Some results on p-adic valuations of Stirling numbers of the second kind

1.   Introduction

Let $n$ and $k$ be nonnegative integers. Let $(x)_k$ and $\langle x\rangle_k$ stand for the falling factorial and the rising factorial, which are defined by $(x)_k: = x(x-1)(x-2)\ldots(x-k+1)$ if $k\ge 1$ and $(x)_0: = 1$, and $\langle x\rangle_k : = x(x+1)\ldots(x+k-1)$ if $k\ge 1$ and $\langle x\rangle_0 : = 1$, respectively. The Stirling number of the first kind, denoted by $s(n, k)$, counts the number of permutations of $n$ elements with $k$ disjoint cycles. One can also characterize $s(n, k)$ by

The Stirling number of the second kind $S(n, k)$ is defined as the number of ways to partition a set of $n$ elements into exactly $k$ nonempty subsets, and we have

where $\binom{k}{i}$ represents the binomial coefficient, which is defined by

One can also characterize the Stirling number of the second kind by

Given a prime $p$ and a nonzero integer $m$, there exist unique integers $a$ and $r$, with $p\nmid a$ and $r\ge 0$, such that $m = ap^r$. The number $r$ is called the $p$-adic valuation of $m$, denoted by $r = v_p(m)$. If $x = \frac{m_1}{m_2}$, where $m_1$ and $m_2$ are integers and $m_2\ne 0$, then we define $v_p(x): = v_p(m_1)-v_p(m_2)$. Let $d_p(m)$ denote the base $p$ digital sum of $m$. Many authors studied the divisibility properties of Stirling numbers of the second kind. For each given $k$, the sequence $\{S(n, k), n\ge k\}$ is known to be periodic modulo prime powers. Carlitz ^[1] and Kwong ^[2] have studied the length of this period. Chan and Manna ^[3] characterized $S(n, k)$ modulo prime powers in terms of binomial coefficients when $k$ is a multiple of prime powers. Let $p$ be a prime. For any positive integer $n$ and $k$, define

By formula (1.1) we can see that $k!S(n, k)-T_p(n, k)$ is divisible at least by $p^n$. In this sense, $T_p(n, k)$ is also known as Stirling-like numbers ^[4]. In 1990, Davis ^[5] gave a method for calculating $v_2(T_2(n, 5))$ and $v_2(T_2(n, 6))$. Then Clarke ^[4] generalized this result by applying Hensel's Lemma on the $p$-adic integers. He also conjectured that $v_p(k!S(n, k)) = v_p(T_p(n, k))$ if $n\ge k$. We note that Hong, Zhao and Zhao ^[6,7,8] have presented some important results on the divisibility properties of Stirling numbers of the second kind. In fact, they proved several conjectures proposed by Amdeberhan, Manna and Moll ^[9] and by Lengyel ^[10]. We refer the readers to ^{[11,12,13,14,15,16]} for other results on the divisibility properties of Stirling numbers of the both kinds.

Let $n$ and $k$ be positive integers. For any positive integer $r$, let $S_r(n, k)$ denote the $r$-associated Stirling number of the second kind, which is defined as the number of ways to partition a set of $n$ elements into $k$ nonempty subsets such that each of the $k$ subsets has at least $r$ elements (see, for example, ^[17]). Define $S_r(0, 0) = 1, S_r(n, 0) = S_r(0, k) = 0$ and $S_r(n, k) = 0$ if $n < rk$. Then $S_1(n, k) = S(n, k)$ and for $r\ge 2$ and $n\ge rk$ one has

Note that

and the first values of $S_2(n, k)$ can be listed as follows (see ^[17]):

We can now state the first main result of this paper as follows.

Theorem 1.1. Let $p$ be a prime number. For any positive integers $n$ and $k$ with $n\ge k+1$, we have

where $t_p(n, k): = 0$ if $k = 1$ and $t_p(n, k): = v_p(f_k(n))-v_p(\langle k+2\rangle_{k-1})$ if $k\ge 2$ with

In particular, we have

and

From Theorem 1.1, we can deduce the following result.

Theorem 1.2. Let $p$ be a prime and $n$ be a positive integer.

(i). If $n\ge 2$, then

(ii). If $n\ge 3$, then

(iii). If $n\ge 4$, then

(iv). If $n\ge 5$, then

(v). If $n\ge 6$, then

(vi). If $n\ge 7$, then

(vii). If $n\ge 8$, then

In 1995, Clarke proposed the following conjecture:

Conjecture 1.3. ^[4] Let $n$ and $k$ be nonnegative integers. If $n\ge k+1$ and $p$ is an odd prime, then

Evidently, Conjecture 1.3 implies that $v_p((n-k)!S(n, n-k)) = v_p(T(n, n-k))$ holds if $n\ge k+1$ and $p$ is an odd prime. For the cases that $1\le k\le 4$, Clarke ^[4] checked the truth of above conjecture numerically. By using Theorem 1.2, we can derive the following result which proves Conjecture 1.3 for the cases $0\le k\le 7$ and is the third main result of this paper.

Theorem 1.4. Let $p$ be an odd prime. Let $k$ be an integer such that $0\le k\le 7$. For any positive integer $n$ with $n\ge k+1$, one has

This paper is organized as follows. First of all, in the next section, we reveal some preliminaries. Subsequently, we prove Theorems 1.1 and 1.2 in Section 3. In Section 4, we give the proof of Theorem 1.4. Finally, we present some interesting consequences of Theorem 1.2 in Section 5.

2.   Some preliminaries

In this section, we present several auxiliary lemmas that are needed.

As introduced before, $S_2(n, k)$ represents the 2-associated Stirling number of the second kind. The exact values of Stirling number of the second kind can be computed by the following result.

Lemma 2.1. ^[17] For any positive integers $n$ and $k$ with $n\ge k$, we have

Lemma 2.2. Let $p$ be an odd prime and $m$ be a positive integer. For positive real number $x$, we define the function $G_{m, p}$ as follows:

Then $G_{m, p}$ is strictly monotonic decreasing on the interval $[2m, +\infty)$ if $p = 3$, and on the interval $[m, +\infty)$ if $p\ge 5$.

Proof. Since $p\ge 3$ and

one gets that $\frac{d}{dx}G_{m, p}(x) < 0$ if $m\le \frac{x}{2}$ with $p = 3$ or $m\le x$ with $p\ge 5$.

For any given real number $y$, let $\lfloor y\rfloor$ stand for the largest integer no more than $y$. We have the following result.

Lemma 2.3. Let $p$ be an odd prime. Let $n$ and $k$ be positive integers such that $k\le \min\{n, 7\}$. We have

Proof. If $\sum_{i = 0}^kv_p(n-i) = 0$, then Lemma 2.3 clearly holds. In the following we let

Then at least one element in the set $\{v_p(n), ..., v_p(n-k)\}$ is nonzero. We now divide rest of the proof into the following cases.

CASE 1. $k\in \{1, 2\}$. In this case, one can easily see that only one of $\{v_p(n), ..., v_p(n-k)\}$ is nonzero. This infers that

CASE 2. $k\in \{3, 4, 5\}$. We need to show that

Since $3\le k\le 5$, one gets that no more than two terms of $\{v_p(n), ..., v_p(n-k)\}$ can be nonzero. If there is only one term being nonzero, then we must have

So (2.1) holds. Otherwise, we can assume that $v_p(n-i)\ge 1$ and $v_p(n-j)\ge 1$ with $0\le i < j\le k$. Then one can deduce that $v_p(j-i)\ge 1$ and so $p\in\{3, 5\}$. But $1\le j-i\le k\le 5$. Hence we must have $v_p(j-i) = 1$. By using the isosceles triangle principle (see, for example, ^[18]), we derive that one of $\{v_p(n-i), v_p(n-j)\}$ must be 1. Therefore

Thus (2.1) is proved.

CASE 3. $k\in \{6, 7\}$. We need to show that

Since $k = 6$ or $k = 7$, one gets that no more than three terms of $\{v_p(n), ..., v_p(n-k)\}$ can be nonzero. If there are at most two terms being nonzero, then

So (2.2) holds.

If three terms of $\{v_p(n), ..., v_p(n-k)\}$ are nonzero, then we must have $p = 3$. Suppose that $v_3(n-i_1)\ge 1$, $v_3(n-i_2)\ge 1$ and $v_3(n-i_3)\ge 1$ with $0\le i_1 < i_2 < i_3\le k\le 7$. Then one can derive that $v_3(i_3-i_2) = v_3(i_2-i_1) = 1$. From this, we can obtain that one of $\{v_3(n-i_1), v_3(n-i_2), v_3(n-i_3)\}$ must be 1. Hence one arrives at

as (2.2) desired.

This completes the proof of Lemma 2.3.

3.   Proofs of Theorems 1.1 and 1.2

This section is dedicated to the proofs of Theorems 1.1 and 1.2. We begin with the proof of Theorem 1.1.

Proof of Theorem 1.1. First of all, Theorem 1.1 is clearly true when $k = 1$, since

So in what follows, we let $n\ge k+1\ge 3$. From Lemma 2.1, one knows that

Let $i$ be a positive integer with $k+2\le i\le 2k$. It is easy to see that

Together with (3.1) give us that

Note that $S_2(i, i-k)$ is an integer. Let

and

Then $f_k(n)$ is a polynomial with integer coefficients and $\deg f_k = k-1$, and by (3.2) one derives that

In the following, we deal with $t_p(n, k)$ for $2\le k\le 7$. We divide this into six cases.

CASE 1. $k = 2$. Since $S_2(4, 2) = 3$, by the definition of $f_2(n)$ we get that

and so

CASE 2. $k = 3$. From $S_2(5, 2) = 10$ and $S_2(6, 3) = 15$ one deduces that

and

CASE 3. $k = 4$. By $S_2(6, 2) = 25$, $S_2(7, 3) = 105$ and $S_2(8, 4) = 105$, we derive that

and

CASE 4. $k = 5$. It follows from $S_2(7, 2) = 56$, $S_2(8, 3) = 490$, $S_2(9, 4) = 1260$ and $S_2(10, 5) = 945$ that

and

CASE 5. $k = 6$. Since $S_2(8, 2) = 119$, $S_2(9, 3) = 1918$, $S_2(10, 4) = 9450$, $S_2(11, 5) = 17325$ and $S_2(12, 6) = 10395$, one deduces that

and

CASE 6. $k = 7$. By $S_2(9, 2) = 246$, $S_2(10, 3) = 6825$, $S_2(11, 4) = 56980$, $S_2(12, 5) = 190575$, $S_2(13, 6) = 270270$ and $S_2(14, 7) = 135135$, we obtain that

and

This concludes the proof of Theorem 1.1.

From the proof of Theorem 1.1, one may see that for any positive integers $n$ and $k$ with $n\ge k+1$, $f_k(n)$ is a polynomial with integer coefficients and $\deg f_k = k-1$. We also note that the expression of $f_k(n)$ depends on the values of the 2-associated Stirling number of the second kind $S_2(n, k)$. However, the computation of the exact value of $S_2(n, k)$ becomes complicated when $k\ge 8$. Therefore we only present the formulas for $f_k(n)$ and $t_p(n, k)$ for the case $1\le k\le 7$.

Subsequently, we give the proof of Theorem 1.2.

Proof of Theorem 1.2. (ⅰ). Since $n\ge 2$, from Theorem 1.1 one knows that

So part (ⅰ) of Theorem 1.2 is proved.

(ⅱ). Let $n\ge 3$. By Theorem 1.1 we derive that

as (ⅱ) desired. Hence part (ⅱ) of Theorem 1.2 is proved.

(ⅲ). Since $n\ge 4$ and $n^2-5n+6 = (n-2)(n-3)$, it follows from Theorem 1.1 that

This completes the proof of part (ⅲ).

(ⅳ). Let $n\ge 5$. By Theorem 1.1 one obtains that

as part (ⅳ) asserted.

(ⅴ). Since $n\ge 6$ and $3n^4-50n^3+305n^2-802n+760 = (n-4)(n-5)(3n^2-23n+38)$, it follows from Theorem 1.1 that

and

Then by using (3.3) and (3.4), we deduce that

So part (ⅴ) is proved.

(ⅵ). Let $n\ge 7$. By Theorem 1.1 we deduce that

Hence the proof of part (ⅵ) is finished.

(ⅶ). Since $n\ge 8$ and

it follows from Theorem 1.1 and (3.5) that

Therefore part (ⅶ) is proved. This completes the proof of Theorem 1.2.

4.   Proof of Theorem 1.4

In this section, we present the proof of Theorem 1.4.

Proof of Theorem 1.4. Let $p$ be an odd prime. First, since $S(n, n) = 1$ and

(1.2) is clearly true when $k = 0$. Now let $1\le k\le 7$. Consider the following cases.

CASE 1. $k = 1$. Since $p\ge 3$, from Theorem 1.2 one knows that

It then follows from Lemma 2.3 that

Note that $n\ge 2$. Hence we derive from Lemma 2.2 that

By (4.1) and (4.2), one then obtains that $v_p((n-1)!S(n, n-1)) < n$ as desired.

CASE 2. $k = 2$. Note that $n\ge 3$ and $p\ge 3$. It follows from Theorem 1.2 that

Thus one deduces that

with

If $v_p(3n-5) = 0$, then by Lemma 2.3 we obtain that

If $v_p(3n-5)\ge 1$, then we must have $p\ge 5$ and $v_p(n-1) = v_p(n-2) = 0$. Also note that $v_p(n) = v_p(3n) = v_p(3n-5+5)$ and $v_p(3n-5) = v_p(\frac{3n-5}{3})\le \log_p n$. This implies that $\min\{v_p(n), v_p(3n-5)\}\le 1$, and so

Since $p\ge 3$, we derive from Lemma 2.2 together with (4.4) to (4.6) that

So (4.3) and (4.7) imply that $v_p((n-2)!S(n, n-2)) < n$.

CASE 3. $k = 3$. Since $p\ge 3$, by Theorem 1.2, we get that

It follows that

with

Note that $v_p(n-2)+v_p(n-3)\le \log_p n$. Hence, by Lemma 2.3 one deduces that

It then follows from (4.9) that

since $p\ge 3$ and $n\ge 4$. So (4.8) and (4.10) give us that $v_p((n-3)!S(n, n-3)) < n$.

CASE 4. $k = 4$. Note that $n\ge 5$ and $p\ge 3$. From Theorem 1.2, we derive that

and

with

First of all, it is easy to check that $D_{4, p}(n) < 0$ for $5\le n\le 9$. In what follows, let $n\ge 10$.

If $v_p(15n^3-150n^2+485n-502) = 0$, then by Lemma 2.3 we obtain that

Now let $v_p(15n^3-150n^2+485n-502)\ge 1$. Note that

and $502 = 2\cdot 251$, $152 = 2^3\cdot 19$, $12 = 2^2\cdot 3$. Then we deduce that $p\neq5$, $v_p(n-3) = v_p(n-4) = 0$ and

since $v_p(15n^3-150n^2+485n-502) = v_p(\frac{15n^3-150n^2+485n-502}{15})+v_p(15)\le \log_p n^3+v_p(15)$. If $p = 3$, then one gets that $v_3(n) = v_3(n-1) = 0$ and $v_3(n-2)\ge1$. Also note that $\min\{v_3(n-2), v_3(15n^3-150n^2+485n-502)\}\le 1$. It follows that

If $p\ge 7$, then one has $v_p(n-2) = 0$ and only one of $\{v_p(n), v_p(n-1)\}$ can be nonzero. Without loss of generality, assume that $v_p(n) = 0$ and $v_p(n-1)\ge 1$. Then it follows that $\min\{v_p(n-1), v_p(15n^3-150n^2+485n-502)\}\le 1$. This gives us that

Since $p\ge 3$ and $n\ge 10$, by (4.12) to (4.14) together with Lemma 2.2 one derives that

Thus (4.11) and (4.15) tell us that $v_p((n-4)!S(n, n-4)) < n$.

CASE 5. $k = 5$. Since $p\ge 3$, from Theorem 1.2, we obtain that

and

with

In what follows, we show that $D_{5, p}(n) < 0$. If $v_p(3n^2-23n+38) = 0$, then by Lemma 2.3 and $v_p(n-4)+v_p(n-5)\le\log_p n$ we deduce that

Now let $v_p(3n^2-23n+38)\ge 1$. Since $38 = 2\cdot 19$, $18 = 2\cdot 3^2$ and

one gets that $v_p(n-2) = v_p(n-3) = v_p(n-5) = 0$ and one of $v_p(n)$ and $v_p(n-1)$ must be zero. If $p = 3$, then we have $v_3(n) = 0$ and

Hence one deduces that

If $p\ge 5$, then one derives that $v_p(n-1) = v_p(n-4) = 0$. Thus

Hence (4.17) to (4.20) imply that

Thus (4.16) and (4.21) give us that $v_p((n-5)!S(n, n-5)) < n$.

CASE 6. $k = 6$. Note that $n\ge 7$ and $p\ge3$. By Theorem 1.2, we deduce that

and

with

and

If $v_p(E_{6, n}) = 0$, then by Lemma 2.3 we derive that

From (4.23) and (4.24) one obtains that

In what follows, we let $v_p(E_{6, n})\ge 1$. Since

and $152696 = 2^3\cdot19087$, $41424 = 2^4\cdot3\cdot 863$, $5304 = 2^3\cdot3\cdot13\cdot17$, $1024 = 2^{10}$, $240 = 2^4\cdot3\cdot5$, $96 = 2^5\cdot3$ and $16 = 2^4$, one can deduce that $v_p(n-3) = v_p(n-6) = 0$. If $p = 3$, then only two of $\{v_3(n), v_3(n-1), v_3(n-2), v_3(n-4), v_3(n-5)\}$ can be nonzero. Suppose that $v_3(n-i)\ge 1$ and $v_3(n-j)\ge 1$ for $0\le i < j\le 6$. Then either $1\le v_3(E_{6, n})\le \log_3(n-5)^5+v_3(63)$ with $v_3(n-i) = v_3(n-j) = 1$ or $v_3(E_{6, n}) = 1$ with $\max\{v_3(n-i), v_3(n-j)\}\ge 2$, this implies that

It is easy to check that $D_{6, 3}(n) < 0$ when $7\le n\le 14$. If $n\ge 15$, then by (4.23) one gets that

If $p\ge 5$, we can obtain that $v_p(n-5) = 0$ and only one of $\{v_p(n), v_p(n-1), v_p(n-2), v_p(n-4)\}$ can be nonzero. Without loss of generality, assume that $v_p(n)\ge 1$ and $v_p(n-1) = v_p(n-2) = v_p(n-4) = 0$. It follows that $\min\{v_p(n), v_p(E_{6, n}\}\le 1$. Hence

Obviously, one has $D_{6, p}(n) < 0$ if $n = 7$. Now let $n\ge 8$. Then by Lemma 2.2 together with (4.23) and (4.27), we arrive at

as desired. Thus (4.22), (4.25), (4.26) and (4.28) tell us that $v_p((n-6)!S(n, n-6)) < n$.

CASE 7. $k = 7$. From Theorem 1.2, we obtain that

and

with

and

In what follows, we show that $D_{7, p}(n) < 0$. If $v_p(E_{7, n}) = 0$, then by Lemma 2.3 we derive that

since $v_p(n-6)+v_p(n-7)\le \log_p n$. It follows from (4.30) that

Now let $v_p(E_{7, n})\ge 1$. Note that

and $6008 = 2^3\cdot751$, $2200 = 2^3\cdot 5^2\cdot 11$, $456 = 2^3\cdot3\cdot 19$, $88 = 2^3\cdot11$, $80 = 2^4\cdot5$, $48 = 2^4\cdot3$, $16 = 2^4$. Thus one deduces that $v_p(n-7) = 0$. We divide rest of the proof into the following subcases.

CASE 7.1. $p = 3$. Then we have $v_3(n) = v_3(n-1) = v_3(n-3) = v_3(n-4) = v_3(n-6) = 0$. Moreover, one can derive that either $1\le v_3(E_{7, n})\le 2+\log_3n^4$ with $v_3(n-2) = v_3(n-5) = 1$ or $v_3(E_{7, n}) = 1$ with $\max\{v_3(n-2), v_3(n-5)\}\ge 2$ and $\min\{v_3(n-2), v_3(n-5)\} = 1$. This implies that

It follows from (4.30) and (4.32) that

CASE 7.2. $p = 5$. One notes that $v_5(n) = v_5(n-2) = v_5(n-3) = v_5(n-5) = 0$ and at most two terms of $\{v_5(n-1), v_5(n-4), v_5(n-6)\}$ can be nonzero. Moreover, if $v_5(n-4)\ge 1$ and $v_5(n-1) = v_5(n-6) = 0$, then we can deduce that

and if $v_5(n-4) = 0$ with $v_5(n-1)\ge 1$ and $v_5(n-6)\ge 1$, then we get that $\min\{v_5(n-1), v_5(n-6)\} = 1$, which infers that

Together with $v_5(E_{7, n})\le\log_5(n-5)^4$, one obtains that

It is easy to check that $D_{7, 5}(n)\le 4(2+\log_5(n-5)^6)-3n-8 = 4\log_5(n-5)^6-3n < 0$ for $8\le n\le 10$. Now let $n\ge 11$. Then it follows from Lemma 2.2 that

CASE 7.3. $p\ge 7$. One has $v_p(n-4) = v_p(n-5) = v_p(n-6) = 0$ and only one of $\{v_p(n), v_p(n-1), v_p(n-2), v_p(n-3)\}$ can be nonzero. This infers that

It follows from (4.30) and (4.35) that

Hence (4.29), (4.31), (4.33), (4.34) and (4.36) give us that $v_p((n-7)!S(n, n-7)) < n$.

This completes the proof of Theorem 1.4.

5.   More results and proofs

From Theorem 1.2, we deduce the following interesting result.

Proposition 5.1. Let $p$ be a prime. Let $n$ and $a$ be positive integers with $(a, p) = 1$.

(i). One has

(ii). If $n\ge 2$, then

and

(iii). If $n\ge 3$, then

and

Proof. Since $n\ge 1$ and $(a, p) = 1$, one may deduce that $v_p(ap^n) = n$ and $v_p(ap^n-1) = 0$.

(ⅰ). From Theorem 1.2, we derive that

Thus the truth of (ⅰ) follows immediately.

(ⅱ). Let $n\ge 2$. One gets that $v_p(ap^n-2) = v_p(2)$ and $v_p(ap^n-3) = v_p(3)$. Note that $v_p(3ap^n-5) = v_p(5)$. By Theorem 1.2, we then obtain that

and

Thus the truth of (ⅱ) follows immediately from (5.5) and (5.6).

(ⅲ). Let $n\ge 3$. It is easy to see that $v_p(ap^n-4) = v_p(4)$, $v_p(ap^n-5) = v_p(5)$, $v_p(ap^n-6) = v_p(6)$ and $v_p(ap^n-7) = v_p(7)$. From Theorem 1.2, one can deduce that

Note that $48 = 2^4\cdot 3$ and $502 = 2\cdot 251$. Thus we derive that

Together with (5.7), one then gets the truth of (5.1).

Subsequently, by Theorem 1.2 we obtain that

Note that $38 = 2\cdot 19$. It follows that

Together with (5.8), one then obtains the truth of (5.2).

Furthermore, from Theorem 1.2 we get that

Since $152696 = 2^3\cdot 19087$ and $v_2(171150) = 1$ and $n\ge 3$, one has

Note that $576 = 2^6\cdot3^2$. Together with (5.9), one derives the truth of (5.3).

Finally, by Theorem 1.2 we can deduce that

Since $6008 = 2^3\cdot 751$ and $v_2(5182) = 1$ and $n\ge 3$, one has

Note that $24 = 2^3\cdot3$. Together with (5.10), one gets the truth of (5.4).

This completes the proof of Proposition 5.1.

Moreover, we can obtain the following result.

Proposition 5.2. Let $p$ be a prime.

(i). One has

(ii). If $p\ge 3$, then

(iii). If $p\ge 5$, then

and

(iv). If $p\ge 7$, then

and

(v). If $p\ge 11$, then

(vi). For any odd prime $p$, one has

and

Proof. It is easy to check that parts (ⅰ) and (ⅱ) are true, so we omit the details.

(ⅲ). Since $p\ge 5$, by Theorem 1.2, we get that

and

Note that $502 = 2\cdot 251$ and $485\not\equiv 2 \pmod{251}$. So one has

Together with (5.11) we arrive at the truth of part (ⅲ).

(ⅳ). Let $p\ge 7$. Since $576 = 2^6\cdot3^2$, from Theorem 1.2, one gets that

and

Note that $38 = 2\cdot19$ and $23\not\equiv2\pmod{19}$, $152696 = 2^3\cdot 19087$ and $171150\not\equiv 2^3\pmod{19087}$. It infers that

and

By (5.12) and (5.13), we can deduce the truth of part (iv).

(ⅴ). Since $p\ge 11$ and $144 = 2^4\cdot3^2$, it follows from Theorem 1.2 that

By $6008 = 2^3\cdot 751$ and $5182\not\equiv 2^3\pmod{751}$, one knows that

Thus the truth of part (ⅴ) follows immediately from (5.14).

(ⅵ). Let $p\ge 3$. By Theorem 1.2, we get that

and

Hence the truth of part (ⅵ) follows from (5.15) to (5.18).

This finishes the proof of Proposition 5.2.

6.   Conclusions

Let $p$ be a prime number. Let $n$ and $k$ be positive integers. The computation of the exact $p$-adic valuation of Stirling numbers is difficult. In Theorem 1.1, we use 2-associated Stirling number $S_2(n, k)$ to represent a formula to calculate the $p$-adic valuation of Stirling numbers of the second kind $S(n, k)$. The formula of $v_p(S(n, n-k))$ depends on the value of $S_2(i, i-k)$, where $k+2\le i\le 2k$. From this, we arrive at a formula to compute $v_p(S(n, n-k))$, which enables us to show that $v_p((n-k)!S(n, n-k)) < n$ with $0\le k\le \min\{7, n-1\}$ and $p\ge 3$. Let $1\le k\le 7$ and $a$ be a positive integer with $(a, p) = 1$. For $n\ge 3$, by Proposition 5.1 we know that $n-3\le v_p(S(ap^n, ap^n-k))\le n+1$, and $v_p(S(ap^n, ap^n-k)) = n$ holds if $p > 19087$. Moreover, for any prime number $p$ with $p > 19087$, Propositions 5.1 and 5.2 also imply that $v_p(S(p, p-k)) = 1$ and $v_p(S(p^2, p^2-k)) = 2$.

Acknowledgments

The authors would like to thank the anonymous referees for careful reading of the manuscript and helpful comments and suggestions.

Conflict of interest

We declare that we have no conflict of interest.