A New Sequence and Its Some Congruence Properties

The aim of this paper is to study the congruence properties of a new sequence, which is closely related to Fubini polynomials and Euler numbers, using the elementary method and the properties of the second kind Stirling numbers. As results, we obtain some interesting congruences for it. This solves a problem proposed in a published paper.


Introduction
Let n ≥ 0 be an integer, the famous Fubini polynomials F n (y) are defined according to the coefficients of following generating function: where F 0 (y) = 1, F 1 (y) = y, and so on.These polynomials are closely related to the Stirling numbers and Euler numbers.For example, if y = − 1 2 , then (1) becomes where E n denotes the Euler numbers.At the same time, the Fubini polynomials with two variables can also be defined by the following identity (see [1,2]): and F n (y) = F n (0, y) for all integers n ≥ 0. Many scholars have studied the properties of F n (x, y), and have obtained many important works.For example, T. Kim et al. proved a series of identities related to F n (x, y) (see [2,3]), one of which is Zhao Jianhong and Chen Zhuoyu [4] studied the computational problem of the sums where the summation in the formula above denotes all k-dimension non-negative integer coordinates They proved the identity where the sequence C(k,i) is defined for positive integer k and i with 0 For clarity, for 1 ≤ k ≤ 9, we list values of C(k, i) in the following Table 1.Meanwhile, Zhao Jianhong and Chen Zhuoyu [4] proposed some conjectures related to the sequence.We believe that this sequence is meaningful because it satisfies some very interesting congruence properties, such as for all odd primes p and integers 0 ≤ i ≤ p − 2. The equivalent conclusion is for all odd primes p and positive integers 1 ≤ i ≤ p − 2. Since some related content can be found in references [5][6][7][8][9][10][11][12][13][14][15], we will not go through all of them here.The aim of this paper is to prove congruence (5) by applying the elementary method and the properties of the second kind Stirling numbers.That is, we will solve the conjectures in [4], which are listed in the following.Theorem 1.Let p be an odd prime.For any integer 1 ≤ i ≤ p − 2, we have congruence From this theorem and (3), we can deduce following three corollaries: Corollary 1.For any positive integer n and odd prime p, we have Peer-reviewed version available at Symmetry 2018, 10, 359; doi:10.3390/sym10090359 Corollary 3.For any odd prime p, we have the congruences 2E p ≡ −1 (mod p), 4E p+2 ≡ 1 (mod p), and 2E p+4 ≡ −1 (mod p).
Note.Since E n is a rational number, we can denote , where U n and V n are integers with Based on this, in our paper, the expression E n ≡ 0 (mod p) means p | U n , while p V n .

Several Lemmas
Lemma 1.For any positive integer k, we have the identity Proof.Taking n = 1 in (3), and noting that F 0 (y) = Lemma 2. For any positive integer n, we have the identity where S(n, k) are the second kind Stirling numbers, which are defined for any integer k, n with 0 ≤ k ≤ n as: where S(0, 0) = 1, S(n, 0) = 0 and S(0, k) = 0 for n, k > 0.
Proof.See Reference [2].Lemma 3.For any positive integers n and k, we have Proof.See Theorem 4.3.12 of [16].Proof.From the definition and properties of S(n, k), we have S(n, k) = 0, if k > n.For any integers 0 ≤ j ≤ p − 1, from the famous Fermat's little theorem, we have the congruence j p ≡ j (mod p).From this congruence and Lemma 3, we have This completes the proof of Lemma 4.

Proof of the Theorem
In this section, we will prove Theorem by mathematical induction.Taking k = p in Lemma 1 and noting that C(p − 1, 0) = 1 and C(p − 1, p − 1) = (p − 1)!, we have: From ( 6), we have the congruence From Lemma 2, we have and n (y) denotes the k-order derivative of F n (y) for variable y.Combining ( 7), ( 9), (10), and (11), we have: or That is, the theorem is true for i = 1.Assume that the theorem is true for all 1 ≤ i ≤ s.That is, then from (7) we have the congruence In congruence (14), taking the (p − s − 1)-order derivative with respect to t, then let y = 0, applying Lemma 2, we have: That is, the theorem is true for i = s +
1, F 1 (y) = y, and the equation a 1 + a 2 + • • • + a k = 1 holds if and only if one of a i is 1, others are 0. The number of the solutions of this equation is ( k 1 ) = k.So, from (3), we have