x),
k!
or more generally
Tn,k
fj (bj)
fj (bj)
; x
bj
bj
1 k
x
= D
fn (bn + x) .
k!
bn + x
Proof. Let (fn (x)) be a sequence of binomial type defined as in the proof of Proposition 4
and xn = nfn−1 (x). We have yn = Dfn (0) and
X i
X i k
X
X
tn
t
tn
t
k
= k! Tn,k (yj ; xyj ) .
exp x yi
D fn (x) =
yi
n!
i!
i!
n!
i≥1
i≥1
n≥k
n≥k
Then
Tn,k yj ; xyj
= Tn,k
1
Dfj (0); xDfj (0) = Dk fn (x).
k!
After that, replace fn (x) by fn (x; b) in the last identity.
Example 9. For fn (x) = xn in Proposition 8 we get
n!
j−1
j−1
(bn + x)n−k−1 (b(n − 1) + x),
Tn,k (bj) ; x(bj)
=
2
(k!)
and for fn (x) = n! nx in Proposition 8 we get
j! bj
n! k
bn + x
j! bj
x
Tn,k
; x
= D
.
n
bj j
bj j
k!
bn + x
Proposition 10. Let b, α, β be real numbers, r be a positive integer and (fn (x)) be a sequence
of binomial type of polynomials. We have
r
βz
Tn,k jDz=0 (e fj−1 (x + z)); αDfj (0)
(kr)!
Tn,kr β1(j=1) + jDfj−1 (0)1(j≥2) ; (kx + α)Dfj (0) ,
=
k!
or more generally,
(x + z)fj−1 (b(j − 1) + x + z) βz
fj (bj)
r
Tn,k jDz=0 (
e ); α
b(j − 1) + x + z
bj
(kr)!
fj (bj)
fj−1 (b(j − 1))
=
.
Tn,kr β1(j=1) + j
1(j≥2) ; (kx + α)
k!
b(j − 1)
bj
6
Proof. Let (fn (x)) be a sequence of binomial type defined as in the proof of Proposition 4
r
and xn = nDz=0
(eβz fn−1 (x + z)). We have yn = Dfn (x)(0) and
i k
k X
1 X ti
t
βz
r
xi
=
iDz=0 e fi−1 (x + z)
k! i≥1 i!
i!
i≥1
tk
r
F (t)kx (Dz=0
(eβz F (t)z ))k
k!
tk
= F (t)kx (β + ln F (t))kr
k!
tk kr βz
= Dz=0
(e F (t)kx+z )
k!
kr
X i
X
ti
1
t
iyi−1
=
βt +
.
exp kx yi
k!
i!
i!
i≥2
i≥1
=
Then
k
X i
n
X
1 X ti
t
t
=
xi
exp α yi
Tn,k xj ; αyj
n!
k! i≥1 i!
i!
i≥1
n≥k
kr
X
X ti
1
ti
=
βt +
iyi−1
exp (kx + α) yi
k!
i!
i!
i≥2
i≥1
n
t
(kr)! X
.
Tn,kr β1(j=1) + jyj−1 1(j≥2) ; (kx + α)yj
=
k! n≥k
n!
Then, we obtain
Tn,k
r
jDz=0
(eβz fj−1 (x
+ z)); αDfj (0)
(kr)!
=
Tn,kr β1(j=1) + jyj−1 1(j≥2) ; (kx + α)Dfj (0) .
k!
To finish the proof, replace fn (x) by fn (x; b) in the last identity.
Example 11. For fn (x) = xn in Proposition 10 we get
r
j−2 βz
j−1
Tn,k jDz=0 ((x + z)(b(j − 1) + x + z) e ); α(bj)
(kr)!
j−2
j−1
.
Tn,kr β1(j=1) + j(b(j − 1)) 1(j≥2) ; (kx + α)(bj)
=
k!
Proposition 12. Let (xn ) and (yn ) be two sequences of real numbers and r be a positive
integer. We have
kr
r
k
Bn,k j+r−1 Tj+r−1,r (xi ; yi ) = n+(r−1)k Tn+(r−1)k,kr (xi ; kyi )
r−1
(r−1)k
7
Proof. Starting with the vertical generating function of Touchard polynomials to get
X
k
X i kr
X i
tn
1
t
t
Tn,r (xi ; yi )
=
exp k yi
xi
k
n!
(r!) i≥1 i!
i!
n≥r
i≥1
(kr)! X
tn
=
T
(x
;
ky
)
.
n,kr
i
i
(r!)k n≥kr
n!
Then
(r!)k k!
Bn,k 0, . . . , 0, Tr,r (xi ; yi ), Tr+1,r (xi ; yi ), . . . ,
Tn,kr (xi ; kyi ) =
| {z }
(kr)!
r−1
and from [5, p. 450] we have
Bn,k 0, . . . , 0, ar , ar+1 , . . . =
i!ai+r−1
n!
.
Bn−(r−1)k,k
(n − (r − 1)k)!
(i + r − 1)!
Then
(r!)k n!k!
j!
Tn,kr (xj ; kyj ) =
Bn−(r−1)k,k
Tr−1+j,r (xi ; yi ) .
(kr)!(n − (r − 1)k)!
(r − 1 + j)!
Change n by n + (r − 1)k to finish the proof.
Example 13. Let r, s, p, q be nonnegative integers with q ≥ 1.
Use Application 7 in Proposition 12 to obtain
r(j − 1) + rs + p
j!(rs + p)
Bn,k
r(j − 1) + rs + p
j−1
q
n!
r(n − k) + k(rs + p)
k(rs + p)
=
.
n−k
k! r(n − k) + k(rs + p)
q
n+s−1
From the identity Tn,r (i!; (i − 1)!s) = n!
given in [5, p. 453] we obtain
r! r+s−1
n! n + k(r + s − 1) − 1
j+r+s−2
=
Bn,k j!
.
r+s−1
k(r + s) − 1
k!
From Theorems 15 and 16 given in [4] we have
n+s
n+s
Tn,r ((i − 1)!; (i − 1)!s) =
and Tn,r (1; s1(i=1) ) =
.
r+s s
r+s s
These identities and Proposition 12 give
r
j+r+s−1
Bn,k j+r−1
=
r+s
s
r−1
r
j+r+s−1
=
Bn,k j+r−1
r+s
s
r−1
8
kr
k
n+(r−1)k
(r−1)k
kr
k
n+(r−1)k
(r−1)k
n + (r + s − 1)k
(r + s)k
n + (r + s − 1)k
(r + s)k
,
ks
ks
.
References
[1] M. Aigner, Combinatorial Theory. Springer, 1979.
[2] H. Belbachir, S. Bouroubi, and A. Khelladi, Connection between ordinary multinomials,
generalized Fibonacci numbers, partial Bell partition polynomials and convolution powers
of discrete uniform distribution. Ann. Math. Inform. 35 (2008), 21–30.
[3] E. T. Bell, Exponential polynomials. Ann. Math. 35 (1934), 258–277.
[4] A. Z. Broder, The r-Stirling numbers. Discrete Math., 49 (1984), 241–259.
[5] C. A. Charalambides, Enumerative Combinatorics. Chapman & Hall/CRC, 2001.
[6] O. Chrysaphinou, On Touchard polynomials. Discrete Math. 54 (1985), 143–152.
[7] L. Comtet, Advanced Combinatorics. D. Reidel Publishing Company, Dordrecht-Holland,
1974.
[8] O. V. Kuzmin and O. V. Leonova, Touchard polynomials and their applications. Disc.
Math. Appl. 10 (2000), 391–402.
[9] M. Mihoubi, Bell polynomials and binomial type sequences. Discrete Math. 308 (2008),
2450–2459.
[10] M. Mihoubi, The role of binomial type sequences in determination identities for Bell polynomials. To appear, Ars Combin. Preprint available at
http://arxiv.org/abs/0806.3468v1.
[11] J. Touchard, Sur les cycles des substitutions, Acta Math. 70 (1939), 243–279.
2010 Mathematics Subject Classification: Primary 05A10; Secondary 05A99, 11B73, 11B75.
Keywords: Touchard polynomials; partial Bell polynomials; polynomials of binomial type.
Received October 21 2010; revised version received February 13 2011. Published in Journal
of Integer Sequences, March 25 2011.
Return to Journal of Integer Sequences home page.
9