Some Congruences for the Partial Bell Polynomials

Miloud Mihoubi^{1} University of Science and Technology Houari Boumediene Faculty of Mathematics P. O. Box 32 16111 El-Alia, Bab-Ezzouar, Algiers Algeria

               miloudmihoubi@hotmail.com

Abstract

Let B_{n,k} and A_{n} = ∑_{j=1}^{n} B_{n,j} with A_{0} = 1 be, respectively, the (n, k)^{th} partial and the n^{th} complete Bell polynomials with indeterminate arguments x_{1}, x_{2}, . . .. Congruences for A_{n} and B_{n,k} with respect to a prime number have been studied by several authors. In the present paper, we propose some results involving congruences for B_{n,k} when the arguments are integers. We give a relation between Bell polynomials and we apply it to several congruences. The obtained congruences are connected to binomial coefficients.

1 Introduction

Let x_{1}, x_{2}, … denote indeterminates. Recall that the partial Bell polynomials B_{n,k} (x_{1}, x_{2}, …) are given by

B_{n,k} (x_{1}, x_{2}, …) = ∑ \frac{n!}{k_{1}!k_{2}! ⋯} (\frac{x_{1}}{1!} )^{k_{1}} (\frac{x_{2}}{2!} )^{k_{2}} … (\frac{x_{n−k+1}}{(n − k + 1)!})^{k_{n−k+1}} ,  (1)

where the summation takes place over all integers k_{1}, k_{2}, … ≥ 0 such that

k_{1} + 2k_{2} + ⋯ + (n − k + 1) k_{n−k+1} = n and k_{1} + k_{2} + ⋯ + k_{n−k+1} = k.

1Research research supported by LAID3 Laboratory of USTHB University.

For references, see Bell [1], Comtet [4] and Riordan [7].

Congruences for Bell polynomials have been studied by several authors. Bell [1] and Carlitz [3] give some congruences for complete Bell polynomials. In this paper, we propose some congruences for partial Bell polynomials when the arguments are integers. Indeed, we give a relation between Bell polynomials, given by Theorem 1 below, and we use it in the first part of the paper, and with connection of the results of Carlitz [3] in the second part, to deduce some congruences for partial Bell polynomials. Some applications to Stirling numbers of the first and second kind and to the binomial coefficients are given.

2 Main results

The next theorem gives an interesting relation between Bell polynomials. We use it to establish some congruences for partial Bell polynomials.

Theorem 1. Let {x_{n}} be a real sequence. Then for n, r, k integers with n, r, k ≥ 1, we have

x_{1}^{k} ∑_{j=1}^{n} B_{n,j} (y_{1}, y_{2}, …) (k − nr)^{j−1} = x_{1}^{nr} \frac{B_{n+k,k} (x_{1}, x_{2}, x_{3}, …)}{k\binom{n+k}{k}}  (2)

with y_{n} = \frac{B_{(r+1)n,nr} (x_{1}, x_{2}, x_{3}, …)}{nr\binom{(r+1)n}{nr}}.

For k = nr + s, Identity (2) becomes

Remark 2. Let {x_{n}} be a real sequence. Then for n, r, s integers with n, r ≥ 1, we get

x_{1}^{s}A_{n} (sy_{1}, sy_{2}, …) = \frac{s}{nr + s} \frac{B_{(r+1)n+s,nr+s} (x_{1}, x_{2}, x_{3}, …)}{\binom{(r+1)n+s}{nr+s}}, s ≥ −nr + 1.  (3)

For s ≥ 0, we obtain Proposition 8 in [5], (see also [6]).

Theorem 3. Let k, s be a nonnegative integers and p be a prime number. Then for any sequence {x_{j}} of integers we have

(k + s + 1) B_{sp,k+s+1} (x_{1}, x_{2}, …) ≡ 0 (mod p).

Application 4. If we denote by s (n, k) and S (n, k) for Stirling numbers of first and second kind respectively, then from the well-known identities

B_{n,k} (0!, −1!, 2!, …) = s (n, k) and B_{n,k} (1, 1, 1, …) = S (n, k)

when x_{n} = 1 or x_{n} = (−1)^{n−1} (n − 1)! in Theorem 3 we obtain

(k + s + 1) S (sp, k + s + 1) ≡ (k + s + 1) s (sp, k + s + 1) ≡ 0 (mod p).

Theorem 5. Let n, k, s be integers with n ≥ k ≥ 1, s ≥ 1 and p be a prime number. Then for any sequence {x_{j}} of integers with x_{1} not a multiple of p we have

\frac{B_{n+sp,k+sp} (x_{1}, x_{2}, x_{3}, …)}{(k + sp) \binom{n+sp}{k+sp}} ≡ x_{1}^{s} \frac{B_{n,k} (x_{1}, x_{2}, x_{3}, …)}{k\binom{n}{k}} (mod p) if p > n − k + 1

x_{1}^{n} \frac{B_{n+sp,sp} (x_{1}, x_{2}, x_{3}, …)}{s\binom{n+sp}{sp}} ≡ x_{1}^{s} \frac{B_{(p+1)n,np} (x_{1}, x_{2}, x_{3}, …)}{n\binom{(p+1)n}{np}} (mod p^{2}) if p > n + 1.

(4)

Application 6. If we consider the cases k = 1 and k = 2 in Theorem 5 we obtain

B_{n+sp,1+sp} (x_{1}, x_{2}, x_{3}, …) ≡ x_{1}^{s}x_{n} (mod p) for p > n,

B_{n+sp,2+sp} (x_{1}, x_{2}, x_{3}, …) ≡ \frac{x_{1}^{s}}{2} ∑_{j=1}^{n−1} \binom{n}{j}x_{j}x_{n−j} (mod p) for p > n − 1.

Then, when x_{n} = 1 or x_{n} = (−1)^{n−1} (n − 1)! we obtain

S (n + sp, 1 + sp) ≡ 1 (mod p) for p > n,
s (n + sp, 1 + sp) ≡ (−1)^{n−1} (n − 1)! (mod p) for p > n,
S (n + sp, 2 + sp) ≡ 2^{n−1} − 1 (mod p) for p > n − 1 and
s (n + sp, 2 + sp) ≡ (−1)^{n−1} \frac{n(n+1)^{2}}{2} (mod p) for p > n − 1,

Theorem 7. Let n, k, s, p be integers with n ≥ k ≥ 1, s ≥ 1, p ≥ 1. Then for any sequence {x_{j}} of integers with x_{1} not a multiple of p we have

\frac{B_{(s+1)n,sn} (x_{1}, 2x_{2}, 3x_{3}, …)}{\binom{(s+1)n}{sn}} ≡ sx_{1}^{n(s−1)} \frac{B_{2n,n} (x_{1}, 2x_{2}, 3x_{3}, …)}{\binom{2n}{n}} (mod n^{2}),

\frac{B_{n+sp,k+sp} (x_{1}, 2x_{2}, 3x_{3}, …)}{(k + sp) \binom{n+sp}{k+sp}} ≡ x_{1}^{s} \frac{B_{n,k} (x_{1}, 2x_{2}, 3x_{3}, …)}{k\binom{n}{k}} (mod p),

x_{1}^{n} \frac{B_{n+sp,sp} (x_{1}, 2x_{2}, 3x_{3}, …)}{s\binom{n+sp}{sp}} ≡ x_{1}^{s} \frac{B_{(p+1)n,np} (x_{1}, 2x_{2}, 3x_{3}, …)}{n\binom{(p+1)n}{np}} (mod n^{2}).

(5)

Application 8. Belbachir et al. [2] have proved that

B_{n,k} (1!, 2!, …, (q + 1)!, 0, …) = \frac{n!}{k!}\binom{k}{n − k}_{q},  (6)

then, for s ≥ 1 and p ∤ j, the two last congruences of (5) and Identity (6) prove that

\binom{k + sp}{j}_{q} ≡ \binom{k}{j}_{q} (mod p) and j\binom{sp}{j}_{q} ≡ s\binom{jp}{j}_{q} (mod p^{2}).

Corollary 9. Let n, k, s be integers with n ≥ k ≥ 1 and p be a prime number. Then for any sequence {x_{j}} of integers with x_{1} not a multiple of p we have

\frac{B_{(p+1)n,np} (x_{1}, x_{2}, …)}{n\binom{(p+1)n}{np}} ≡ x_{1}^{n−1} \frac{B_{n+p,p} (x_{1}, x_{2}, x_{3}, …)}{\binom{n+p}{p}} (mod p^{2}) if p > n + 1,

\frac{B_{(p+1)n,np} (x_{1}, 2x_{2}, 3x_{3}, …)}{n\binom{(p+1)n}{np}} ≡ x_{1}^{n−1} \frac{B_{n+p,p} (x_{1}, 2x_{2}, 3x_{3}, …)}{\binom{n+p}{p}} (mod p^{2}).

Application 10. As in Application 8, we have

\binom{jp}{j}_{q} ≡ j\binom{p}{j}_{q} (mod p^{2}).

Theorem 11. Let k ≥ 2, j ≥ 1 be integers and p be an odd prime number. Then for any sequence of integers {x_{j}} we have

\frac{B_{p^{j}+k,k} (x_{1}, 2x_{2}, 3x_{3}, …)}{k\binom{p^{j}+k}{k}} ≡ x_{1}^{k−1}x_{p^{j}+1} (mod p) if p ∤ kx_{1},

\frac{B_{(r+1)p^{j},p^{j}r} (x_{1}, 2x_{2}, 3x_{3}, …)}{p^{j}r\binom{(r+1)p^{j}}{p^{j}r}} ≡ x_{1}^{r−1} (x_{p^{j}+1} − x_{p^{j−1}+1}) (mod p) if p ∤ x_{1}.

(7)

Application 12. As in Application (8), let j = 1 in the second congruence of Theorem 11. Then

\frac{(p − 1)!}{r}\binom{pr}{p}_{q} ≡ −1 (mod p).

Theorem 13. Let k ≥ 2, j ≥ 1 be integers and p be an odd prime number. Then for any sequence of integers {x_{j}} we have

\frac{B_{2p^{j}+k,k} (x_{1}, 2x_{2}, 3x_{3}, …)}{k\binom{2p^{j}+k}{k}} ≡ x_{1}^{k−2} ((k − 1) x_{p^{j}+1}^{2} + x_{1}x_{2p^{j}+1}) (mod p) if p ∤ kx_{1}

\frac{B_{2(r+1)p^{j},2p^{j}r} (x_{1}, 2x_{2}, 3x_{3}, …)}{2p^{j}r\binom{2(r+1)p^{j}}{2p^{j}r}} ≡ x_{1}^{2r−2} (x_{1}x_{2p^{j}+1} − x_{p^{j}+1}^{2}) (mod p) if p ∤ x_{1}.

Remark 14. Similarly to the last proofs, one can exploit the results of Carlitz [3] with connection to Theorem 1 to obtain more congruences for partial Bell polynomials.

3 Proof of the main results

Proof of Theorem 1. Let {x_{n}} be a sequence of real numbers with x_{1} := 1 and let {f_{n} (x)} be a sequence of polynomials defined by

f_{n} (x) = ∑_{j=1}^{n} B_{n,j} (\frac{x_{2}}{2} , \frac{x_{3}}{3} , \frac{x_{4}}{4} …) (x)_{j} ,

with f_{0} (x) = 1, (x)_{j} := x (x − 1) ⋯ (x − j + 1) for j ≥ 1 and (x)_{0} := 1.

We have nf_{n−1} (1) = n ∑_{j=1}^{n−1} B_{n−1,j} (\frac{x_{2}}{2} , \frac{x_{3}}{3} , …) (1)_{j} = x_{n} and D_{x=0}f_{1} (0) = 1 ≠ 0.

It is well known that {f_{n} (x)} presents a sequence of binomial type, see [4]. Then, from Proposition 1 in [5] we have

y_{n} = \frac{1}{nr\binom{(r+1)n}{nr}}B_{(r+1)n,nr} (1, x_{2}, x_{3}, …) = \frac{f_{n} (nr)}{nr} = D_{x=0}f_{n} (x; r) ,  (8)

where {f_{n} (x; a)} is a sequence of binomial type defined by

f_{n} (x; a) := \frac{x}{an + x}f_{n} (an + x)  (9)

with a is a real number, see [5]. From Proposition 1 in [5] we have also

\frac{B_{n+k,k} (1, x_{2}, x_{3}, …)}{k\binom{n+k}{k}} = \frac{f_{n} (k)}{k} = \frac{f_{n} (k − nr; r)}{k − nr},  (10)

but from [8] we can write f_{n} (k − nr; r) as

f_{n} (k − nr; r) = ∑_{j=1}^{n} B_{n,j} (D_{x=0}f_{1} (x; r) , D_{x=0}f_{2} (x; r) , …) (k − nr)^{j} .  (11)

Then, by substitution (11) in (10) and by using (8) we obtain

\frac{B_{n+k,k} (1, x_{2}, x_{3}, …)}{k\binom{n+k}{k}} = ∑_{j=1}^{n} B_{n,j} (y_{1}, y_{2}, …) (k − nr)^{j−1} .  (12)

We can verify that Identity (2) is true for x_{1} = 0, and, for x_{1} ≠ 0 it can be derived from (12) by replacing x_{n} by \frac{x_{n}}{x_{1}} and by using the well known identities

\begin{matrix}B_{n,k} (xa_{1}, xa_{2}, xa_{3}, …) = x^{k}B_{n,k} (a_{1}, a_{2}, a_{3}, …) and \\ B_{n,k} (xa_{1}, x^{2}a_{2}, x^{3}a_{3}, …) = x^{n}B_{n,k} (a_{1}, a_{2}, a_{3}, …) , \end{matrix}  (13)

where {a_{n}} is any real sequence. Proof of Theorem 3. We prove that kB_{sp,k} ≡ 0 (mod p), k ≥ s + 1. From the identities

\binom{sp}{j} ≡ 0 (mod p), for p ∤ j and \binom{sp}{pj} ≡ \binom{s}{j} (mod p),  (14)

and from the recurrence relation given by

kB_{n,k} = ∑_{j}\binom{n}{j}x_{j}B_{n−j,k−1}

with B_{n,k} := B_{n,k} (x_{1}, x_{2}, …) and x_{j} = 0 for j ≤ 0, we obtain

(k + 1) B_{sp,k+1} = ∑_{j}\binom{sp}{j}x_{j}B_{sp−j,k} ≡ ∑_{j=1}^{s} \binom{s}{j}x_{jp}B_{(s−j)p,k} (mod p).

Then, for s = 0, we get kB_{0,k} ≡ 0 (mod p), k ≥ 0. For s = 1, we get (k + 1) B_{p,k+1} ≡ x_{p}B_{0,k} ≡ 0 (mod p), k ≥ 1. For s = 2, the last congruences imply that (k + 1) B_{2p,k+1} ≡ 2x_{p}B_{p,k} + x_{2p}B_{0,k} = 0 (mod p), k ≥ 2 and p ∤ k. The induction on s proves that kB_{sp,k} ≡ 0 (mod p) when k ≥ s + 1. Proof of Theorem 5. From [4] we have

B_{n,k} (x_{1}, x_{2}, …) = \frac{n!}{(n − k)!} ∑_{j=0}^{k} B_{n−k,k−j} (\frac{x_{2}}{2} , \frac{x_{3}}{3} , \frac{x_{4}}{4} , …) \frac{x_{1}^{j}}{j!} , n ≥ k ≥ 1.  (15)

Then, for i ∈ {1, …, n} , the last identity and Identities (13) imply

\begin{matrix}t_{i} = ((n + 1)!)^{i} y_{i} = \frac{((n + 1)!)^{i}}{ir\binom{(r+1)i}{ir}} B_{(r+1)i,ir} (x_{1}, x_{2}, …, x_{i−j+1}) = \\ ∑_{j=1}^{i} \frac{(ir − 1)!}{(ir − j)!}x_{1}^{ir−j}B_{i,j} (\frac{(n + 1)!}{2}x_{2}, \frac{((n + 1)!)^{2}}{3} x_{3}, …, \frac{((n + 1)!)^{i−j}}{i − j + 1} x_{i−j+1}) , \end{matrix}

from which we deduce that t_{1}, …, t_{n} are integers, and then B_{n,1} (t_{1}, t_{2}, …) , …, B_{n,n} (t_{1}, t_{2}, …) are also integers. Therefore, by using the second identity of (13), Identity (2) becomes

x_{1}^{k} ∑_{j=1}^{n} B_{n,j} (t_{1}, t_{2}, …) (k − nr)^{j−1} = x_{1}^{nr} \frac{((n + 1)!)^{n} B_{n+k,k} (x_{1}, x_{2}, x_{3}, …)}{k\binom{n+k}{k}}.

Hence, when we replace k by α + sp in the last identity we obtain

x_{1}^{α+sp} ∑_{j=1}^{n} B_{n,j} (t_{1}, t_{2}, …) (α + sp − nr)^{j−1} = x_{1}^{nr} \frac{((n + 1)!)^{n} B_{n+α+sp,α+sp} (x_{1}, x_{2}, x_{3}, …)}{(α + sp) \binom{n+α+sp}{α+sp}},

and when we reduce modulo p in the last identity we obtain

x_{1}^{nr} \frac{((n + 1)!)^{n} B_{n+α+sp,α+sp} (x_{1}, x_{2}, x_{3}, …)}{(α + sp) \binom{n+α+sp}{α+sp}} ≡ x_{1}^{α+s} ∑_{j=1}^{n} B_{n,j} (t_{1}, t_{2}, …) (α − nr)^{j−1} (mod p).

But from (2) we have

x_{1}^{α} ∑_{j=1}^{n} B_{n,j} (t_{1}, t_{2}, …) (α − nr)^{j−1} = ((n + 1)!)^{n} x_{1}^{α} ∑_{j=1}^{n} B_{n,j} (y_{1}, y_{2}, …) (α − nr)^{j−1}

= ((n + 1)!)^{n} x_{1}^{nr} \frac{B_{n+α,α} (x_{1}, x_{2}, x_{3}, …)}{α\binom{n+α}{α}},

from which the last congruence becomes

\frac{((n + 1)!)^{n} B_{n+α+sp,α+sp} (x_{1}, x_{2}, x_{3}, …)}{(α + sp) \binom{n+α+sp}{α+sp}} ≡ x_{1}^{s} ((n + 1)!)^{n} \frac{B_{n+α,α} (x_{1}, x_{2}, x_{3}, …)}{α\binom{n+α}{α}} (mod p).

Now, when we replace n by n − α, the last congruence becomes

\frac{((n − α + 1)!)^{n} B_{n+sp,α+sp} (x_{1}, x_{2}, x_{3}, …)}{(α + sp) \binom{n+sp}{α+sp}} ≡ x_{1}^{s} ((n − α + 1)!)^{n} \frac{B_{n,α} (x_{1}, x_{2}, x_{3}, …)}{α\binom{n}{α}} (mod p).

Then, if p > n − α + 1 we obtain

\frac{B_{n+sp,α+sp} (x_{1}, x_{2}, …)}{(α + sp) \binom{n+sp}{α+sp}} ≡ x_{1}^{s} \frac{B_{n,α} (x_{1}, x_{2}, x_{3}, …)}{α\binom{n}{α}} (mod p).

For the second part of theorem, when we replace k by sr in (2) we get

x_{1}^{sr} ∑_{j=1}^{n} B_{n,j} (t_{1}, t_{2}, …) r^{j−1} (s − n)^{j−1} = x_{1}^{nr} \frac{((n + 1)!)^{n} B_{n+sr,sr} (x_{1}, x_{2}, x_{3}, …)}{sr\binom{n+sr}{sr}},

and, because B_{n,j} (t_{1}, t_{2}, …) (1 ≤ j ≤ n) are integers, the last identity proves that

x_{1}^{nr} \frac{((n + 1)!)^{nr} B_{n+sr,sr} (x_{1}, x_{2}, x_{3}, …)}{sr\binom{n+sr}{sr}} ≡ x_{1}^{sr}z_{n}

≡ x_{1}^{sr} \frac{((n + 1)!)^{nr} B_{(r+1)n,nr} (x_{1}, x_{2}, x_{3}, …)}{nr\binom{(r+1)n}{nr}} (mod r).

Let r = p > n + 1 be a prime number. Now, because the expressions

\frac{((n + 1)!)^{np} B_{n+sp,sp} (x_{1}, x_{2}, x_{3}, …)}{sp\binom{n+sp}{sp}} and \frac{((n + 1)!)^{np} B_{(p+1)n,np} (x_{1}, x_{2}, x_{3}, …)}{np\binom{(p+1)n}{np}}

are integers, we obtain

x_{1}^{n} \frac{B_{n+sp,sp} (x_{1}, x_{2}, x_{3}, …)}{s\binom{n+sp}{sp}} ≡ x_{1}^{s} \frac{B_{(p+1)n,np} (x_{1}, x_{2}, x_{3}, …)}{n\binom{(p+1)n}{np}} (mod p^{2}).

Proof of Theorem 7. From Identity (15) we get

\frac{B_{n+k,k} (x_{1}, 2x_{2}, 3x_{3}, …)}{k\binom{n+k}{k}} = ∑_{j=1}^{k} \frac{(k − 1)!}{(k − j)!}B_{n,j} (x_{2}, x_{3}, x_{4}, …) x_{1}^{k−j}, n, k ≥ 1  (16)

and this implies that the numbers

z_{n} = \frac{B_{(r+1)n,nr} (x_{1}, 2x_{2}, 3x_{3}, …)}{nr\binom{(r+1)n}{nr}}, n ≥ 1  (17)

are integers, and then, the numbers B_{n,j} (z_{1}, z_{2}, …) (1 ≤ j ≤ n) are also integers. From Identity (3), when we replace r by 1 and s by n (s − 1) we obtain

B_{(s+1)n,sn} (x_{1}, 2x_{2}, 3x_{3}, …) = x_{1}^{n(s−1)}ns\binom{(s + 1) n}{sn} ∑_{j=1}^{n} B_{n,j} (\overline{z}_{1}, \overline{z}_{2}, …) ((s − 1) n)^{j−1} , (18)

with \overline{z}_{n} := _{\frac{1}{n\binom{2n}{n}}}B_{2n,n} (x_{1}, 2x_{2}, …) .

Furthermore, from (18), we have

\frac{B_{(s+1)n,sn} (x_{1}, 2x_{2}, 3x_{3}, …)}{\binom{(s+1)n}{sn}} = nx_{1}^{n(s−1)}s ∑_{j=1}^{n} B_{n,j} (z_{1}, z_{2}, …) ((s − 1) n)^{j−1}

≡ n {x_{1}^{n(s−1)}sz_{n}}

≡ n \{x_{1}^{n(s−1)}s_{\frac{1}{n\binom{2n}{n}}}B_{2n,n} (x_{1}, 2x_{2}, …)\}

≡ x_{1}^{n(s−1)}s\frac{1}{\binom{2n}{n}}B_{2n,n} (x_{1}, 2x_{2}, …) (mod n^{2}), i.e.,

\frac{B_{(s+1)n,sn} (x_{1}, 2x_{2}, 3x_{3}, …)}{\binom{(s+1)n}{sn}} ≡ sx_{1}^{n(s−1)} \frac{B_{2n,n} (x_{1}, 2x_{2}, 3x_{3}, …)}{\binom{2n}{n}} (mod n^{2}).

For the second part of (5), when we replace k by α + sp in (2), we obtain

x_{1}^{α+sp} ∑_{j=1}^{n} B_{n,j} (z_{1}, z_{2}, …) (α + sp − nr)^{j−1} = x_{1}^{nr} \frac{B_{n+α+sp,α+sp} (x_{1}, 2x_{2}, 3x_{3}, …)}{(α + sp) \binom{n+α+sp}{α+sp}}

with z_{n} is given by (17). Because the numbers B_{n,j} (z_{1}, z_{2}, …) , 1 ≤ j ≤ n, are integers, then when we reduce modulo p in the last identity we get

x_{1}^{nr} \frac{B_{n+α+sp,α+sp} (x_{1}, 2x_{2}, 3x_{3}, …)}{(α + sp) \binom{n+α+sp}{α+sp}} ≡ x_{1}^{α+s} ∑_{j=1}^{n} B_{n,j} (z_{1}, z_{2}, …) (α − nr)^{j−1} (mod p)

and by (2) the last congruence becomes

\frac{B_{n+α+sp,α+sp} (x_{1}, 2x_{2}, 3x_{3}, …)}{(α + sp) \binom{n+α+sp}{α+sp}} ≡ x_{1}^{s} \frac{B_{n+α,α} (x_{1}, 2x_{2}, 3x_{3}, …)}{α\binom{n+α}{α}} (mod p).

To terminate, it suffices to replace n by n − α in the last congruence.

For the third part of (5), when we replace k by kr in (2), we obtain

x_{1}^{kr} ∑_{j=1}^{n} B_{n,j} (z_{1}, z_{2}, …) (kr − nr)^{j−1} = x_{1}^{nr} \frac{B_{n+kr,kr} (x_{1}, 2x_{2}, 3x_{3}, …)}{rk\binom{n+kr}{kr}}

and because the numbers B_{n,j} (z_{1}, z_{2}, …) , 1 ≤ j ≤ n, are integers, then when we reduce modulo r in the last identity we get

x_{1}^{nr} \frac{B_{n+kr,kr} (x_{1}, 2x_{2}, 3x_{3}, …)}{rk\binom{n+kr}{kr}} ≡ x_{1}^{kr}z_{n} ≡ \frac{x_{1}^{kr}B_{(r+1)n,nr} (x_{1}, 2x_{2}, 3x_{3}, …)}{nr\binom{(r+1)n}{nr}} (mod p).

Now, because

\frac{B_{n+kr,kr} (x_{1}, 2x_{2}, 3x_{3}, …)}{rk\binom{n+kr}{kr}} and \frac{x_{1}^{kr}B_{(r+1)n,nr} (x_{1}, 2x_{2}, 3x_{3}, …)}{nr\binom{(r+1)n}{nr}}

are integers and x_{1}^{p} ≡ x_{1} (mod p) for any prime number p, then when we put r = p, the last congruence becomes

x_{1}^{n} \frac{B_{n+kp,kp} (x_{1}, 2x_{2}, 3x_{3}, …)}{kp\binom{n+kp}{kp}} ≡ x_{1}^{k} \frac{B_{(p+1)n,np} (x_{1}, 2x_{2}, 3x_{3}, …)}{np\binom{(p+1)n}{np}} (mod p).

To complete this proof, it suffices to multiply the two sides of the last congruence by p.

Proof of Corollary 9. From the first congruence of (4) when we replace s by s−1, n by n+p and k by p we get

\frac{B_{n+sp,sp} (x_{1}, x_{2}, x_{3}, …)}{s\binom{n+sp}{sp}} ≡ x_{1}^{s−1} \frac{B_{n+p,p} (x_{1}, x_{2}, x_{3}, …)}{\binom{n+p}{p}} (mod p^{2}), p > n + 1, s ≥ 1,

and by combining the last congruence and the second congruence of (4) we obtain

\frac{B_{(p+1)n,np} (x_{1}, x_{2}, x_{3}, …)}{n\binom{(p+1)n}{np}} ≡ x_{1}^{n−1} \frac{B_{n+p,p} (x_{1}, x_{2}, x_{3}, …)}{\binom{n+p}{p}} (mod p^{2}), p > n + 1.

Similarly, we use the second and the third congruences of (5) to get the second part of the corollary.

Proof of Theorem 11. Identity (2) can be written as

\begin{matrix}x_{1}^{pr} (k − p) \frac{B_{p+k,k} (x_{1}, 2x_{2}, 3x_{3}, …)}{k\binom{p+k}{k}} = x_{1}^{k}A_{p} ((k − p) z_{1}, (k − p) z_{2}, …) , \\ with z_{n} = \frac{B_{(r+1)n,nr} (x_{1}, 2x_{2}, 3x_{3}, …)}{nr\binom{(r+1)n}{nr}}, k ≥ 1. \end{matrix}  (19)

Bell [1] showed, for any indeterminates x_{1}, x_{2}, …, that

A_{p} (x_{1}, x_{2}, x_{3}, …) ≡ x_{1}^{p} + x_{p} (mod p).  (20)

Therefore, from (20) and (19), we obtain

x_{1}^{pr} (k − p) \frac{B_{p+k,k} (x_{1}, 2x_{2}, 3x_{3}, …)}{k\binom{p+k}{k}} ≡ x_{1}^{k} \{(k − p)^{p} z_{1}^{p} + (k − p) z_{p}\} (mod p),

and Identity (16) shows that \frac{B_{p+k,k} (x_{1}, 2x_{2}, 3x_{3}, …)}{k\binom{p+k}{k}} and the terms of the sequence \{z_{n}; n ≥ 1\}

are integers. Now, because z_{1} = x_{1}^{r−1}x_{2}, then, when k is not a multiple of p, the last congruence and Fermat little Theorem prove that

x_{1}^{r} \frac{B_{p+k,k} (x_{1}, 2x_{2}, 3x_{3}, …)}{k\binom{p+k}{k}} ≡ x_{1}^{k−1} \{x_{1}^{r}x_{2} + x_{1}y_{p}\} (mod p).

For k = 1 in the last congruence we have

y_{p} ≡ x_{1}^{r−1}x_{p+1} − x_{1}^{r−1}x_{2} (mod p).

The proof for j = 1 results from the two last congruences. Assume now that the congruences given by (7) are true for the index j. Carlitz [1] showed, for any indeterminates x_{1}, x_{2}, …, that

A_{p^{j}} ≡ x_{1}^{p^{j}} + x_{p}^{p^{j−1}} + x_{p^{2}}^{p^{j−2}} + ⋯ + x_{p^{j}} (mod p).

For x_{1}, x_{2}, … integers we obtain

A_{p^{j}} ≡ x_{1} + x_{p} + x_{p^{2}} + ⋯ + x_{p^{j}} (mod p).

Then, when we use Identity (19) and the fact that the sequence {z_{n}; n ≥ 1} is a sequence of integers, we obtain when p ∤ kx_{1}

x_{1}^{r} \frac{B_{p^{j+1}+k,k} (x_{1}, 2x_{2}, 3x_{3}, …)}{k\binom{p^{j+1}+k}{k}} ≡ x_{1}^{k} (z_{1} + z_{p} + z_{p^{2}} + ⋯ + z_{p^{j+1}})

≡ x_{1}^{k} (z_{1} + z_{p} + z_{p^{2}} + ⋯ + z_{p^{j}}) + x_{1}^{k}z_{p^{j+1}}

≡ x_{1}^{r} \frac{B_{p^{j}+k,k} (x_{1}, 2x_{2}, 3x_{3}, …)}{k\binom{p^{j}+k}{k}} + x_{1}^{k}z_{p^{j+1}}

≡ x_{1}^{k−1}x_{p^{j}+1} + x_{1}^{k}z_{p^{j+1}} (mod p).

For k = 1 in the last congruence we have

x_{1}^{r}x_{p^{j+1}+1} ≡ x_{p^{j}+1} + x_{1}z_{p^{j+1}} (mod p).

From the two last congruences we deduce that

\frac{B_{p^{j+1}+k,k} (x_{1}, 2x_{2}, 3x_{3}, …)}{k\binom{p^{j}+k}{k}} ≡ x_{1}^{k−1}x_{p^{j+1}+1} (mod p) if p ∤ kx_{1},

\frac{B_{(r+1)p^{j+1},p^{j+1}r} (x_{1}, 2x_{2}, 3x_{3}, …)}{p^{j+1}r\binom{(r+1)p^{j+1}}{p^{j+1}r}} ≡ x_{1}^{r−1} (x_{p^{j+1}+1} − x_{p^{j}+1}) (mod p) if p ∤ x_{1},

which completes the proof.

Proof of Theorem 13. Carlitz [1] showed, for any indeterminates x_{1}, x_{2}, …, that

A_{2p^{j}} ≡ A_{p^{j}}^{2} + x_{2p^{j}} (mod p).

Then, for x_{1}, x_{2}, … integers we get

A_{2p^{j}} ≡ (x_{1}^{p^{j}} + x_{p}^{p^{j−1}} + ⋯ + x_{p^{j}})^{2} + x_{2p^{j}} ≡ (x_{1} + x_{p} + ⋯ + x_{p^{j}})^{2} + x_{2p^{j}} (mod p), and, when we use Identity (19), we obtain

x_{1}^{2p^{j}r} (k − 2p^{j}r) \frac{B_{2p^{j}+k,k} (x_{1}, 2x_{2}, 3x_{3}, …)}{k\binom{2p^{j}+k}{k}} = x_{1}^{k}A_{2p^{j}} ((k − 2p^{j}r) z_{1}, (k − 2p^{j}r) z_{2}, …) ,

and because {z_{n}; n ≥ 1} is a sequence of integers, the last identity gives

\begin{matrix}x_{1}^{2p^{j}r} (k − 2p^{j}r) \frac{B_{2p^{j}+k,k} (x_{1}, 2x_{2}, 3x_{3}, …)}{k\binom{2p^{j}+k}{k}} ≡ \\ x_{1}^{k} ((k − 2p^{j}r)^{2} (z_{1} + z_{p} + z_{p^{2}} + ⋯ + z_{p^{j}})^{2} + (k − 2p^{j}r) z_{2p^{j}}) (mod p). \end{matrix}

From the proof of Theorem 11, the last congruence gives when p ∤ kx_{1}

x_{1}^{k} (x_{1}^{2r} \frac{B_{2p^{j}+k,k} (x_{1}, 2x_{2}, 3x_{3}, …)}{k\binom{2p^{j}+k}{k}}) ≡ kx_{1}^{2k} (z_{1} + z_{p} + z_{p^{2}} + ⋯ + z_{p^{j}})^{2} + x_{1}^{2k}z_{2p^{j}}

≡ k (x_{1}^{r} \frac{B_{p^{j}+k,k} (x_{1}, 2x_{2}, 3x_{3}, …)}{k\binom{p^{j}+k}{k}})^{2} + x_{1}^{2k}z_{2p^{j}}

≡ k (x_{1}^{r+k−1}x_{p^{j}+1})^{2} + x_{1}^{2k}z_{2p^{j}} (mod p), i.e.,

x_{1}^{2r} \frac{B_{2p^{j}+k,k} (x_{1}, 2x_{2}, 3x_{3}, …)}{k\binom{2p^{j}+k}{k}} ≡ x_{1}^{k} (kx_{1}^{2r−2}x_{p^{j}+1}^{2} + z_{2p^{j}}) (mod p).

For k = 1 in the last congruence we get x_{1}^{2r−1}x_{2p^{j}+1} ≡ x_{1}^{2r−2}x_{p^{j}+1}^{2} + z_{2p^{j}}, i.e.,

z_{2p^{j}} ≡ x_{1}^{2r−2} (x_{1}x_{2p^{j}+1} − x_{p^{j}+1}^{2}) (mod p).

Then

x_{1}^{2} \frac{B_{2p^{j}+k,k} (x_{1}, 2x_{2}, 3x_{3}, …)}{k\binom{2p^{j}+k}{k}} ≡ x_{1}^{k} ((k − 1) x_{p^{j}+1}^{2} + x_{1}x_{2p^{j}+1}) (mod p).

4 Acknowledgements

The author thanks the anonymous referee for his/her careful reading and valuable comments and suggestions.

References

[1] E. T. Bell, Exponential polynomials. Ann. Math. 35 (1934), 258–277.

[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. Annales Mathematicae et Informaticae 35 (2008), 21–30.

[3] L. Carlitz, Some congruences for the Bell polynomials. Pacific J. Math. 11 (1961), 1215– 1222.

[4] L. Comtet, Advanced Combinatorics. D. Reidel Publishing Company, Dordrecht-Holland, 1974.

[5] M. Mihoubi, Bell polynomials and binomial type sequences. Discrete Math. 308 (2008), 2450–2459.

[6] M. Mihoubi, The role of binomial type sequences in determination identities for Bell polynomials. To appear, Ars Combinatoria. Preprint available at

   http://arxiv.org/abs/0806.3468v1.

[7] J. Riordan, Combinatorial Identities. R. E. Krieger, 1979.

[8] S. Roman, The Umbral Calculus. Academic Press, 1984.

2000 Mathematics Subject Classification: Primary 05A10; Secondary 11B73, 11B75, 11P83. Keywords: Bell polynomials, congruences, Stirling numbers, binomial coefficients.

Received January 10 2009; revised version received April 29 2009. Published in Journal of Integer Sequences, May 12 2009.

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