On some properties of the number of permutations being products of pairwise disjoint $d$-cycles

Let $d\geq 2$ be an integer. In this paper we study arithmetic properties of the sequence $(H_d(n))_{n\in\N}$, where $H_{d}(n)$ is the number of permutations in $S_{n}$ being products of pairwise disjoint cycles of a fixed length $d$. In particular we deal with periodicity modulo a given positive integer, behaviour of the $p$-adic valuations and various divisibility properties. Moreover, we introduce some related families of polynomials and study they properties. Among many results we obtain qualitative description of the $p$-adic valuation of the number $H_{d}(n)$ extending in this way earlier results of Ochiai and Ishihara, Ochiai, Takegehara and Yoshida.


Introduction
We let N denote the set of non-negative integers, N + the set of positive integers, P the set of prime numbers and finally we write N ≥k for the set {n ∈ N : n ≥ k}. In the sequel we will also need the notion of the p-adic valuation of an integer, where p ∈ P is fixed. More precisely, if n ∈ Z then the number ν p (n) := max{k ∈ N : p k | n} is called the p-adic valuation of n. We also adopt the standard convention that ν p (0) = +∞. From the definition we easily deduce that for each n 1 , n 2 ∈ Z the following properties hold: ν p (n 1 n 2 ) = ν p (n 1 ) + ν p (n 2 ) and ν p (n 1 + n 2 ) ≥ min{ν p (n 1 ), ν p (n 2 )}.
If ν p (n 1 ) = ν p (n 2 ), then the inequality can be replaced by equality. Moreover, one can easily extend the notion of p-adic valuation to rational numbers x = a/b, b = 0, in the following way: ν p (x) = ν p (a) − ν p (b).
Let d ∈ N ≥2 , n ∈ N and consider the number H d (n) of those σ ∈ S n which are products of pairwise disjoint cycles of length d. As usual, the identity permutation is also counted as a product of 0 cycles of length d. In particular, if d = p ∈ P is a prime number, then H p (n) counts the number of solutions in S n of the equation σ p = id.
In the paper [1] the authors initiated the study of arithmetic properties of the sequence (H d (n)) n∈N for d = 2. In this case, the number H 2 (n) is the number of involutions in the group S n . In particular, they obtained several combinatorial identities, presented description of the 2-adic valuation of H 2 (n) and gave precise information about the rates of growth of H 2 (n). The mentioned paper can be seen as a complement of case p = 2 of the papers [5,10,6,7] concerning p-adic valuations of numbers H p (n), where p is a prime number. The study of the sequence (H p (n)) n∈N with p-prime, is quite natural because the number H p (n) counts the elements of order p in the group S n . However, it seems that there are no results concerning the general sequence (H d (n)) n∈N , where d ∈ N ≥2 is not necessarily a prime. In particular, it should be stressed that there is no results on p-adic valuations of numbers H d (n), where p is a prime number greater than d (even in the case when d is a prime). Our first aim is to fill this gap and present broad spectrum of results concerning various arithmetic properties of the sequence (H d (n)) n∈N and also some families of polynomials related to them.
Let us introduce the content of the paper. Section 2 is devoted to basic properties of the general sequence (H d (n)) n∈N . In particular, we recall the standard recurrence relation and closed formula for our sequence. Moreover, we present the expression for the exponential generating function n! x n .
With the help of the function H d (x) we obtain several interesting identities involving elements of our sequence and related sequences (such as roots of unity or Bernoulli numbers). In Section 3 we study periodicity properties of the sequences of remainders of numbers H d (n) modulo a given positive integer c. We start with the results for c ∈ {d + 1, d + 2}. We continue the work with the case of c being of the form d + k, where k is a fixed positive integer. Next we show that if c is a power of a prime number not equal to d, then c is a period of the sequence (H d (n) (mod c)) n∈N . At the end of the section, the result for composite d and arbitrary c or for c co-prime to d is given.
Section 4 begins with a revision of a lower estimate of the p-adic valuation of the number H p (n): where the prime number p is fixed. We give a new proof of this known fact in order to obtain some results on periodicity of sequences Hp(n) p βn (mod p r ) n∈N , where r ∈ N + and β n is mentioned lower bound of ν p (H p (n)) or β n = ν p (H p (n)). Moreover, we show that for each prime number p and j ∈ {0, ..., p − 1} there exists a b j ∈ {0, ..., p − 1} such that H p (kp 2 + jp + b j ) = k(p − 1) + j for any k ∈ N. These results can be seen as a complement of the series of papers [5,10,6,7]. In Section 5 we describe the p-adic valuations of the numbers H d (n), where p > d. In order to do this, we show that the sequence (H d (n)) n∈N is a restriction of a differentiable function f d : Z p → Z p , where Z p is a ring of p-adic integers. Then, we prove that there exists a function g d : Z p × pZ p → Q p such that for any x ∈ Z p and h ∈ pZ p we have . At last, we obtain a qualitative description of the sequence (ν p (H d (n))) n∈N using Hensel's lemma.
Section 6 is devoted to study of a family of polynomials W d,m (x) which are closely related to mth derivative of the generating function H d (x). More precisely, we have H (m) We show that W d,m (x) is a monic polynomial with integral coefficients and we give a formula for its coefficients in case of odd d. These coefficients are expressed in terms of (shifted) elements of the sequence (H d (n)) n∈N . Next, we compute the exponential generating function W d (x, t) of the sequence (W d,m (x)) m∈N and prove identities and congruences involving these polynomials. We finish the section with the proof of periodicity of the sequence (x ⌊ m p ⌋p(1−p) W p,m (x) (mod p)) m∈N . In Section 7 we deal with divisors of the number H d (n). The section is divided into three parts. In the first part we show that for each d ≥ 2 the set of prime divisors of numbers H d (n) is infinite. In the second part we compute the greatest common divisor of numbers H d (n) − 1 and n. As a consequence, we get that the set prime divisors of numbers H d (n) − 1, n ∈ N, is whole P. The last part is devoted to the study of values 2 ≤ a < b, n ∈ N and c ∈ N ≥2 such that c divides both H a (n) and H b (n).
In Section 8 we introduce the sequence of polynomials (H d (n, x)) n∈N being a generalizations of the sequence (H d (n)) n∈N . Namely, H d (n, 1) = H d (n) and the coefficient near the x k in the polynomial H d (n, x) is the number of those permutations in S n which are products of pairwise disjoint d-cycles and have exactly k fixed points. After that, we compute the remainders of polynomials H d (n, x) modulo d and study coefficients of the polynomials H 2 (n − 1, x)H 2 (n + 1, x) − H 2 (n, x) 2 and H d (n − d, x)H d (n + d, x) − H d (n, x) 2 . In particular, we prove that if U (n, x) = H 2 (n − 1, x)H 2 (n + 1, x) − H 2 (n, x) 2 , then U (n, x) = V (n, x 2 ), where V ∈ Z[x] has degree n − 1 and all the coefficients of V near x i , i = 0, n − 1 are positive. Moreover, among some identities and congruences the most interesting result of this section is the result which says that the congruence H d (n − d, x)H d (n + d, x) ≡ H d (n, x) 2 (mod d!) is true if and only if d is a prime number or a square of a prime number. Section 9 collects some observations, questions and conjectures related to objects introducted in previous sections.

First results
For d ∈ N ≥2 we have H d (i) = 1 for i ∈ {0, 1, . . . , d − 1} by definition and for n ≥ d the standard combinatorial argument shows that (1) H d (n) = H d (n − 1) where as usual, for m ∈ N the symbol (u) (m) denotes the falling factorial defined as Indeed, let us fix σ ∈ S d,n . If σ(n) = n then σ = σ 1 for some σ 1 ∈ S d,n−1 and it can be chosen by H d (n − 1) ways. If σ(n) = n then σ = σ 1 • π, where π is a d-cycle containing n and σ 1 is a product of pairwise disjoint d-cycles on a set {1, ..., n}\supp(π). Then σ 1 can be treated as a member of the set S d,n−d . Thus, it can be chosen by H d (n − d) ways. Moreover, π can be chosen in (n − 1) (d−1) ways. Furthermore, in the sequel we will also use the convention H d (n) = 0 for n < 0. As a simple consequence of the recurrence formula we see that if q ∈ N, with q | (d − 1)!, then H d (n) ≡ 1 (mod q) for each n ∈ N. In particular, if d is a composite number > 4, then H d (n) ≡ 1 (mod d). Moreover, if p is a prime number < d, then ν p (H d (n)) = 0 for each n ∈ N.
It is well known that the exponential generating function for the sequence (H d (n)) n∈N is given by In consequence, the exact expression for H d (n) is given by The knowledge of the closed form of exponential generating function allows us to obtain some identities involving numbers H d (n).
Theorem 2.1. The following identities are true. (1) (2) For n ∈ N + and d ∈ N ≥3 an odd number we have (3) For n ∈ N + and d ∈ N ≥2 an even number we have if n = md.
(4) Let ζ d be a d-th root of unity. Then (5) Let ζ 2d be a 2d-th root of unity satisfying ζ d 2d = −1. Then (6) Let d be an even number and (B n ) n∈N denote the sequence of Bernoulli numbers. Then (7) For n ∈ N we have the identity Proof. The identities (1)- (6) can be easily proved with the help of generating functions technique. Indeed, it is well known that for In order to get the first identity we write f (x) = H d (x) and g(x) = e −x and get f (x)g(x) = e x d d . Comparing the coefficients on the both sides of this identity we get the result.
In order to get the second identity we put is odd then f (x)g(x) = 1 and hence the result.
The third identity can be proved in the same way. Taking d even and = e 2 x d d and hence the result. In order to get the fourth identity we put and the identity follows.
The fifth identity follows by taking In this case f (x)g(x) = e (1+ζ 2d )x and we get required identity.
In order to get the sixth identity we put i.e., f is the exponential generating function of the sequence (2 n B n ) n∈N , and where A n = 0 for n odd and A n = H d (n + 1)/(n + 1) for n even. With f, g chosen in this way we get f (x)g(x) = H d (−x) and comparing the coefficients near x n on both sides of this identity we get the result. The last identity is a consequence of the recurrence relation (1). Indeed, rewriting (1) as and summing from m = 1 to n on both sides we get telescoping sum on the left and get The result follows.

Periodicity of the numbers H d (n) modulo a given positive integer
We know that if c < d or c = d > 4 is a composite number then the sequence (H d (n) (mod c)) n∈N is constant and equal to 1. In this section we present some results concerning periodicity of the sequences (H d (n) (mod c)) n∈N when c > d. We start with cases c ∈ {d + 1, d + 2}. As we will see, the behaviour of the sequence (H d (n) (mod c)) n∈N depends mainly on the question whether c is a prime or not.
Proposition 3.1. Let p be an odd prime number. Then for each n ∈ N we have In particular, ν p (H p−1 (n)) = 0 for any n ∈ N.
Proof. We start with the computation of H p−1 (n) (mod p). We proceed by induction on n ∈ N. For n ∈ {0, ..., p − 2} we have H p−1 (n) = 1. Let us fix now N ≥ p − 1 and assume that congruence (3) is satisfied for each n < N . If N ≡ −1, 0 (mod p) then by (1) we have since in the product of numbers N − 1, ..., N − p+ 2 there is a number divisible by p and N − 1 ≡ −1 (mod p), which means that H p−1 (N − 1) ≡ 1 (mod p). If N ≡ 0 (mod p), then since N − 1 and N − p + 1 are not congruent to −1 modulo p and (p − 2)! ≡ 1 (mod p) as an immediate consequence of Wilson's theorem.
The proof of congruence (4) is analogous. We proceed by induction on n ∈ N. For n ∈ {0, ..., p − 3} we have H p−2 (n) = 1. Let us fix now N ≥ p − 2 and assume that congruence (4) is satisfied for each n < N . If N ≡ −2, −1, 0 (mod p), then by (1) we have since in the product of numbers N − 1, ..., N − p + 3 there is a number divisible by p and N − 1 ≡ −2, −1 (mod p), which means that as an immediate consequence of Wilson's theorem. If N ≡ −2 (mod p), then  According to the last proposition and numerical computations, we prove that for a fixed positive integer k and sufficiently large positive integer d such that d+k is a composite number, the sequence (H d (n) (mod d + k)) n∈N is constant and equal to 1. Theorem 3.3. Let us fix k ∈ N + . Then for each positive integer d ≥ max{k + 2, 3 + 2 √ k + 2} such that d + k is a composite number there holds H d (n) ≡ 1 (mod d + k) for all n ∈ N.
Proof. Similarly as in the proof of the previous proposition, it suffices to show that d + k | (d − 1)!. Let q be the least divisor of d + k greater than 1. Then d+k If q < d+k q , then q and d+k q are two distinct positive integers appearing as factors in (d − 1)!, For our next result, we need the following auxiliary lemma.
Lemma 3.4. Let p be a prime number and d be a composite positive integer divisible by p. Assume that (p, d) = (2,4).
Our lemma is proved.
We are ready to prove the main result of the section. It concerns periodicity of the sequence (H d (n) (mod p r )) n∈N , where p is a prime number not equal to d and r is any positive integer. More precisely, we have the following Theorem 3.5. Let d, r be positive integers with d ≥ 2. Let p be a prime number not equal to d. Assume that (d, p, r) = (4, 2, 2). Then the sequence (H d (n) (mod p r )) n∈N is periodic of period p r . Moreover, if p > d, then p r is the basic period of the sequence (H d (n) (mod p r )) n∈N .
Proof. Let us notice that for the proof that p r is a period of (H d (n) (mod p r )) n∈N we only need to show that (5) H d (p r ) ≡ H d (0) = 1 (mod p r ).
Then, using (1) and a simple induction on n ∈ N, we prove that H d (n + p r ) ≡ H d (n) (mod p r ).
In order to prove (5) we will use the exact formula and show that for k > 1, the k-th summand in the above sum has the p-adic valuation greater than r. We write this summand: It suffices to show, that ν p We estimate ν p ((dk − 1) ((d−1)k−1) ) from below by number of multiples of p among factors dk − 1, dk − 2, ..., k + 1 and use Lemma 3.4 to obtain the following.
This implies ν p and finishes the proof of (5).
We need to show that p r is the basic period of the sequence (H d (n) (mod p r )) n∈N provided that p > d. Let us assume by contrary that p r−1 is a period of (H d (n) (mod p r )) n∈N . If r = 1 then we directly obtain a contradiction, since the sequence (H d (n) (mod p)) n∈N is not constant. Indeed, H d (d) = 1 + (d − 1)! ≡ 1 (mod p) as d < p. If r > 1, then we consider the congruences , we get the following: and equivalently After simplification (where we divide by (d−1)! and d!, respectively, and these numbers are coprime to p) we obtain two congruences Subtracting the second congruence from the first one, we get 1 − 1 d ≡ 0 (mod p), or equivalently d−1 d ≡ 0 (mod p). This is impossible since 0 < d − 1 < p. Hence p r must be the basic period of the sequence (H d (n) (mod p r )) n∈N . Remark 3.6. In the case of (d, p, r) = (4, 2, 2) the sequence (H 4 (n) (mod 4)) n∈N has the basic period 8. More precisely, we have Proof. We prove by induction on m ∈ N that We check that congruence (6) holds for m = 0 and i ∈ {0, 1, 2, 3, 4, 5, 6, 7}. Assume now that it is satisfied by some m ∈ N. By direct calculations we show (6) for m + 1.
Example 3.7. If p < d, then it is possible that p r is not the basic period of the sequence (H d (n) (mod p r )) n∈N . For example, if d = 5, p = 2, r = 4, then we obtain a sequence (H 5 (n) (mod 16)) n∈N , which has a period 8 < 2 4 .
Based on the above considerations we can give precise value of the period of the sequence (H d (n) (mod c)) n∈N provided that d is a composite number or d ∤ c. 4. The behaviour of the p-adic valuation of H p (n) for some p Amdeberhan and Moll were able to compute the exact value of the 2-adic valuation of H 2 (n) for n ∈ N. Quite unexpected they proved that for given i ∈ {0, 1, 2, 3} the sequence (ν 2 (H 2 (4n + i)) − n) n∈N is constant. This raises natural question whether the sequence of p-adic valuations of H p (n) can share similar property. In fact, this question was studied earlier by Ochiai in [10]. In the cited paper the reader can find the formulae for the p-adic valuation of the numbers H p (n), n ∈ N, where p is a prime number at most equal to 11. Actually, Ochiai in his paper gave a qualitative description of the sequence (ν p (H p (n))) n∈N for any prime number p ≤ 23. Further, Ishihara, Ochiai, Takegahara and Yoshida extend this result for any prime number p in [6]. In particular, the following inequality holds true for all n ∈ N, where the equality holds if p | n p . This inequality was proved for p = 2 by Chowla, Herstein and Moore in [3]. In general, this inequality was proved by Grady and Newman in [5]. Moreover, in 2010 Kim and Kim gave a combinatorial proof of the above inequality. Despite this facts we would like to give the proof of (7) different than the ones presented by mentioned authors. In consequence, we will be able to prove the following two results which have not appeared yet in the literature.
has basic period equal to 2p 2 .
We start with one lemma. We omit its proof as it can be found in [6, Lemma 2.1]. Lemma 4.3. If p is an odd prime number and k is a positive integer then At this moment, we give a result on congruences involving numbers H p (n), n ∈ N. This result is crucial in the proof of Theorem 4.1.
Proof. We prove the congruences concerning numbers H p (n) for n ∈ N. We proceed by induction on n. We write n = kp Let us assume that n = kp 2 for some k ∈ N + . By induction hypothesis for (k − 1)p 2 + (p − 1)p + p − 1 and (k − 1)p 2 + (p − 1)p, Wilson's theorem and Lemma 4.3 we obtain Assume now that n = kp 2 + jp for k ∈ N and j ∈ {1, ..., p − 1}. By the recurrence (1), induction hypothesis for n = kp 2 + (j − 1)p + p − 1, Lemma 4.3 and Wilson's theorem we have At this moment we assume that n = kp 2 + jp + i for k ∈ N and i, j ∈ {0, ..., p − 1}. If i = 0, then the third congruence obviously is true. For i > 0, by (1), induction hypothesis for n = kp 2 +jp+i−1 and Wilson's theorem we have We are ready to prove the lower bound for the p-adic valuation of H p (n).
Corollary 4.5. For odd prime number p, non-negative integer k and integers i, where the inequality is in fact equality for j = 0. In other words, Proof. We only need to prove that ν p (H p (kp 2 + jp + i)) = k(p − 1) + j for j = 0. We proceed by induction on i ∈ {0, ..., p − 1}. The equality ν p (H p (kp 2 + i)) = k(p − 1) holds for i = 0 by Theorem 4.4. If i > 0, then we write Then ν p ((kp 2 + i − 1) (p−1) ) ≥ 2, since kp 2 appears as a factor in (kp 2 Now we give the proofs of the main results of this section. Proof of Theorem 4.1. Let us define γ n = n p 2 ·(p−1)+ n p (mod p) for simplicity of notation. At first, we show that for each n ∈ N. We write n = kp 2 + jp + i, k ∈ N, i, j ∈ {0, 1, ..., p − 1}, and proceed by induction on jp + i. By Theorem 4.4 we see that the congruence (8) is satisfied for jp + i = 0. If jp + i > 0 and i > 0 then If jp + i > 0 and i = 0 then we perform similar computations as in the proof of Theorem 4.4.
We are left with proving that p 2 is a basic period of the sequence (−1) It suffices to show that p is not a period of the sequence (−1) In order to do this, we assume the contrary. In particular, we have . Using recurrence relation for numbers H p (n) we obtain the congruences or equivalently Since p is a period of the sequence (−1) which means that However, the number (p − 1) (p−2) is not divisible by p. A contradiction shows that p is not a period of the sequence (−1) The sequence is thus periodic with period 2p 2 . We know, that p 2 is not a period of this sequence, as The number 2 also cannot be a period of this sequence. If we assume the contrary then, since . This stays in contradiction with the fact, that . It suffices to show that 2p is not a period of the sequence In order to do this, we assume the contrary. In particular, we have Hp(2p+2) p 2 ≡ H p (2) = 1 (mod p). Using recurrence relation for numbers H p (n) we obtain the congruences or equivalently Since 2p is a period of the sequence and The numbers 2(p − 1) (p−2) and 2(2p + 1)(2p − 1) (p−3) are both not divisible by p. Hence, the numbers H p (p + 1) and H p (p + 2) are divisible by p 2 . On the other hand, is not divisible by p 2 . This is why the numbers H p (p + 1) and H p (p + 2) cannot be both divisible by p 2 .
A contradiction shows that p is not a period of the sequence Proof of Proposition 4.2. If p = 2, then the result is well known (see for example [1,Theorem 4.1]). From Corollary 4.5 we know that ν p (H p (p − 1)) = 0. Hence we put b 0 = p − 1. Let us fix j ∈ {1, ..., p − 1} and assume by induction hypothesis that we we infer that at least one of the numbers H p (jp + b j−1 ), H p (jp + b j−1 − 1) has the p-adic valuation equal to j. We thus take b j as one of the values b j−1 − 1, b j−1 such that ν p (H p (jp + b j )) = j. In particular, b j ≥ p − j − 1. At last, we use Theorem 4.1 to obtain In Theorem 4.1 we proved periodicity of the sequence (H p (n)/p γp(n) ) n∈N , where In general, we ask the following.
Question 4.6. For which p ∈ P and w ∈ N + the sequence For each prime number p and positive integer w the sequence  ). Let us take any n ∈ N and write n = kp 2 + r for some k ∈ N and r ∈ {0, ..., p 2 − 1}. Then, by the above fact we have the following chain of congruences for any t ∈ N: where we use Euler's theorem as p ∤ λ(l) for any l ∈ N.
Let us define a particular subset of the set of prime numbers: Then we are able to give a partial answer to Question 4.6.
Corollary 4.8. If p ∈ A then for each w ∈ N + the sequence Proof. By the previous theorem we have for each t ∈ N. After division by p αr(p) , where r = n (mod p 2 ), we get Since α(p) − α r (p) ≥ 0, we thus obtain as we wanted to prove.
In case of p ∈ A we predict that the sequence is not periodic for any w ∈ N + but proving this fact seems to be out of reach for us.

The behaviour of the p-adic valuation of
On the beginning of the paper we marked that if p is a prime number < d, then ν p (H d (n)) = 0 for each n ∈ N. The problem of computation of the p-adic valuations of the numbers H d (n) in case p = d was considered in the previous section and it was studied in several publications. In the opposition to this fact the question concerning the computation of the p-adic valuations of the numbers H d (n), where p > d, is almost unexplored. It was first considered by Amdeberhan and Moll in [1]. According to results on periodicity of the sequence (H 2 (n) (mod c)) n∈N they claimed that if p is an odd prime number not dividing the values H 2 (n) for n ∈ {0, ..., p − 1}, then ν p (H d (n)) = 0 for each n ∈ N. The situation is more complicated if p divides some of those values. It happens for p = 5. Based on numerical computations they stated a conjecture which equivalent version is as follows Conjecture 5.1. If p is an odd prime number such that p | H 2 (n) for some n ∈ {0, ...p − 1} then for each k ∈ N + there exists a unique n k modulo p k such that p k | H 2 (n) if and only if n ≡ n k (mod p k ).
The above conjecture is not true for two reasons. The first reason is that there exist prime numbers p and positive integers n 1 such that ν p (H 2 (n)) = 1 for each n ≡ n 1 (mod p), for example (p, n 1 ) = (19, 6). The second one is that for some prime numbers p, for example 59 and 61, there exist more than one value n 1 modulo p such that p | H 2 (n) for n ≡ n 1 (mod p).
The aim of this section is to shed some light on the behaviour of ν p (H d (n)) in the case when p > d.
We use the exact formula for the number H d (n) and make some modifications: The above notation suggests us to define the function where the last value in the above inequalities tend to +∞ when k → +∞. The following three lemmas show that • f d is a p-adic differentiable function, • there exists a function g d : By the above facts we will be able to describe behaviour of the p-adic valuations of numbers H d (n), n ∈ N, using classical Hensel's lemma.
Proof. We compute the derivative from the very definition: where we put 0≤j1<...<jt≤dk−1 ..·(x−jt) from below. We will do this by estimating the p-adic valuation of the product ( where we assumed that ν p (h) ≥ 1 and use the Legendre formula ν p (k!) = k−sp(k) p−1 . Here, s p (k) is the sum of digits of the number k written in the (unique) p-ary expansion. Hence, the p-adic valuation of is at least −2 and it tends to +∞ when k → +∞. Thus, the series is convergent and its p-adic valuation is at least −2. We infer that This series is convergent, as when k → +∞, where we used the fact that j, dk − j − 1 < dk, so j, dk − j − 1 have at most ⌊log p dk⌋ + 1 p-ary digits. Moreover, as at least one of the numbers j, dk − j − 1 is greater than or equal to k. Thus the p-adic valuation of +∞ k=1 Proof. Let us fix x, h ∈ Z p with ν p (h) ≥ 1. We then have: For the change of the order of summation, we will show that Let us estimate the p-adic valuation of the number 0≤j1<...<j dk−l ≤dk−1 (x − j 1 ) · ... · (x − j dk−l ) from below. We have the inequality Since p > d ≥ 2, we thus have Then we change summation in the last expression in (9) and get the following: We check that − 2 ≥ 0 for p > d ≥ 3 and k ≥ p + 1. If d ≥ 3 and k = p, then either l ≥ 3 and then ν p (h l−2 ) ≥ l − 2 ≥ 1 = ν p (p!) or l = 2 an then in each product of the form (x − j 1 ) · ... · (x − j dk−2 ), where 0 ≤ j 1 < ... < j dk−2 ≤ dk − 1 there is a factor with the p-adic valuation at least equal to 1.
Proof. By the previous lemma we can write . After adding the above two equalities and simplifying we obtain At this moment we are prepared to state the theorem on qualitative description of the sequence (ν p (H d (n)) n∈N .
Theorem 5.5. Let d be a positive integer at least equal to 2, p > d be a prime number and k be a positive integer. Assume that H d (n k ) ≡ 0 (mod p k ) for some integer n k . Then the number of solutions n modulo p k+1 of the congruence H d (n) ≡ 0 (mod p k+1 ), satisfying the condition n ≡ n k (mod p k ), is equal to: In particular, if p | H d (n 1 ) for some n 1 ∈ N and ν p (f ′ d (n 1 )) = 0, then for each k ∈ N + there exists a unique value n k modulo p k such that if n ≡ n 1 (mod p), then p k | H d (n) if and only if n ≡ n k (mod p k ). In other words there exists a unique p-adic integer b ≡ n 1 (mod p) such that ν p (H d (n)) = ν p (n − b) for each non-negative integer n ≡ n 1 (mod p).

Proof. We have
for congruent p-adic integers x 1 , x 2 modulo p. These all facts make the proof of our theorem exactly the same as the proof of classical Hensel's lemma from [9, p. 44].
Remark 5.6. Let us compute f ′ d (n) (mod p) for n ∈ N: That is a way to compute the value f ′ d (n) (mod p) by adding only finitely many summands. However, in many cases, having d ∈ N ≥2 , a prime number p and n 1 ∈ {d, ..., p − 1} such that p | H d (n 1 ) (recall that H d (n) = 1 for n ∈ {0, ..., d − 1}), we do not need to compute f ′ d (n) (mod p). Instead of this, it suffices to compute ν p (H d (n 1 )). If ν p (H d (n 1 )) = 1, then we check p-adic valuations of numbers H d (n 1 + pt) for t ∈ {1, ..., p − 1}. If ν p (H d (n 1 + pt)) ≥ 2 for some t ∈ {1, ..., p − 1} then for each k ∈ N + there exists a unique value n k modulo p k such that if n ≡ n 1 (mod p) then p k | H d (n) if and only if n ≡ n k (mod p k ). Otherwise, ν p (H d (n)) = 1 for any n ≡ n 1 (mod p). If ν p (H d (n 1 )) ≥ 2 then we only check p-adic valuation of the number H d (n 1 + p). If ν p (H d (n 1 + p)) = 1 then for each k ∈ N + there exists a unique value n k modulo p k such that if n ≡ n 1 (mod p) then p k | H d (n) if and only if n ≡ n k (mod p k ). Otherwise, ν p (H d (n)) ≥ 2 for any n ≡ n 1 (mod p).
Remark 5.7. We found all the triples (d, p, n 1 ) for d ∈ {2, 3, 4, 5}, first 25 prime numbers p and numbers n 1 ∈ {d, ..., p− 1} such that p | H d (n 1 ). We checked the behaviour of the p-adic valuations of numbers H d (n) where n ≡ n 1 (mod p). The only triples (d, p, n 1 ) such that ν p (H d (n)) = 1 for all n ≡ n 1 (mod p) are (2, 19, 6), (3,13,7). For any other triple and a positive integer k we have the existence of a unique n k modulo p k such that if n ≡ n 1 (mod p) then p k | H d (n) if and only if n ≡ n k (mod p k ).

Properties of polynomials attained during differentiation the exponential generating function of the sequence (H d (n)) n∈N
Let m ∈ N and for m- . We have W d,0 (x) = 1 and due to the identity . We summarize the basic properties of the sequence (W d,m (x)) m∈N in the following: Theorem 6.1. The following properties hold: (1) W d,n (x) is unitary polynomial with non-negative integer coefficients and deg W d,n (x) = (d − 1)n; (3) For d ∈ N ≥2 and m ∈ N let us write If d ∈ N ≥3 is an odd number then for n > (d − 1)m ; (4) The exponential generating function for the sequence (W d,n (x)) n∈N takes the form Proof.
(1) This is clear. Because W d,0 (x) = 1 is a polynomial and on the right hand side of the In consequence, from the identity

Multiplying both sides by
and hence the result.
As an immediate consequence of the shape of exponential generating function for (W d,n (x)) n∈N we get the following: Corollary 6.2. The sequence (W d,n (x)) n∈N satisfies the following recurrence relation: W d,n (x) = 0 for n < 0, W d,0 (x) = 1 and for n ≥ 1.
Proof. By comparing the coefficients near t n in the identity we get the equality Multiplying both sides by n!, writing n!/(n − i)! as (n) (i−1) and replacing n by n − 1 and i by j − 1 we get the result (with the convention W d,n (x) ≡ 0 for n < 0). Remark 6.3. In the proof of Corollary 6.2 we proved recurrence relation for the sequence (W d,n (x)) n∈N by manipulating of exponential generating functions. One can also consider the (formal) ordinary generating function of the sequence (W d,n (x)) n∈N and get a relation of a different kind. Indeed, it is an easy exercise to prove that the (formal) power series S d (x, t) = ∞ n=0 W d,n (x)t n satisfies the differential equation Multiplying both sides of the above identity by (1 − (1 + x d−1 )t) −1 and comparing coefficients near t n we easily deduce the following relation This gives us an expression of W d,n (x) in terms of the sequence (W ′ d,n (x)) n∈N .
In particular, (W 2,n (1)) n∈N is a binomial transform of the sequence (H 2 (n)) n∈N . Moreover, as was observed by Donaghey in [4], this sequence enumerates the general switchboard problem, i.e., it enumerates the states of a telephone exchange with n subscribers which is provided with means to connect subscribers singly to outside lines and in pairs internally (no conference circuits).
The last results of this section are devoted to the behaviour of sequences (W d,m (x) (mod c)) m∈N where c = d − 1 or c = d = p is a prime number.
Theorem 6.5. The following congruence holds for each d ≥ 2 and m ∈ N: In particular, if p is a prime number then Proof. We proceed by induction on m ∈ N. If m = 0 then W d,0 (x) = 1, hence the statement of our theorem is true. If m > 0 then we use Corollary 6.2 and the induction hypothesis for m − 1 and get Theorem 6.6. For each prime p we have W p,p (x) ≡ x p(p−1) (mod p).
Proof. We easily check that we have W 2,2 (x) = x 2 + 2x + 2, so assertion holds for p = 2. Now we assume that p is an odd prime number. We will show that all the coefficients of the polynomial W p,p (x) except for the leading one are divisible by p. We use the formula from the third part of the statement of Theorem 6.1. It suffices to prove that for each n ∈ is periodic and the length of the period is p.
In particular Proof. First, we prove by induction on m ∈ N that The above congruence is satisfied for m = 0 by the Theorem 6.6. Assume now that (12) holds for some m ∈ N and we will show that it holds for m + 1. By (11) we get . Now we have congruence (12) proved. Let us fix m ∈ N and write m = jp + i. Then, using (12) j times, we get W p,m (x) ≡ x p(p−1)j W p,i (x) (mod p).
After multiplication by x p(1−p)j and substitution j = m p and i = m (mod p) we obtain and get the result 7. Divisors of numbers H d (n) In this section we study various properties of the numbers H d (n) and H d (n) − 1. We are interested in prime divisors and various GCD properties. 7.1. Prime divisors of numbers H d (n). Let us denote P d = {p ∈ P : p | H d (n) for some n ∈ N}.
Theorem 7.1. For each d ∈ N ≥2 the set P d is infinite.
Proof. Assume by the contrary that p 1 < ... < p s are all the elements of the set P d . We consider two cases. The first case is when d is a composite number. We put P = p 1 · ... · p s . By Corollary 3.8 we know that the sequence (H d (n) (mod P )) n∈N is periodic of period P , as 2 < d and thus 2 is not a divisor of any number H d (n). In particular, the numbers H d (P m), m ∈ N, are congruent to 1 modulo P . Hence these numbers have no prime divisors. This combined with the fact that H d (P m) > 0 implies H d (P m) = 1 for any m ∈ N. However, this is a contradiction with the fact that the sequence (H d (n)) n∈N is an ultimately strictly increasing sequence of positive integers.
The second case is when d is a prime number. Then . We put P = p 2 1 · p 2 · ... · p s and Q = p 2 · ... · p s . By Corollary 3.8 we know that the sequence (H d (n) (mod Q)) n∈N is periodic of the period Q, as d is a prime number not dividing Q. In particular, since Q | P , we have H d (P m) ≡ H d (P m + 1) ≡ 1 (mod p i ) for m ∈ N and i ∈ {2, ..., s}. According to Corollary 4.5, ν p1 (H d (P m)) = ν p1 (H d (P m + 1)) = Qm(p 1 − 1). Since H d (P m) > 0 and p 1 , ..., p s are all possible divisors of the numbers H d (n), n ∈ N, thus H d (P m) = H d (P m + 1) = p Qm(p1−1) 1 for any m ∈ N.
Conjecture 7.2. For each d ∈ N ≥2 the set P\P d is infinite and its asymptotic density in P is equal to 1 e . The following heuristic reasoning allows us to claim the second statement in the conjecture above. If we fix a prime number p and randomly choose a sequence (a n ) n∈N such that the sequence (a n (mod p)) n∈N has the period p, then the probability that p does not divide any term of this sequence is equal to 1 − 1 p p . The probability tends to 1 e with p → +∞. Note that p > d belongs to the set P\P d if and only if p does not divide any number H d (n) for n ∈ {d, ..., p − 1}. Moreover, the sequence (H d (n) (mod p)) n∈N is periodic of the period p. Therefore we suppose that the probability that p ∈ P\P d equals to 1 − 1 p p−d . Thus, this probability tends to 1 e with p → +∞ and hence the asymptotic density of the set P\P d in the set P is expected to be equal to 1 e . Remark 7.3. We performed small numerical search in order to check whether our expectations are likely to be true. More precisely, for d ∈ {2, . . . , 10}, k ≤ 4000 and n ∈ {d, . . . , p k } we computed the number of primes p i such that p i |H d (n). In the table below we present the result of our computations.  Table 1. The number of integers k ≤ 4000 such that H d (n) ≡ 0 (mod p k ) for some positive integer n ≤ p k and d ∈ {2, . . . , 10}.
The numbers in the last row in the table above are quite close to the number 1 − 1/e ≃ 0.63212, which is the conjectured density of the set P d in the set P. We have just proven that is an integer for each k ∈ 1, ..., n d . Thus Z and, as a result, n | H d (n) − 1 in the case of d > 4 composite.
Let us consider now the case of d = 4. By (13) we know that H 4 (n)−1 = n ⌊ n 4 ⌋ k=1 n−1 4k−1 · (4k−1)! k!4 k . If p is an odd prime number, then ν p In the above computation we used the Legendre formula ν 2 (m!) = m − s 2 (m) and its corollary is an integer for k ≥ 3.
Meanwhile for k ∈ {1, 2} we compute ν 2 which is equal to 0 if and only if 4 | n. For k = 2 we have which is equal to 0 if and only if 8 | n. If ν 2 (n) < 2 then ν 2 is a prime number not equal to d then 7.3. Nontrivial common divisors of numbers H a (n) and H b (n). Let us fix two distinct integers a, b ≥ 2. It is natural to ask about existence of indices n for which the numbers H a (n) and H b (n) have a common divisor greater than 1. The above question can be answered immediately. Let d ∈ N ≥2 , n ∈ N ≥d and p be a prime divisor of H d (n). Then p | GCD(H d (n + p), H p (n + p)). Indeed, it is a simple consequence of periodicity of the sequence (H d (n) (mod p)) n∈N and divisibility properties of the sequence (H p (n)) n∈N for prime number p. Moreover, we give the following. . We compute and we see that the divisibility of H 2d+1 (3d+ 2) by 2d+ 1 is equivalent to the divisibility of (d!) 2 − 1 by 2d + 1.
In other words, if (a, b, n, c) = (d, 2d where d is an odd positive integer such that 2d + 1 is a prime number and we use the Iverson bracket Let us observe that all the examples of quadruples (a, b, n, c) with the property that c > 1 divides both H a (n) and H b (n) are such that c = b is a prime number and n ≥ b. This is a motivation to formulate the following.
Question 7.8. Are there infinitely many triples (a, b, c) ∈ N 3 , a, b, c ≥ 2, such that c > 1 is coprime to ab and c | GCD(H a (n), H b (n)) for some n ∈ N?
Moreover, it is quite natural to ask the following. Question 7.9. Are there infinitely many pairs (a, b) ∈ N 2 , a, b ≥ 2, such that GCD(H a (n), H b (n)) = 1 for each n ∈ N?
Then H d (n, 1) = H d (n) and we have exact formula Proposition 8.1. The exponential generating function of the sequence (H d (n, x)) n∈N takes the form Proof. Let us write e xt+ t d d = +∞ n=0 a d (n,x) n! t n By comparing the coefficients near t n−1 in the identity and multiplying by (n − 1)! we get the recurrence The above recurrence combined with the fact that Theorem 8.2. Let Fix n (σ) and S d,n denote the set of fixed points of the permutation σ ∈ S n and the set of permutations being product of pairwise disjoint d-cycles, respectively. Then, we have the following equality x #Fixn(σ) .
Proof. Let us write the right hand side of the above equality as g d,n (x). We will show that the sequence (g d,n (x)) n∈N satisfies the same recurrence as the sequence (H d (n, x)) n∈N . This fact combined with obvious equality g d,n (x) = x n for n ∈ {0, ..., d − 1} will give us the statement of our theorem. Let us fix σ ∈ S d,n . If σ(n) = n then σ = σ 1 for some σ 1 ∈ S d,n−1 . The number of fixed points of σ is equal to the number of fixed points of σ 1 increased by 1, since n is an extra fixed point of σ. Hence the summand x #Fixn(σ) in g d,n (x) reduces with the summand x #Fixn(σ1)+1 in xg d,n−1 (x). If σ(n) = n then σ = σ 1 • π, where π is a d-cycle containing n and σ 1 is a product of pairwise disjoint d-cycles on a set {1, ..., n}\supp(π). Then σ 1 can be treated as a member of the set S d,n−d and #Fix n−d (σ 1 ) = #Fix n (σ). Moreover, π can be chosen in (n − 1) (d−1) ways. Thus the summand x #Fixn(σ) in g d,n (x) reduces with the summand x #Fixn(σ1) in (n − 1) (d−1) g d,n−d (x). Summing over all σ ∈ S d,n we obtain g d,n (x) = xg d,n−1 (x) + (n − 1) (d−1) g d,n−d (x), which completes the proof. For given d ∈ N ≥2 , the sequence of polynomials (H d (n, x)) n∈N contains more precise information concerning the number of permutations in S n which are product of disjoint d-cycles. However, because our primary object of study is the sequence (H d (n)) n∈N , we offer only few results concerning this natural polynomial generalization. First we prove the following: . Then the following congruence holds for any n ∈ N: Proof. From the expression for H 2 (n, x) we see that it is an odd polynomial if n ≡ 1 (mod 2) and an even polynomial in case of n ≡ 0 (mod 2). We thus get that the polynomial U (n, x) = H 2 (n − 1, x)H 2 (n + 1, x) − H 2 (n, x) 2 is an even polynomial and for each n ∈ N + we can write U (n, x) = V (n, x 2 ) for some polynomial V (n, x). The expressions for V (n, x) for n ∈ {1, 2, 3} are clear from the shape of H 2 (n, x) for n ∈ {1, 2, 3, 4}.
Let us observe that the sequence (H 2 (n, x)) n∈N is a holonomic sequence of polynomials. Indeed, this is consequence of the fact that the generating function H d (x, t) (as a function in variable t) is Dfinite for each d, i. e. is a solution of some linear differential equation with polynomial coefficients. We recall that a sequence is holonomic if satisfies the recurrence relation with coefficients being polynomials in n. It is well known that if (f n ) n∈N , (g n ) n∈N are holonomic sequences then the sequences (f n ± g n ) n∈N , (f n g n ) n∈N are holonomic too. It is straightforward (but a bit tedious) exercise to check that the sequence (U (n, x)) n∈N+ satisfies the relation given in the statement of our theorem (with x replaced by x 2 ) and hence the sequence (V (n, x)) n∈N+ satisfies exactly the relation presented above.
t n , then it is an easy task to produce the differential equation satisfied by the function V(x, t). Indeed, applying standard methods and the relation (15) we check that The initial condition V(x, 0) = 1 allow us to compute the unique solution of the above differential equation in the form which is exactly the expression presented in the statement of our theorem. Now, it is an easy task to compute the coefficient a(i, n) for i = 0, 1. We have a(0, 1) = 1, a(0, 2) = −1, a(1, 2) = 1, a(0, 3) = 3, a(1, 3) = 0 and our formulas are true in case of n = 1, 2, 3. We see that the sequence (a(0, n)) n∈N satisfies the recurrence relation of degree 3 given by a(0, n) = (n − 3)a(0, n − 1) + (n − 2)na(0, n − 2) − (n − 3)(n − 2) 2 a(0, n − 3).
A direct check confirms that the expressions for a (1, 2n), a(1, 2n + 1) given in the statement are correct. We omit simple details. Having proved the expressions for a(0, n), a(1, n) we are ready to prove positivity of a(i, n) for i ∈ {2, . . . , n − 1}. First, we get rid of the 0-th coefficient in V (n + 1, x) by differentiating this polynomial with respect to x. Thus, by differentiating the generating function V(x, t) with respect to x, we arrive to the identity Dividing the last equality by t, comparing expansions on both sides of the above identity, we see that the polynomial V ′ (n + 2, x)/(n + 1) is just the binomial convolution of the integer sequence (h n ) n∈N and the sequence of polynomials (R n (x)) n∈N , where More precisely, The power series expansion of F 1 (t) is well known and we get that h n = (2n + 1)! 2 2n n! .
In consequence, we see that the coefficients of the power series expansion of F 1 (t 2 ) are non-negative. Next, from the definition of F 2 (t) we see that R 0 (x) = 1. Moreover, we have that The above equality, together with the power series expansion 1/(1 − t) 2 = ∞ n=0 (n + 1)t n allow us to get the recurrence relation satisfied by the sequence (R n (x)) n∈N . Indeed, by comparing the coefficients near t n we easily get Let us observe that R n (0) = 0 for n ≥ 1 and x = 0 is a simple root. Indeed, the simplicity of the root x = 0 is clear from the relation Thus, R ′ n (0) = 1 for n ≥ 1. Moreover, from the recurrence relation satisfied by the sequence (R n (x)) n∈N , it is clear that deg R n = n and by writing R n (x) = n i=0 b(i, n)x i we get b(i, n) > 0 for i ∈ {1, . . . , n}. Finally, by comparing the coefficients near x i , i ∈ {1, . . . , n − 1}, on both sides of the formula (17) we get the identity (18) (i + 1)a(i + 1, n + 2) In consequence a(i, n) > 0 for each i ∈ {1, ..., n − 1}.
Remark 8.5. Let us note that the sequence of polynomials (R n (x)) n∈N coming from the power series expansion is also an interesting object. Indeed, if we write R n (x) = n i=0 b(i, n)x i , then the double array of coefficients (b(i, n)), where n ∈ N and i ∈ {0, . . . , n}, can be easily found in OEIS database as the sequence A271703 [11]. The sequence is called the unsigned Lah numbers and has closed form b(i, n) = n−1 i−1 n!/i! for n ∈ N + and i ∈ {1, . . . , n − 1}. It has many combinatorial interpretations. More precisely, the number b(i, n) counts: (1) partially ordered sets on n elements that consist entirely of k chains; (2) number of ways to split the set {1, . . . , n} into an ordered collection of n + 1 − i non-crossing nonempty sets; and (3) Dyck n-paths with n + 1 − i peaks labeled 1, 2, . . . , n + 1 − i in some order, see [8]. It would be nice to have combinatorial explanation of the identity (18).
Remark 8.6. The positivity of coefficients of the polynomial V (n, x) is a bit unexpected property and the question arises whether similar phenomena hold for the case of d = 2. It seems that the direct generalization does not work. However, based on some numerical experiments we are able to formulate the following general and all coefficients of V d (n, x) are positive. In particular, for each d ∈ N ≥2 and n ≥ d we have It is possible to prove the above conjecture in the case d = 2 by using the same approach as in the proof of Theorem 8.4. More precisely, one can prove the identity and careful analysis of the convolution form of the polynomial U 2 (n, √ x) allows to prove that if n ≡ 1 (mod 2), then all coefficients but 0th are positive; in case of n ≡ 0 (mod 2) all coefficients are positive. However, it seems that this approach can not be used in order to get the positivity of the coefficients of the polynomial V d (n, x) (or equivalently, the non-negativity of coefficients of the polynomial U d (n, x)).
It is clear that our discussion is connected with the concept of concavity of sequences. More precisely, let us recall that the sequence a = (a n ) n∈N is called logarithmically concave sequence, or a log-concave sequence for short, if a 2 n ≥ a n−1 a n+1 holds for all n ≥ k for some k. If the opposite inequality is satisfied for n ≥ k, then the sequence is called logarithmically convex (log-convex for short). Let us observe that if the sequence a is positive, then the log-concavity of a implies the log-convexity of the sequence a −1 = (a −1 n ) n∈N (and vice-versa of course). As an immediate consequence of Theorem 8.4, we get the following ν p (n + 1) = 1 then we write n + 1 = kp 2 + lp for some k, l ∈ N + with l < p and compute 0≤j1<...<j c−2k ≤c−1 (n − j 1 ) · ... · (n − j c−2k ) ≡ 0 (mod d).
Assume now that d is a composite number not of the form 2p, 2p 2 for some odd prime number p or 2 k for some integer k ≥ 2. Then there is an odd prime number p such that d = tp s , where s ∈ N + and t ≥ 3 is an integer with p-adic valuation less than s. Taking c = 2p s and n = p d − 1 in the expression 2 c! ⌊ c 2 ⌋ k=1 d 2k 0≤j1<...<j c−2k ≤c−1 (n − j 1 ) · ... · (n − j c−2k ), we obtain 2 (2p s )! 0≤j1<...<j c−2k ≤c−1 (p d − 1 − j 1 )·...·(p d − 1 − j c−2k ) has p-adic valuation less than ν p (d), which implies that it cannot be divisible by d -a contradiction.
We are checking now the numbers d of the form 2p, 2p 2 for some odd prime number p. We compute the value 2d−1 d−1 (mod p). For shortening the proof we will perform the computation for numbers d = 2p 2 as for the remaining case the computations are simpler.  Then 2 4p 2 −1 2p 2 −1 ≡ 6 (mod p) but, on the other hand, there should be 2 4p 2 −1 2p 2 −1 ≡ 2 (mod p). This implies the congruence 6 ≡ 2 (mod p) and, as a result, p | 4 -a contradiction, as p is an odd prime number.
Let us recall the following useful result. We are ready to prove the following Theorem 9.2.
(1) We have x 0 e t d d dt.
Proof. In order to get the exponential generating function for the sequence (G d (n)) n∈N we apply Lemma 9.1 with A(x) = H d (x) and get the result. The recurrence relation for the sequence (G d (n)) n∈N can be easily proved by noting the identity H d (n) = G d (n) − G d (n − 1). Indeed, from the relation (1) we get and hence the result.
We can apply the previous result and get the following general fact (with special case for d = 2 obtained by Amdeberhan and Moll). Comparing the coefficients on both sides of the first and the last expression we get the result.
Let d ∈ N ≥2 be fixed and for n ∈ N + define the matrix M d (n, x) = [H d (i + j, x)] 0≤i,j≤n−1 .
Based on numerical calculations we formulate the following Conjecture 9.5. Let d ∈ N ≥2 .
Let a, b ∈ N ≥2 with a < b. It seems to be interesting to find out if there are some integers n ≥ a and m ≥ b such that We only know that if a is a prime number then the equation (24) has no solutions, as a divides the left hand side and does not divide the right hand side.
Question 9.6. Is there any quadruple (a, b, n, m) ∈ N 4 such that 2 ≤ a < b, n ≥ a, m ≥ b and H a (n) = H b (m)? If yes, are there infinitely many such quadruples?