The 2-adic valuations of Stirling numbers of the second kind

TL;DR

This paper studies the 2-adic valuations of Stirling numbers of the second kind, confirming two conjectures related to their divisibility properties and binary digit sums.

Contribution

It proves two conjectures by Amdeberhan, Manna, and Moll regarding 2-adic valuations of Stirling numbers of the second kind.

Findings

Established the condition for equality of valuations $v_2(S(4i, 5))$ and $v_2(S(4i+3, 5))$

Proved that $v_2(S(2^n+1, k+1))= s_2(n)-1$ for all positive integers $n$

Confirmed conjectures on divisibility and binary digit sum relations of Stirling numbers

Abstract

In this paper, we investigate the 2-adic valuations of the Stirling numbers $S(n, k)$ of the second kind. We show that $v_2(S(4i, 5))=v_2(S(4i+3, 5))$ if and only if $i\not\equiv 7\pmod {32}$. This confirms a conjecture of Amdeberhan, Manna and Moll raised in 2008. We show also that $v_2(S(2^n+1, k+1))= s_2(n)-1$ for any positive integer $n$, where $s_2(n)$ is the sum of binary digits of $n$. It proves another conjecture of Amdeberhan, Manna and Moll.

Equations

$$(x{)}_n=\sum _{k=0}^{n}s(n,k)x^k\mathrm{and}{x}^n=\sum _{k=0}^{n}S(n,k)(x{)}_k$$

$$\sum _{l=0}^{\max (j,k)}s(l,j)\cdot S(k,l)={\delta }_{jk}\mathrm{and}\sum _{l=0}^{\max (j,k)}S(l,j)\cdot s(k,l)={\delta }_{jk},$$

$$v_2(S(2^n,k))=s_2(k)-1,$$

$$v_2(S(2^n+1,k+1))=s_2(k)-1.$$

$$S(n,k)=\frac{1}{k!}\sum _{j=1}^k(-1{)}^{k-j}(\genfrac{}{}{0ex}{}{k}{j})j^n.$$

$$\frac{1}{(1-x)(1-2x)...(1-kx)}=\sum _{n=1}^{\infty }S(n,k)x^n.$$

$$S(n,k)=S(n-1,k-1)+kS(n-1,k).$$

$$v_2((2^{N}r\pm 1{)}^{2^{m}i}-1)=m+N.$$

$$(2^{N}r\pm 1{)}^{2^{m}i}-1=\pm 2^{m+N}\cdot r\cdot i+\sum _{j=2}^{2^{m}i}(\pm 1{)}^j\cdot (2^{N}r{)}^j\cdot (\genfrac{}{}{0ex}{}{2^{m}i}{j}).$$

$$v_2((2^{N}r{)}^j\cdot \left(\genfrac{}{}{0ex}{}{2^{m}i}{j}\right))\ge Nj+m-v_2(j!)\ge Nj+m-j+1\ge m+N+1.$$

$$C_{m,j}:=\{2^{m}i+j:2^{m}i+j\ge k\mathrm{and}i\in \mathbb{N}\}$$

$$v_2(C_{m,j}):=\{v_2(S(2^{m}i+j,k):2^{m}i+j\ge k\mathrm{and}i\in \mathbb{N}\}.$$

$$C_{1,0}=\{2i:i\in \mathbb{N}\}\mathrm{and}{C}_{1,1}=\{2i+1:i\in \mathbb{N}\}.$$

$$C_{m-1,i_1},...,C_{m-1,i_e},\mathrm{where}1\le i_1<...<i_e.$$

$$S(n,5)=\frac{1}{24}(5^{n-1}-4^n+2\cdot 3^n-2^{n+1}+1).$$

$$24S(2^{5}t+12,5)$$

$$\equiv 5^{11}+2\cdot 3^{12}+1$$

$$\equiv 2^6(\mathrm{mod}{2}^7)$$

$$24S(2^{5}r+15,5)$$

$$\equiv 5^{14}+2\cdot 3^{15}+1$$

$$\equiv 2^6(\mathrm{mod}{2}^7).$$

$$S(2^{5}t+12,5)\equiv 8(\mathrm{mod}{2}^4)$$

$$S(2^{5}r+15,5)\equiv 8(\mathrm{mod}{2}^4).$$

$$24S(2^5{t}_1+28,5)-24S(2^5{t}_2+12,5)$$

$$\equiv 5^{27}+2\cdot 3^{28}-(5^{11}+2\cdot 3^{12})$$

$$\equiv 5^{11}(5^{2^4}-1)+2\cdot 3^{12}(3^{2^4}-1)$$

$$\equiv 5^{11}{2}^6{\alpha }_1+2\cdot 3^{12}{2}^6{\alpha }_2$$

$$\equiv 2^6(\mathrm{mod}{2}^7),$$

$$24S(2^5{t}_1+28,5)\equiv 0(\mathrm{mod}{2}^7).$$

$$24S(2^5{r}_1+31,5)-24S(2^5{r}_2+15,5)$$

$$\equiv 5^{30}+2\cdot 3^{31}-(5^{14}+2\cdot 3^{15})$$

$$\equiv 5^{14}(5^{2^4}-1)+2\cdot 3^{15}(3^{2^4}-1)$$

$$\equiv 5^{14}{2}^6{\alpha }_3+2\cdot 3^{15}{2}^6{\alpha }_4$$

$$\equiv 2^6(\mathrm{mod}{2}^7),$$

$$24S(2^5{r}_2+31,5)\equiv 0(\mathrm{mod}{2}^7).$$

$$24S(2^{6}t+60,5)$$

Full text

The research of Hong was supported partially by the National Science Foundation of China Grant # 10971145, by the Ph.D. Programs Foundation of Ministry of Education of China Grant #20100181110073 and by Program for New Century Excellent Talents in University Grant # NCET-06-0785

1. Introduction

Divisibility properties of integer sequences have long been objects of interest in number theory. $p$-Adic valuation is the modern language of divisibility. Given a prime $p$ and a positive integer $m$, there exist unique integers $a$ and $n$, with $a$ not divisible by $p$ and $n\geq 0$, such that $m=ap^{n}$. The number $n$ is called the $p$-adic valuation of $m$, denoted by $n=v_{p}(m)$. Stirling numbers are common topics in number theory and combinatorics. Let $\mathbb{N}$ denote the set of natural numbers. The Stirling numbers of the first kind, denoted by $s(n,k)$ (with a lower-case ”$s$”), count the number of permutations of $n$ elements with $k$ disjoint cycles. The Stirling numbers of the second kind $S(n,k)$ (with a capital ”$S$”) is defined for $n\in\mathbb{N}$ and positive integer $k\leq n$ as the number of ways to partition a set of $n$ elements into exactly $k$ nonempty subsets. One can characterize the Stirling numbers of the first kind and the Stirling numbers of the second kind by

[TABLE]

respectively, where $(x)_{n}$ is the falling factorial $(x)_{n}=x(x-1)(x-2)...(x-n+1)$. The Stirling numbers of the first and second kind can be considered to be inverses of one another:

[TABLE]

where $\delta_{jk}$ is the Kronecker delta. See [1]-[9] for some results on this topic.

Amdeberhan, Manna and Moll [1] studied the 2-adic valuations of Stirling numbers of second kind. Actually, they computed the 2-adic valuation $v_{2}(S(n,k))$ for $k\leq 4$. Furthermore, they found that the 2-adic valuation of the Stirling numbers of second kind of order 5 is the first nontrivial case. In fact, they showed that $v_{2}(S(4n+1,5))=v_{2}(S(4n+2,5))=0$ for all natural numbers $n$. It was also observed in [1] that $v_{2}(S(4n,5))=v_{2}(S(4n+3,5))$ for most indices. Consequently, Amdeberhan, Manna and Moll proposed a conjecture describing those indices $n$ such that $v_{2}(S(4n,5))\neq v_{2}(S(4n+3,5))$. That is,

Conjecture 1.1. [1] $v_{2}(S(4n,5))\neq v_{2}(S(4n+3,5))$* if and only if $n\in\{32j+7:j\in\mathbb{N}\}$.

On the other hand, Lengyel [7] conjectured, proved by Wannemacker [9], a special case of the 2-adic valuation of $S(n,k)$:

[TABLE]

independently of $n$, where $s_{2}(k)$ means the base $2$ digital sum of $k$. In [1], Amdeberhan, Manna and Moll also posed the following conjecture.

Conjecture 1.2. [1] For all $k$ and $1\leq k\leq 2^{n}$, we have

[TABLE]

In this paper, we concern on the $2$-adic valuations of the Stirling numbers of the second kind. We provide detailed analysis to the $k$-level of the Stirling numbers of the second kind of order 5 where $5\leq k\leq 8$ (Note that this analysis was given in [1] when $k=4$). Using this we then show that $v_{2}(S(4i,5))\neq v_{2}(S(4i+3,5))$ if and only if $i\in\{32j+7:j\geq 1\}$. This proves that Conjecture 1.1 is true. We show also that $v_{2}(S(2^{n}+1,k+1))=s_{2}(k)-1$ for any positive integer $n$, where $s_{2}(n)$ means the sum of binary digits of $n$. This confirms Conjecture 1.2.

Throughout the paper, we will use several elementary properties of $S(n,k)$, listed below:

[TABLE]

The generating function

[TABLE]

The recurrence

[TABLE]

2. Proof of Conjecture 1.1

In the present section, we show the truth of Conjecture 1.1. We begin with the following lemma.

Lemma 2.1. Let $N\geq 2$ be an integer and $r,i$ be odd numbers. For any $m\in\mathbb{Z}^{+}$,

[TABLE]

In particular, we have $v_{2}(3^{2^{m}}-1)=v_{2}(5^{2^{m}}-1)=m+2$.

Proof.

First we have

[TABLE]

On the other hand, for any integer $2\leq j\leq 2^{m}i$, since $N\geq 2$, we have

[TABLE]

Then it follows from (2.2) and (2.3) that (2.1) is true. ∎

Let $k\in\mathbb{N}$ be fixed and $m\in\mathbb{N}$. Then for $0\leq j<2^{m}$, we define

[TABLE]

and

[TABLE]

Evidently, the classes $C_{m,j}$ form a partition of $\mathbb{N}$ into classes modulo $2^{m}$ and each class $C_{m,j}$ splits into exactly two classes modulo $2^{m+1}$: $C_{m+1,j},C_{m+1,j+2^{m}}$. The class $C_{m,j}$ is called constant if $v_{2}(C_{m,j})$ consists of a singe value. This single value is called constant of the class $C_{m,j}$. Otherwise the class $C_{m,j}$ is called non-constant. We can now define inductively $m$-level. The 1-level consists of two classes:

[TABLE]

Assume that for $m\geq 2$, the $(m-1)$-level has been defined and that it consists of the $e$ classes

[TABLE]

Each class $C_{m-1,i_{j}}$ splits into two classes modulo $2^{m}$, i.e. $C_{m,i_{j}}$ and $C_{m,i_{j}+2^{m}}$. Then the $m$-level is formed by the nonconstant classes modulo $2^{m}$.

Lemma 2.2. [1] Each of the following is true:

(i). $v_{2}(C_{4,4})=v_{2}(C_{4,7})=2$.

(ii). $v_{2}(C_{4,12})>2$* and $v_{2}(C_{4,15})>2$.*

(iii). *The 4-level of $S(n,5)$ is $\{C_{4,12},C_{4,15}\}$.

*Lemma 2.3. Each of the following is true:

(i). $v_{2}(C_{5,12})=v_{2}(C_{5,15})=3.$

(ii). $v_{2}(C_{5,28})>3$* and $v_{2}(C_{5,31})>3$.*

(iii). The 5-level of $S(n,5)$ is $\{C_{5,28},C_{5,31}\}$.

Proof.

By Lemma 2.2, we know that the 4-level of $S(n,5)$ is $\{C_{4,12},C_{4,15}\}$ and $v_{2}(C_{4,12})>2$ and $v_{2}(C_{4,15})>2$. It is clear that $C_{4,12}=C_{5,12}\sqcup C_{5,28}$ and $C_{4,15}=C_{5,15}\sqcup C_{5,31}$. Then we only need to consider the four classes: $C_{5,12},C_{5,28},C_{5,15},$ $C_{5,31}$. By (1.1), we know that the Stirling number $S(n,5)$ is given by

[TABLE]

First we prove that (i) holds. By Lemma 2.1 we have $5^{2^{5}}\equiv 3^{2^{5}}\equiv 1\pmod{2^{7}}$. It then follows from (1.1) that for any nonnegative integers $r$ and $t$, we have

[TABLE]

and

[TABLE]

Thus we get

[TABLE]

and

[TABLE]

These imply that $v_{2}(C_{5,12})=3$ and $v_{2}(C_{5,15})=3$. Part (i) is proved.

Now we consider $C_{5,28},C_{5,31}$. By (2.6) we have

[TABLE]

where $t_{1},t_{2}\in\mathbb{N}$ and $\alpha_{1}$, $\alpha_{2}\in 1+2\mathbb{N}$. By (2.7) and (2.9) we get

[TABLE]

Clearly $v_{2}(S(2^{5}t_{1}+28,5))>3$, i.e., $v_{2}(C_{5,28})>3$. Similarly we compute that

[TABLE]

where $r_{1},r_{2}\in\mathbb{N}$ and $\alpha_{3}$, $\alpha_{4}\in 1+2\mathbb{N}$. Then by (2.8) and (2.10) we obtain that

[TABLE]

Hence $v_{2}(C_{5,31})>3$. Part (ii) is proved.

It is easy to check that $v_{2}(S(28,5))=6$, $v_{2}(S(60,5))=4$, $v_{2}(S(31,5))=7$ and $v_{2}(S(61,5))=4$. Since $28,60\in C_{5,28}$ and $31,63\in C_{5,31}$, $C_{5,28}$ and $C_{5,31}$ are non-constant classes. The proof of Lemma 2.3 is complete. ∎

Lemma 2.4. Each of the following is true:

(i). $v_{2}(C_{6,60})=v_{2}(C_{6,63})=4$.

(ii). $v_{2}(C_{6,28})>4$* and $v_{2}(C_{6,31})>4$.*

(iii). The 6-level of $S(n,5)$ is $\{C_{6,28},C_{6,31}\}$.

Proof.

By Lemma 2.3, we have 5-level of $S(n,5)$ is $\{C_{5,28},C_{5,31}\}$ and $v_{2}(C_{5,28})>3$ and $v_{2}(C_{5,31})>3$. It is easy to see that $C_{5,28}=C_{6,28}\sqcup C_{6,30}$ and $C_{5,31}=C_{6,31}\sqcup C_{6,63}$. So it remains to deal with $C_{6,28},C_{6,60},C_{6,31},$ $C_{6,63}$. First we consider $C_{6,28}$ and $C_{6,60}$. By Lemma 2.1 we have $5^{2^{6}}\equiv 3^{2^{6}}\equiv 1\pmod{2^{8}}$. Then

[TABLE]

where $t\in\mathbb{N}$. Hence we conclude that

[TABLE]

So $v_{2}(S(2^{6}t+60,5))=4$. This implies that $v_{2}(C_{6,60})=4$. Similarly we have

[TABLE]

Thus $v_{2}(S(2^{6}t+28,5))>4$. That is $v_{2}(C_{6,28})>4$. We can compute that $v_{2}(S(28,5))=6$ and $v_{2}(S(92,5))=5$. So $C_{6,28}$ is non-constant. In the same way, we obtain that $v_{2}(C_{6,63})=4$, $v_{2}(C_{6,31})>4$ and $C_{6,31}$ is non-constant. This completes the proof of Lemma 2.4. ∎

Lemma 2.5. Each of the following is true:

(i). $v_{2}(C_{7,92})=v_{2}(C_{7,95})=5$.

(ii). $v_{2}(C_{7,28})>5$* and $v_{2}(C_{7,31})>5$.*

(iii). *The 7-level of $S(n,5)$ is $\{C_{7,28},C_{7,31}\}$. *

Proof.

By Lemma 2.4, we get 6-level of $S(n,5)$ is $\{C_{6,28},C_{6,31}\}$ and $v_{2}(C_{6,28})>4$ and $v_{2}(C_{6,31})>4$. Clearly $C_{6,28}=C_{7,28}\sqcup C_{7,92}$ and $C_{6,31}=C_{7,31}\sqcup C_{6,95}$. First we consider $C_{7,28}$ and $C_{7,92}$. Since $5^{2^{7}}\equiv 3^{2^{7}}\equiv 1\pmod{2^{9}}$ by Lemma 2.1, then we have

[TABLE]

where $t\in\mathbb{N}$. Hence $v_{2}(C_{7,92})=5$. Similarly

[TABLE]

Thus $v_{2}(C_{7,28})>5$. Since $v_{2}(S(28,5))=6$ and $v_{2}(S(156,5))=11$, $C_{7,28}$ is non-constant. In the same way, we obtain that $v_{2}(C_{7,95})=5$, $v_{2}(C_{7,31})>5$ and $C_{7,31}$ is a non-constant class. This completes the proof of Lemma 2.5. ∎

Lemma 2.6. Each of the following is true:

(i). $v_{2}(C_{8,28})=v_{2}(C_{8,159})=6$.

(ii). $v_{2}(C_{8,156})>6$* and $v_{2}(C_{8,31})>6$.*

(iii). The 8-level of $S(n,5)$ is $\{C_{8,156},C_{8,31}\}$.

Proof.

By Lemma 2.5, we have 7-level of $S(n,5)$ is $\{C_{7,28},C_{7,31}\}$, $v_{2}(C_{7,28})>5$ and $v_{2}(C_{7,31})>5$. Clearly $C_{7,28}=C_{8,28}\sqcup C_{8,156}$ and $C_{7,31}=C_{8,31}\sqcup C_{8,159}$. Now we consider $C_{8,28}$ and $C_{8,156}$. Since $5^{2^{8}}\equiv 3^{2^{8}}\equiv 1\pmod{2^{10}}$ by Lemma 2.1, we have

[TABLE]

where $t\in\mathbb{N}$. Thus $v_{2}(C_{8,28})=6$. Similarly

[TABLE]

This implies that $v_{2}(C_{8,28})>6$. Since $v_{2}(S(156,5))=11$ and $v_{2}(S(412,5))=7$, $C_{8,156}$ is a non-constant class. In the same way, we get $v_{2}(C_{8,159})=6$, $v_{2}(C_{8,31})>6$ and $C_{8,31}$ is a non-constant class. This completes the proof of Lemma 2.6. ∎

We can now show that Conjecture 1.1 is true.

Theorem 2.7. We have that $v_{2}(S(4i,5))\neq v_{2}(S(4i+3,5))$ if and only if $i\in\{32j+7:j\in\mathbb{N}\}$.

Proof.

First let $i\in\{32j+7:j\geq 0\}$. Then $4i=2^{7}j+28\in C_{7,28}$ and $4i+3=2^{7}j+31\in C_{7,31}$ for some nonnegative integer $j$. Clearly $C_{7,28}=C_{8,28}\sqcup C_{8,156}$ and $C_{7,31}=C_{8,31}\sqcup C_{8,159}$. But by Lemma 2.6 (i) and (ii), we have $v_{2}(C_{8,156})>6$, $v_{2}(C_{8,31})>6$ and $v_{2}(C_{8,28})=v_{2}(C_{8,159})=6$. Then we obtain that

[TABLE]

and

[TABLE]

for any integer $t\geq 0$. This implies that

[TABLE]

Now we give a partition of $\{4i:i\in\mathbb{N}\}=C_{2,0}$:

[TABLE]

We also get a partition of $\{4i+3:i\in\mathbb{N}\}=C_{2,3}$:

[TABLE]

By Lemmas 2.2-2.6, we have $v_{2}(C_{3,0})=v_{2}(C_{3,3})=1$, $v_{2}(C_{4,4})=v_{2}(C_{4,7})=2$, $v_{2}(C_{5,12})=v_{2}(C_{5,15})=3$ ,$v_{2}(C_{6,60})=v_{2}(C_{6,63})=4$ and $v_{2}(C_{7,92})=v_{2}(C_{7,95})=5$. Thus $v_{2}(S(4i,5))=v_{2}(S(4i+3,5))$ except that $i\in\{32j+7:j\geq 0\}$. This completes the proof of Theorem 2.7. ∎

By Theorem 2.7, we know immediately that $v_{2}(S(4i,5))=v_{2}(S(4i+3,5))$ if and only if $i\not\equiv 7\pmod{32}$.

3. The 2-adic valuations of $S(2^{n}+1,k+1)$ and $S(2^{n}+2,k+2)$

In this section, we investigate the 2-adic valuations of $S(2^{n}+1,k+1)$ and $S(2^{n}+2,k+2)$. In 2005, Wannemacker computed the 2-adic valuation of $S(2^{n},k)$ as follows.

Lemma 3.1. [9] For all $k\in\mathbb{N}$ such that $1\leq k\leq 2^{n}$, we have

[TABLE]

First we calculate the 2-adic valuation of $S(2^{n}+1,k+1)$. For this purpose, we define $u(k)$ as follows: If $k=\sum_{i=0}^{\infty}a_{i}2^{i}$, where $a_{i}\in\{0,1\}$, then $u(k):=0$ if $a_{0}=0$, and $u(k):=t+1$ if $a_{0}=a_{1}=...=a_{t}=1$ and $a_{t+1}=0$. That is, $u(k)$ is the first index $i$ such that $a_{i}=0$.

Theorem 3.2. For all $k\in\mathbb{N}$ such that $1\leq k\leq 2^{n}$, we have

[TABLE]

Proof.

Let $k=\sum_{i=0}^{n}a_{i}2^{i}$, where $a_{i}\in\{0,1\}$. Then $s_{2}(k)=\sum_{i=0}^{n}a_{i}$. We claim that

[TABLE]

If $u(k)=0$, then (3.2) is clearly true. If $u(k)\geq 1$, then $a_{0}=1$. We may write $k=1+2+...+2^{u(k)-1}+a_{u(k)+1}2^{u(k)+1}+...+a_{n}2^{n}$. Then

[TABLE]

Thus $s_{2}(k)=u(k)+\sum_{i=u(k)+1}^{n}a_{i}$ and $s_{2}(k+1)=1+\sum_{i=u(k)+1}^{n}a_{i}$. It infers that $s_{2}(k+1)-1=s_{2}(k)-u(k).$ The claim (3.2) is proved.

By (1.3) and Lemma 3.1 we have

[TABLE]

where $\beta_{1}$, $\beta_{2}\in 1+2\mathbb{N}$.

If $2|k$, then $u(k)=0$. Hence by (3.2) and (3.4) we get

[TABLE]

It follows immediately that

[TABLE]

as desired.

If $2\nmid k$, then $u(k)\geq 1$. By (3.2) and (3.4) we have

[TABLE]

where $k+1=2^{u(k)}\beta$ for some $\beta\in 1+2\mathbb{N}$ by (3.3). In this case we also have

[TABLE]

as required. This completes the proof of Theorem 3.2. ∎

By Theorem 3.2, we know that Conjecture 1.2 is true. Finally, we calculate the 2-adic valuation of $S(2^{n}+2,k+2)$.

Theorem 3.3 Let $k\in\mathbb{Z}$ and $1\leq k\leq 2^{n}$.

(i). If $u(k)=0$, then $v_{2}(S(2^{n}+2,k+2))=s_{2}(k)-1.$

(ii). If $u(k)=1$, then $v_{2}(S(2^{n}+2,k+2))\geq s_{2}(k).$

(iii). If $u(k)\geq 2$, then $v_{2}(S(2^{n}+2,k+2))=s_{2}(k)-u(k).$

Proof.

By (1.3) we have

[TABLE]

By Theorem 3.2 we get

[TABLE]

and

[TABLE]

If $u(k)=0$, then $a_{0}=0$. By (3.2) we get $s_{2}(k+1)-1=s_{2}(k).$ Thus by (3.5), (3.6) and (3.7) we deduce that part (i) is true.

If $u(k)\geq 1$, then $a_{0}=1$ and $v(k+2)=0$. Then by (3.2) and (3.7) we have

[TABLE]

Hence by (3.5), (3.6) and (3.8) we know that parts (ii) and (iii) are true. The proof of Theorem 3.3 is complete. ∎

**Acknowledgments

**

The authors would like to thank the anonymous referee for his/her helpful comments which improved its presentation.