3 47 6 3 Journal of Integer Seuences, Vol. 009), Article 09.4. Some Congruences for the Partial Bell Polynomials Miloud Mihoubi University of Science and Technology Houari Boumediene Faculty of Mathematics P. O. Box 3 6 El-Alia, Bab-Ezzouar, Algiers Algeria miloudmihoubi@hotmail.com Abstract Let B n, and A n = n j= B n,j with A 0 = be, respectively, the n, ) th partial and the n th complete Bell polynomials with indeterminate arguments x, x,.... Congruences for A n and B n, 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, 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. Introduction Let x,x,... denote indeterminates. Recall that the partial Bell polynomials B n, x,x,...) are given by B n, x,x,...) = n!!! x! ) x ) ) n + xn +..., )! n + )! where the summation taes place over all integers,,... 0 such that + + + n + ) n + = n and + + + n + =. Research research supported by LAID3 Laboratory of USTHB University.
For references, see Bell [], Comtet [4] and Riordan [7]. Congruences for Bell polynomials have been studied by several authors. Bell [] 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 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 ind and to the binomial coefficients are given. Main results The next theorem gives an interesting relation between Bell polynomials. We use it to establish some congruences for partial Bell polynomials. Theorem. Let {x n } be a real seuence. Then for n,r, integers with n,r,, we have x j= B n,j y,y,...) ) j = x B n+, x,x,x 3,...) ) ) n+ with y n = B r+)n, x,x,x 3,...) ). r+)n For = + s, Identity ) becomes Remar. Let {x n } be a real seuence. Then for n,r,s integers with n,r, we get x s A n sy,sy,...) = s + s B r+)n+s,+s x,x,x 3,...) ), s +. 3) r+)n+s +s For s 0, we obtain Proposition 8 in [5], see also [6]). Theorem 3. Let,s be a nonnegative integers and p be a prime number. Then for any seuence {x j } of integers we have + s + )B sp,+s+ x,x,...) 0 mod p). Application 4. If we denote by s n,) and S n,) for Stirling numbers of first and second ind respectively, then from the well-nown identities B n, 0!,!,!,...) = s n,) and B n,,,,...) = S n,) when x n = or x n = ) n n )! in Theorem 3 we obtain + s + )S sp, + s + ) + s + )ssp, + s + ) 0 mod p).
Theorem 5. Let n,,s be integers with n, s and p be a prime number. Then for any seuence {x j } of integers with x not a multiple of p we have B,+sp x,x,x 3,...) + sp) ) x s B n, x,x,x 3,...) ) n mod p) if p > n + +sp x n B,sp x,x,x 3,...) s ) x s B p+)n, x,x,x 3,...) sp n ) mod p ) if p > n +. p+)n 4) Application 6. If we consider the cases = and = in Theorem 5 we obtain B,+sp x,x,x 3,...) x s x n mod p) for p > n, B,+sp x,x,x 3,...) xs n ) n x j x n j mod p) for p > n. j j= Then, when x n = or x n = ) n n )! we obtain S n + sp, + sp) mod p) for p > n, s n + sp, + sp) ) n n )! mod p) for p > n, S n + sp, + sp) n mod p) for p > n and n nn+) s n + sp, + sp) ) mod p) for p > n, Theorem 7. Let n,,s,p be integers with n, s, p. Then for any seuence {x j } of integers with x not a multiple of p we have B s+)n,sn x, x, 3x 3,...) ) sx ns ) s+)n sn B n,n x, x, 3x 3,...) n n ) mod n ), B,+sp x, x, 3x 3,...) + sp) ) x s B n, x, x, 3x 3,...) ) n mod p), +sp x n B,sp x, x, 3x 3,...) s ) x s B p+)n, x, x, 3x 3,...) sp n ) mod n ). p+)n 5) Application 8. Belbachir et al. [] have proved that B n,!,!,..., + )!, 0,...) = n! ), 6)! n then, for s and p j, the two last congruences of 5) and Identity 6) prove that ) ) ) ) + sp sp jp mod p) and j s mod p ). j j j j 3
Corollary 9. Let n,,s be integers with n and p be a prime number. Then for any seuence {x j } of integers with x not a multiple of p we have B p+)n, x,x,...) n ) x n B n+p,p x,x,x 3,...) p+)n n+p ) mod p ) if p > n +, p B p+)n, x, x, 3x 3,...) n ) x n B n+p,p x, x, 3x 3,...) p+)n n+p ) mod p ). p Application 0. As in Application 8, we have ) ) jp p j mod p ). j j Theorem. Let, j be integers and p be an odd prime number. Then for any seuence of integers {x j } we have Application.. Then B p j +, x, x, 3x 3,...) ) x p j + x p j + mod p) if p x, B r+)p j,p j r x, x, 3x 3,...) p j r ) ) x r+)p j r xp j + x p j + mod p) if p x. p j r As in Application 8), let j = in the second congruence of Theorem ) p )! pr mod p). r p Theorem 3. Let, j be integers and p be an odd prime number. Then for any seuence of integers {x j } we have B p j +, x, x, 3x 3,...) ) x p j + ) x p j + + x x p +) j mod p) if p x B r+)p j,p j r x, x, 3x 3,...) p j r ) ) x r+)p j r x x p j + x p j + mod p) if p x. p j r Remar 4. Similarly to the last proofs, one can exploit the results of Carlitz [3] with connection to Theorem to obtain more congruences for partial Bell polynomials. 3 Proof of the main results Proof of Theorem. Let {x n } be a seuence of real numbers with x := and let {f n x)} be a seuence of polynomials defined by f n x) = j= B n,j x, x 3 3, x 4 4... ) x) j, 4 7)
with f 0 x) =, x) j := x x ) x j + ) for j and x) 0 :=. n x We have nf n ) = n B n,j, x ) 3 3,... ) j = x n and D x=0 f 0) = 0. j= It is well nown that {f n x)} presents a seuence of binomial type, see [4]. Then, from Proposition in [5] we have y n = )B r+)n r+)n,,x,x 3,...) = f n ) where {f n x;a)} is a seuence of binomial type defined by f n x;a) := = D x=0 f n x;r), 8) x an + x f n an + x) 9) with a is a real number, see [5]. From Proposition in [5] we have also B n+,,x,x 3,...) ) = f n ) n+ but from [8] we can write f n ;r) as f n ;r) = = f n ;r), 0) B n,j D x=0 f x;r),d x=0 f x;r),...) ) j. ) j= Then, by substitution ) in 0) and by using 8) we obtain B n+,,x,x 3,...) ) = n+ B n,j y,y,...) ) j. ) j= We can verify that Identity ) is true for x = 0, and, for x 0 it can be derived from ) by replacing x n by x n x and by using the well nown identities B n, xa,xa,xa 3,...) = x B n, a,a,a 3,...) and B n, xa,x a,x 3 a 3,...) = x n B n, a,a,a 3,...), 3) where {a n } is any real seuence. Proof of Theorem 3. We prove that B sp, 0 mod p), s +. From the identities ) ) ) sp sp s 0 mod p), for p j and mod p), 4) j pj j and from the recurrence relation given by B n, = j ) n x j B n j, j 5
with B n, := B n, x,x,...) and x j = 0 for j 0, we obtain + )B sp,+ = ) sp s ) s x j B sp j, x jp B s j)p, mod p). j j j j= Then, for s = 0, we get B 0, 0 mod p), 0. For s =, we get + )B p,+ x p B 0, 0 mod p),. For s =, the last congruences imply that + )B p,+ x p B p, + x p B 0, = 0 mod p), and p. The induction on s proves that B sp, 0 mod p) when s +. Proof of Theorem 5. From [4] we have B n, x,x,...) = n! n )! x B n, j, x 3 3, x ) 4 x j 4,..., n. 5) j! Then, for i {,...,n}, the last identity and Identities 3) imply i j= t i = n + )!) i y i = ir )! ir j)! xir j B i,j j=0 n + )!)i ir ) B r+)i r+)i,ir x,x,...,x i j+ ) = ir n + )! x, n + )!) n + )!)i j x 3,..., 3 i j + x i j+ from which we deduce that t,...,t n are integers, and then B n, t,t,...),...,b n,n t,t,...) are also integers. Therefore, by using the second identity of 3), Identity ) becomes ), x j= B n,j t,t,...) ) j = x n + )!) n B n+, x,x,x 3,...) ). n+ Hence, when we replace by α + sp in the last identity we obtain x j= B n,j t,t,...) α + sp ) j = x n + )!) n B n+, x,x,x 3,...) α + sp) ) n+, and when we reduce modulo p in the last identity we obtain x n + )!) n B n+, x,x,x 3,...) α + sp) ) n+ x α+s But from ) we have B n,j t,t,...) α ) j mod p). j= 6
x α j= B n,j t,t,...) α ) j = n + )!) n x α from which the last congruence becomes B n,j y,y,...) α ) j j= = n + )!) n x B n+α,α x,x,x 3,...) α ) n+α, α n + )!) n B n+, x,x,x 3,...) α + sp) ) n+ x s n + )!) n B n+α,α x,x,x 3,...) α ) n+α mod p). α Now, when we replace n by n α, the last congruence becomes n α + )!) n B, x,x,x 3,...) α + sp) ) x s n α + )!) n B n,α x,x,x 3,...) α ) n mod p). α Then, if p > n α + we obtain B, x,x,...) α + sp) ) x s B n,α x,x,x 3,...) α ) n mod p). α For the second part of theorem, when we replace by sr in ) we get x sr j= B n,j t,t,...) r j s n) j = x n + )!) n B n+sr,sr x,x,x 3,...) sr ) n+sr, sr and, because B n,j t,t,...) j n) are integers, the last identity proves that x n + )!) B n+sr,sr x,x,x 3,...) sr ) n+sr x sr sr z n n + )!) B x sr r+)n, x,x,x 3,...) ) mod r). r+)n Let r = p > n + be a prime number. Now, because the expressions n + )!) B,sp x,x,x 3,...) sp ) and n + )!) B p+)n, x,x,x 3,...) sp ) p+)n are integers, we obtain x n B,sp x,x,x 3,...) s ) x s B p+)n, x,x,x 3,...) sp n ) mod p ). p+)n Proof of Theorem 7. From Identity 5) we get B n+, x, x, 3x 3,...) ) = n+ j= )! j)! B n,j x,x 3,x 4,...) x j, n, 6) 7
and this implies that the numbers z n = B r+)n, x, x, 3x 3,...) ), n 7) r+)n are integers, and then, the numbers B n,j z,z,...) j n) are also integers. From Identity 3), when we replace r by and s by n s ) we obtain ) B s+)n,sn x, x, 3x 3,...) = x ns ) s + )n ns B n,j z,z,...) s )n) j, 8) sn with z n := n n n) B n,n x, x,...). Furthermore, from 8), we have B s+)n,sn x, x, 3x 3,...) ) = nx ns ) s+)n sn { n n s j= x ns ) j= B n,j z,z,...) s ) n) j } sz n { x ns ) s n n n) B n,n x, x,...) x ns ) s n n) B n,n x, x,...) mod n ), i.e., B s+)n,sn x, x, 3x 3,...) ) sx ns ) s+)n sn } B n,n x, x, 3x 3,...) n n ) mod n ). For the second part of 5), when we replace by α + sp in ), we obtain x B n,j z,z,...) α + sp ) j = x B n+, x, x, 3x 3,...) α + sp) ) n+ j= with z n is given by 7). Because the numbers B n,j z,z,...), j n, are integers, then when we reduce modulo p in the last identity we get x B n+, x, x, 3x 3,...) α + sp) ) n+ x α+s B n,j z,z,...) α ) j mod p) and by ) the last congruence becomes B n+, x, x, 3x 3,...) α + sp) ) n+ x s B n+α,α x, x, 3x 3,...) α ) n+α mod p). α To terminate, it suffices to replace n by n α in the last congruence. For the third part of 5), when we replace by r in ), we obtain x r B n,j z,z,...) r ) j = x B n+r,r x, x, 3x 3,...) r ) n+r r j= 8 j=
and because the numbers B n,j z,z,...), j n, are integers, then when we reduce modulo r in the last identity we get x Now, because B n+r,r x, x, 3x 3,...) r ) x r n+r z n xr B r+)n, x, x, 3x 3,...) r ) mod p). r+)n B n+r,r x, x, 3x 3,...) r ) and xr B r+)n, x, x, 3x 3,...) n+r r ) r+)n are integers and x p x mod p) for any prime number p, then when we put r = p, the last congruence becomes x n B n+p,p x, x, 3x 3,...) p ) x B p+)n, x, x, 3x 3,...) n+p p ) mod p). p+)n 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, n by n+p and by p we get B,sp x,x,x 3,...) s ) x s B n+p,p x,x,x 3,...) n+p ) mod p ), p > n +, s, sp p and by combining the last congruence and the second congruence of 4) we obtain B p+)n, x,x,x 3,...) n ) x n B n+p,p x,x,x 3,...) p+)n n+p ) mod p ), p > n +. p Similarly, we use the second and the third congruences of 5) to get the second part of the corollary. Proof of Theorem. Identity ) can be written as p) B p+, x, x, 3x 3,...) ) = x p+ A p p)z, p)z,...), x pr with z n = B r+)n, x, x, 3x 3,...) ),. r+)n Bell [] showed, for any indeterminates x,x,..., that 9) A p x,x,x 3,...) x p + x p mod p). 0) Therefore, from 0) and 9), we obtain p) B p+, x, x, 3x 3,...) ) x p+ { p) p z p + p)z p } mod p), x pr 9
and Identity 6) shows that B p+, x, x, 3x 3,...) ) and the terms of the seuence {z p+ n ;n } are integers. Now, because z = x r x, then, when is not a multiple of p, the last congruence and Fermat little Theorem prove that x r B p+, x, x, 3x 3,...) ) x p+ {x r x + x y p } mod p). For = in the last congruence we have y p x r x p+ x r x mod p). The proof for j = results from the two last congruences. Assume now that the congruences given by 7) are true for the index j. Carlitz [] showed, for any indeterminates x,x,..., that For x,x,... integers we obtain A p j x pj + x pj p + x pj p + + x p j mod p). A p j x + x p + x p + + x p j mod p). Then, when we use Identity 9) and the fact that the seuence {z n ;n } is a seuence of integers, we obtain when p x B p j+ +, x, x, 3x 3,...) z + z p + z p + + z p j+) x r p j+ + ) x ) x z + z p + z p + + z p j + x z p j+ B x r p j +, x, x, 3x 3,...) ) + x p j + z p j+ x x p j + + x z p j+ mod p). For = in the last congruence we have x r x p j+ + x p j + + x z p j+ mod p). From the two last congruences we deduce that B p j+ +, x, x, 3x 3,...) ) x p j + x p j+ + mod p) if p x, B r+)p j+,p j+ r x, x, 3x 3,...) p j+ r ) ) x r+)p j+ r xp j+ + x p j + mod p) if p x, p j+ r which completes the proof. 0
Proof of Theorem 3. Carlitz [] showed, for any indeterminates x,x,..., that Then, for x,x,... integers we get A p j A p j A p + x j pj mod p). x pj + x pj p + + x p j) + xp j x + x p + + x p j) + xp j mod p), and, when we use Identity 9), we obtain x pj r p j r ) B p j +, x, x, 3x 3,...) ) = x p j + A p j p j r ) z, p j r ) z,... ), and because {z n ;n } is a seuence of integers, the last identity gives x x pj r p j r) B p j +, x, x, 3x 3,...) ) p j + p j r) ) z + z p + z p + + z p j) + p j r) z p j mod p). From the proof of Theorem, the last congruence gives when p x ) x x B r p j +, x, x, 3x 3,...) ) ) x p j + z + z p + z p + + z p j + x z p j ) x B r p j +, x, x, 3x 3,...) ) + x p j + z p j ) x r+ x p j + + x z p j mod p), i.e., x r B p j +, x, x, 3x 3,...) ) x p j + x r x p j + + z p j ) mod p). For = in the last congruence we get x r x p j + x r x p j + + z pj, i.e., ) z p j x r x x p j + x p j + mod p). Then x B p j +, x, x, 3x 3,...) ) x p j + ) x p j + + x x p j +) mod p). 4 Acnowledgements The author thans the anonymous referee for his/her careful reading and valuable comments and suggestions.
References [] E. T. Bell, Exponential polynomials. Ann. Math. 35 934), 58 77. [] 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 35 008), 30. [3] L. Carlitz, Some congruences for the Bell polynomials. Pacific J. Math. 96), 5. [4] L. Comtet, Advanced Combinatorics. D. Reidel Publishing Company, Dordrecht-Holland, 974. [5] M. Mihoubi, Bell polynomials and binomial type seuences. Discrete Math. 308 008), 450 459. [6] M. Mihoubi, The role of binomial type seuences in determination identities for Bell polynomials. To appear, Ars Combinatoria. Preprint available at http://arxiv.org/abs/0806.3468v. [7] J. Riordan, Combinatorial Identities. R. E. Krieger, 979. [8] S. Roman, The Umbral Calculus. Academic Press, 984. 000 Mathematics Subject Classification: Primary 05A0; Secondary B73, B75, P83. Keywords: Bell polynomials, congruences, Stirling numbers, binomial coefficients. Received January 0 009; revised version received April 9 009. Published in Journal of Integer Seuences, May 009. Return to Journal of Integer Seuences home page.