Distribution of Nonzero Digits in a Greedy Sequence of Powers of Two

TL;DR

This paper investigates the digit distribution in a specially constructed sequence of powers of two, revealing how trailing digits can be preserved and zeroes maximized between nonzero digits using elementary methods.

Contribution

It introduces a method to construct powers of two that preserve trailing digits and maximize zeroes between nonzero digits, providing new insights into digit distribution patterns.

Findings

Sequence can preserve trailing digits

Maximizes zeroes between nonzero digits

Provides heuristic for trailing digit patterns

Abstract

Understanding the distribution of digits in the expansions of perfect powers in different bases is difficult. Rather than consider the asymptotic digit distributions, we consider the base-10 digits of a restricted sequence of powers of two. We apply elementary methods to show that this sequence of powers of two can be constructed to preserve trailing digits while locally maximizing the number of zeroes between nonzero digits. We also provide a heuristic description of the trailing digits of these powers of two.

Tables

Table 1. Table 1: First and second quartiles for frequency versus number of zeros. The third column gives a more accurate sense of how much each data point contributes to the average gap. The data were collected by running the algorithm until 1013 nonzero digits were found. That is, d 1013 subscript 𝑑 1013 d_{1013} was computed.

Gap Size	Frequency	Weighted Sum
0	0	0
1	0	0
2	98	196
3	81	243
4	34	136
5	18	90
6	11	66
7	5	35
8	2	16
9	1	9
10	1	10
11	0	0
12	1	12
Average	3.226190476

Table 2. Table 2: Third and fourth quartiles for frequency versus number of zeros. The third column gives a more accurate sense of how much each data point contributes to the average gap.

Gap Size	Frequency	Weighted Sum
0	0	0
1	0	0
2	293	586
3	235	705
4	112	448
5	62	310
6	26	156
7	21	147
8	7	56
9	1	9
10	1	10
11	1	11
12	1	12
Average	3.223684211

Table 3. Table 3: Frequency of nonzero digits. Note the higher than expected frequency of 5 and the lower than expected frequencies of 1 and 9.

Digit	1	3	5	7	9	Total
Frequency	55	128	139	124	67	513
	0.107	0.250	0.271	0.242	0.131
Digit	1	3	5	7	9	Total
Frequency	113	260	272	244	124	1013
	0.112	0.257	0.269	0.241	0.122

Table 4. Table 4: The number of zeros between consecutive nonzero digits for 1013 nonzero digits.

	Number of zeros
Digit	2	3	4	5	6	7	8	9	10	11	12
1	0	46	35	19	8	2	1	2	0	0	0
3	140	72	23	9	6	2	3	0	1	2	1
5	133	66	24	23	11	11	3	0	0	0	1
7	129	54	27	18	5	8	1	2	0	0	0
9	0	56	36	17	11	3	1	0	0	0	0

Table 5. Table 5: The corresponding probabilities for gaps of size n 𝑛 n following the nonzero digits, computed directly from Table 5 .

	Probabilities
Digit	2	3	4	5	6	7	8	9	10	11	12
1	0.000	0.407	0.310	0.168	0.071	0.018	0.009	0.018	0.000	0.000	0.000
3	0.541	0.278	0.089	0.035	0.023	0.008	0.012	0.000	0.004	0.008	0.004
5	0.489	0.243	0.088	0.085	0.040	0.040	0.011	0.000	0.000	0.000	0.004
7	0.529	0.221	0.111	0.074	0.020	0.033	0.004	0.008	0.000	0.000	0.000
9	0.000	0.452	0.290	0.137	0.089	0.024	0.008	0.000	0.000	0.000	0.000

Equations

$$2^{\chi }=b_1\cdot 10^{d_1}+b_2\cdot 10^{d_2}+\cdots ,$$

$$2^{p_n}\equiv b_1\cdot 10^{d_1}+b_2\cdot 10^{d_2}+\cdots +b_n\cdot 10^{d_n}(\mathrm{mod}10^{d_n+1}),$$

$$2^p\equiv b_1\cdot 10^{d_1}+b_2\cdot 10^{d_2}+\cdots +b_n\cdot 10^{d_n}(\mathrm{mod}10^e),$$

$$2^{p_2}=8\cdot 10^0+3\cdot 10^3+\cdots .$$

$$2^{p_3}=8\cdot 10^0+3\cdot 10^3+1\cdot 10^6+\cdots$$

$$2^{p_4}=8\cdot 10^0+3\cdot 10^3+9\cdot 10^6+1\cdot 10^{13}+\cdots .$$

$$2^{p_n}\equiv b_1\cdot 10^{d_1}+b_2\cdot 10^{d_2}+\cdots +b_n\cdot 10^{d_n}(\mathrm{mod}10^{d_n+1}),$$

$$2^p\equiv b_1\cdot 10^{d_1}+b_2\cdot 10^{d_2}+\cdots +b_n\cdot 10^{d_n}(\mathrm{mod}10^e).$$

$$2^{i+4j{5}^{m-1}}\equiv x+a_{j}10^m(\mathrm{mod}10^{m+1}),$$

$$2^{p_n}\equiv b_1+b_{2}10^{d_2}+b_{3}10^{d_3}+\cdots +b_{n}10^{d_n}(\mathrm{mod}10^{d_n+1}),$$

$$2^{p_{n-1}+4j{5}^{d_n}}\equiv x+a_{j}10^{d_n}(\mathrm{mod}10^{d_n+1}).$$

$$2^{p_{n+1}}\equiv b_1\cdot 10^{d_1}+b_2\cdot 10^{d_2}+\cdots +b_{n+1}\cdot 10^{d_{n+1}}(\mathrm{mod}10^{d_{n+1}+1}),$$

$$2^{p_n+4j\cdot 5^{d_n-1}+4k\cdot 5^{d_n}}\equiv b_1\cdot 10^{d_1}+b_2\cdot 10^{d_2}+\cdots +a_j\cdot 10^{d_n}+c_{j,k}\cdot 10^{d_n+1}(\mathrm{mod}10^{d_n+2}),$$

$$\left(2^{4(j-1)5^{d_n-1}+4k\cdot 5^{d_n}}-1\right)2^{p_n+4\cdot 5^{d_n-1}}\equiv (a_j-a_1)\cdot 10^{d_n}+(c_{j,k}-c_{1,0})\cdot 10^{d_n+1}(\mathrm{mod}10^{d_n+2}).$$

$$2^{4(j-1)5^{d_n-1}+4k\cdot 5^{d_n}}-1$$

$$2^{2}N\equiv (a_j-a_1)+10(c_{j,k}-c_{1,0})(\mathrm{mod}100),$$

$$2^{p_n}=b_1\cdot 10^{d_1}+b_2\cdot 10^{d_2}+\cdots +b_n\cdot 10^{d_n}+\cdots .$$

$$2^{p_2+0\cdot 5^2}$$

$$2^{p_2+4\cdot 5^2}$$

$$2^{p_2+8\cdot 5^2}$$

$$2^{p_2+12\cdot 5^2}$$

$$2^{p_2+16\cdot 5^2}$$

$$2^{p_2+0\cdot 5^2+16\cdot 5^3}$$

$$2^{p_2+12\cdot 5^2+16\cdot 5^3}$$

$$2^{p_n}$$

$$2^{p_n^{\prime }}$$

$$2^{p_n}$$

$$2^{p_n^{\prime }}$$

$$2^{p_n}$$

$$2^{p_n^{\prime }}$$

$$\mathbb{E}[\text{\# of zeros between nonzero digits}]=\frac{13}{4}$$

$$\frac{3}{4}\mathbb{E}[\text{Starting digit 3, 5, or 7}]+\frac{1}{4}\mathbb{E}[\text{Starting digit 1 or 9}].$$

$$N=2+\sum _{n=1}^{\infty }\frac{n}{2^{n+1}}.$$

$$\sum _{n=0}^{\infty }n{r}^n=\frac{r}{(1-r{)}^2}.$$

Full text

1 Introduction

Let’s consider the distribution of the base-10 digits of powers of two. A natural heuristic argues that the fraction of nonzero digits should be $9/10$. Previous results proved bounds on the density of nonzero digits, such as Stewart’s lower bound of $\ln(x)/(C+\ln\ln(x))$ nonzero digits [1] or Radcliffe’s elementary lower bound of $\log_{4}(x)$ nonzero digits [2]. Instead of proving a lower bound over all powers of two, we consider the problem of locally maximizing the sparsity of nonzero digits in a 10-adic sequence of powers of $2$. We use a very greedy algorithm that has surprising properties, and is well-suited to analyzing 10-adic powers of $2$.

Recall that every power of two (and every natural number) can be uniquely represented in the form

[TABLE]

where $d_{1}<d_{2}<\cdots$ and $1\leq b_{i}\leq 9$. In other words, this representation singles out the exponents $d_{n}$ that correspond to nonzero digits in the base-10 expansion of $2^{\chi}$. Our goal is construct a sequence of exponents $p_{n}$ of $2$ such that the the sequence $d_{1},d_{2},\cdots$ is locally sparse. That is, we try to locally maximize $d_{n}-d_{n-1}$, the gaps between adjacent nonzero digits. We explain the algorithm formally and provide a running example.

We proceed as follows. Suppose we have constructed $p_{n}$ satisfying

[TABLE]

where $1\leq b_{i}\leq 9$ and $0=d_{1}<d_{2}<\cdots$.

Then, for each $p_{n+1}$, repeat this process to find $p_{n+2}$. If some value of $p_{n+1}$ yields a value of $p_{n+2}$ that is smaller than the value of $p_{n+2}$ by some other $p_{n+1}$, discard the smaller $p_{n+1}$ and its descendants $p_{n+2}$, etc.

Consider all powers $2^{p}$ with

[TABLE]

for $e>d_{n}$ and retain only those exponents $p$ that give the largest value of $e-d_{n}$. We denote each of these values of $p$ as $p_{n+1}$. For example, let $p_{1}=3$ so $2^{p_{1}}\equiv 8\cdot 10^{0}\pmod{10^{1}}$. In our example, we look at $p>3$ and find that $p\equiv 3\pmod{100}$ gives $e-d_{1}=3$. We can show that this is the largest possible $e-d_{1}$. Hence, we can take $p_{2}$ to be the residue class $3\pmod{100}$.

In our example, let’s take $p_{2}=103$. Then

[TABLE]

Then we look for $p>103$ such that $e-d_{2}$ (here, we can see that $d_{2}=3$ is maximal). We find that if $p\equiv 2403\pmod{12500}$, then $e-d_{2}=2$. However, if $p\equiv 2013\pmod{12500}$ then $e-d_{2}=3$, and it can be shown that this is the maximum possible $e-d_{2}$. Therefore, we take $p_{3}$ to be the residue class $2013\pmod{12500}$ and discard all other residue classes such as $p\equiv 2403\pmod{12500}$.

We illustrate the next two iterations. Taking $p_{3}=2103$, we find

[TABLE]

Then we search for $p>2103$ and find $p_{4}=670414603$. This gives

[TABLE]

Hence we only preserve the residue class $670414603\pmod{4\cdot 5^{12}}$. The process continues thereafter.

This yields a sequence of exponents $p_{1},p_{2},p_{3},\ldots$ of $2$ and corresponding exponents $d_{1},d_{2},d_{3},\ldots$ of $10$. In other words, we are using a greedy strategy to construct a $10$-adic power of $2$ whose infinite $10$-adic expansion has many zeros.

In other words, when constructing $p_{n+1}$, we preserve $b_{i}$ and $d_{i}$ for $1\leq i\leq n$. Suppose we have constructed $p_{n}$ satisfying

[TABLE]

where all $b_{i}\neq 0$ and $0=d_{1}<d_{2}<\cdots$. We only consider powers $2^{p}$ with

[TABLE]

With this restriction, we show in Section 2 that there are very few possible values of $p$. More specifically, suppose we have constructed $p_{n}\pmod{4\cdot 5^{d_{k}-1}}$. Then if $e=d_{n}+1$, the only valid exponents $p$ are $p\equiv p_{n}+i\cdot 4\cdot 5^{d_{n}-1}\pmod{4\cdot 5^{e}}$ for $1\leq i\leq 5$. Pseudocode for the algorithm can be found in the Appendix.

Our strategy does not necessarily give powers of $2$ with the highest density of zeros. For example, $2^{103}$ has only $5$ zeros among its 32 digit, while $2^{102}$ has seven zeros among its $31$ digits. However, we obtain many zeros in the early terms of the expansion.

In Section 2, we show that we always have $d_{n+1}-d_{n}\geq 2$, so we can always obtain at least $2$ consecutive zeros after each nonzero digit. Moreover, we show that our constraints create a unique sequence of $d_{n}$. We also investigate various other phenomena with digit distributions that occur. In Section 3, we provide a heuristic that predicts that the density $\#\{d_{n}\leq x\}/x\rightarrow 13/4$ as $x\rightarrow\infty$. This agrees well with the data we computed in Tables 1 and 2.

2 Properties of the greedy sequence

We first state and prove a lemma.

Lemma 1**.**

Let $i\geq j\geq m\geq 1$. Then $2^{i}\equiv 2^{j}\pmod{10^{m}}$ if and only if $i\equiv j\pmod{4\cdot 5^{m-1}}$.

Proof.

Since $2$ is a primitive root mod $5^{n}$, we have $2^{i}\equiv 2^{j}\pmod{5^{m}}$ if and only if $i\equiv j\pmod{4\cdot 5^{m-1}}$. Now since $i\geq j\geq m$, we have $2^{i}\equiv 2^{j}\equiv 0\pmod{2^{m}}$. The result follows from Chinese Remainder Theorem. ∎

Lemma 2**.**

Let $i>m\geq 1$. Suppose $2^{i}\equiv x\pmod{10^{m}}$. Write

[TABLE]

with $0\leq a_{j}\leq 9$. As $j\geq 0$ runs through the integers $\pmod{5}$, the values of $a_{j}$ run through all the integers in either $\{1,3,5,7,9\}$ or $\{0,2,4,6,8\}$.

Proof.

Since $i>m$, we know that $2^{i+4j5^{m-1}}\equiv 2^{i}\equiv 0\pmod{2^{m+1}}$. Since $2^{m}$ divides $x$, we can divide through by $2^{m}$, which shows that $x+a_{j}\equiv 0\pmod{2}$ Therefore the parity of $a_{j}$ is determined by $x$. Lemma 1 says that distinct values of $j\pmod{5}$ give distinct values of $a_{j}$. The lemma follows easily. ∎

For $n=1,2,3,\ldots,$ choose integers $p_{n}$ and write

[TABLE]

where $1\leq b_{i}\leq 9$ for each $i$ and where $d_{1}=0<d_{2}<d_{3}<\cdots<d_{n}$. Assume that the $p_{n}$ are chosen by the “greedy strategy”. That is, $p_{1},p_{2},\ldots,p_{n-1}$ are chosen so that $p_{n}>d_{n}$ and $d_{n}-d_{n-1}$ is as large as possible.

Theorem 1**.**

If the $p_{n}$ are chosen by the greedy strategy, then the $b_{i}$ are odd for $i\geq 2$.

Proof.

In Lemma 2, let $x=b_{1}+\cdots+b_{n-1}10^{d_{n-1}}$, let $i=p_{n-1}$, and let $m=d_{n}$. Write

[TABLE]

For some $j=j_{0}$, we have $a_{j_{0}}=b_{n}$. Lemma 1 shows that as $j$ runs through the integers $\pmod{5}$, $a_{j}$ will run through the integers between [math] and $9$ that have the same parity as $b_{n}$. Therefore, if $b_{n}$ is even, we can pick $j$ such that $a_{j}=0$. This means that $d_{n}$ is not maximal, so $b_{n}$ must be odd, as desired. ∎

Theorem 2**.**

Let

[TABLE]

where $p_{n}$ is chosen by the greedy strategy described in the introduction. Then $d_{n+1}-d_{n}\geq 2$.

Proof.

Suppose $d_{n+1}-d_{n}=1$. Then

[TABLE]

with $1\leq a_{j},c_{j,k}\leq 9$ for $j,k=0,1,2,3,4$. Theorem 1 says that $a_{j}$ and $c_{j,k}$ must be odd. When $j\geq 1$,

[TABLE]

Since $5^{d_{n}}$ divides

[TABLE]

and $p_{n}+4\cdot 5^{d_{n}-1}\geq d_{n}+2$ (this is why we used $j\geq 1$ instead of $j\geq 0$), we can divide by $10^{d_{n}}$ to obtain

[TABLE]

for some integer $N$. Since $c_{j,k}$ and $c_{1,0}$ are odd, we find that $a_{j}\equiv a_{1}\pmod{4}$. This allows at most three values of $a_{j}$. But Lemma 2 says that $a_{j}$ runs through all values congruent to $a_{1}\pmod{2}$, and we have omitted only $j=0$ from consideration, so there are $4$ remaining distinct values of $a_{j}$. This contradiction implies that $d_{n+1}-d_{n}\geq 2$. ∎

We now illustrate an example of Lemma 2.

Example 2.1**.**

We can compute $2^{103}=8+3\cdot 10^{3}+4\cdot 10^{4}+\cdots$. Here the third nonzero digit is $4$, which is even. As Lemma 2 claims, we can construct the modified exponent $p^{\prime}=103+4\cdot\varphi(5^{4})$ such that $2^{p^{\prime}}=8+3\cdot 10^{3}+1\cdot 10^{6}+\cdots$. The effect of adding $4\cdot\varphi(5^{4})$ to the exponent is that the nonzero even digit $4$ corresponding to $10^{4}$ was changed to a [math]. Furthermore, note that aside from the units digit, all of the nonzero digits in $2^{p^{\prime}}\pmod{10^{7}}$ are odd.

Now suppose that we have greedily constructed $p_{1},p_{2},\ldots,p_{n}$. Recall that

[TABLE]

We claim that there is a unique choice of $b_{n}$ that corresponds with $p_{n+1}$.

Definition**.**

For $1\leq k\leq 9$, We say that $b_{n}=k$ is $\beta$-forceable if there exists $e$ such that $2^{p_{n}}\equiv 2^{e}\pmod{10^{d_{n}+\beta}}$. In other words, we can find $e$ such that $2^{e}$ has at least $\beta$ zeros between $b_{n}$ and $b_{n+1}$. Equivalently, we require that $d_{n+1}-d_{n}\geq\beta+1$. On the other hand, we say that $b_{n}$ is $\beta$–unforceable if there does not exist $e$ such that $2^{p_{n}}\equiv 2^{e}\pmod{10^{d_{n}+\beta}}$.

Example 2.2**.**

We consider $p_{1}=3,p_{2}=103$, with $d_{1}=0$ and $d_{2}=3$. So $2^{p_{1}}=8\cdot 10^{0}$ and $2^{p_{2}}=8\cdot 10^{0}+3\cdot 10^{3}+\cdots$. Consider the values of $2^{p_{2}+4k\cdot 5^{2}}$:

[TABLE]

Therefore, by looking at the parity of $b_{3}$ for each possible $b_{2}$, we see that $b_{2}=1$, $b_{2}=5$, and $b_{2}=9$ are [math]-forceable (equivalently, they are $1$-unforceable). Therefore, we only need to consider $b_{2}=3$ and $b_{2}=7$.

Next, we calculate $2^{p_{2}+0\cdot 5^{2}+4k\cdot 5^{3}}$ and $2^{p_{2}+12\cdot 5^{2}+4k\cdot 5^{3}}$ for $0\leq k\leq 4$ and find

[TABLE]

Note that there is a unique choice of $k$ that forces a zero after $b_{2}$. Therefore, we see that $b_{2}=3$ is $2$-forceable, while $b_{2}=7$ is $1$-forceable.

Corollary 1**.**

Consider the sequence $p_{1}$, $p_{2}$, $\cdots$, $p_{n-1}$, $p_{n}$. Suppose $b_{n}$ is $2$-forceable. If $b_{n}=3$ then it is the unique $2$-forceable $b_{n}$. The same result holds if $b_{n}=7$. That is, there is no alternate sequence $p_{1}$, $p_{2}$, $\cdots$, $p_{n-1}$, $p_{n}^{\prime}$ such that $b_{n}^{\prime}\neq b_{n}$ and $b_{n}^{\prime}$ is $2$-forceable.

Proof.

Suppose that $b_{n}$ is $2$-forceable and $b_{n}=3$; the proof for $b_{n}=7$ is analogous. By definition, we can construct $p_{n}$ such that $2^{p_{n}}\equiv x+3\cdot 10^{d_{n}}\pmod{10^{d_{n}+3}}$, where $x$ is a $d_{n}$ digit integer. We split the proof into two cases.

Case 1: We replace $b_{n}$ with $1$, $5$, or $9$.

This case is simple; working modulo $4$ as in the proof of Theorem 2, we must have $b_{n}-3\equiv 0\pmod{4}$. So $b_{n}$ cannot be $1$, $5$, or $9$.

Case 2: We replace $b_{n}$ with $7$.

If we replace $b_{n}$ with $7$, we can force at least one more zero by using the modulo $4$ argument as in the proof of Theorem 2. Now we have

[TABLE]

where $0\leq a\leq 9$ and $p_{n}^{\prime}=p_{n}+4k\cdot 5^{d_{n}-1}$ for some $1\leq k\leq 4$. Now we have $p_{n}\geq d_{n}$ and $5^{d_{n}}\mid 2^{p_{n}^{\prime}-p_{n}}-1=2^{4k\cdot 5^{d_{n}-1}}-1$. Therefore, we can divide out by $10^{d_{n}}$; we arrive at $4+100a\equiv 0\pmod{1000}$. Thus $a$ must be odd, which implies that $b_{n}=7$ is $2$-unforceable, as desired. ∎

Corollary 2**.**

Suppose that $b_{n}$ is $2$-forceable. If $b_{n}=5$, then it is the unique $2$-forceable $b_{n}$.

Proof.

The proof is entirely analogous to the proof of Corollary 1. ∎

The same property does not hold if either $1$ or $9$ happen to be $2$-forceable because their difference is a multiple of $8$. However, the following two properties do hold.

Corollary 3**.**

$b_{n}=1$* is $2$-forceable if and only if $b_{n}=9$ is $2$-forceable.*

Proof.

Suppose that $p_{1}$, $p_{2}$, $\ldots$, $p_{n}$ has been constructed such that $b_{n}=1$. Then the modified sequence $p_{1}$, $p_{2}$, $\ldots$, $p_{n}^{\prime}$ has $b_{n}^{\prime}=9$.

Applying the modulo $4$ argument as in Theorem 2, we find that if $b_{n}=1$ is $1$-forceable, then $b_{n}^{\prime}=9$ is also $1$-forceable. The reverse implication also follows. Now assume that $b_{n}=1$ is $1$-forceable. So we have

[TABLE]

for $0\leq a,b\leq 9$ and $p_{n}^{\prime}=p_{n}+4k\cdot 5^{d_{n}-1}$ for $1\leq k\leq 4$.

Subtracting and dividing by $10^{d_{n}}$, we obtain $8+100(a-b)\equiv 0\pmod{1000}$. Therefore, $a\equiv b\pmod{2}$. By hypothesis, $b_{n}=1$ is $2$-forceable, so $a$ is even. Therefore $b$ is also even, which means that $b_{n}=9$ is also $2$-forceable, as desired. ∎

Corollary 4**.**

Suppose that $b_{n}$ is $3$-forceable. If $b_{n}=1$, then it is the unique $3$-forceable $b_{n}$. The same result holds if $b_{n}=9$. That is, exactly one of $b_{n}=1$ and $b_{n}=9$ is $3$-forceable; the other $b_{n}$ is $3$-unforceable.

Proof.

Suppose that $b_{n}=1$ or $b_{n}=9$ is $2$-forceable. Then by Corollary 3, there exist $p_{n}$ and $p_{n}^{\prime}$ such that

[TABLE]

where again $0\leq a,b\leq 9$ and $p_{n}^{\prime}=p_{n}+4k\cdot 5^{d_{n}-1}$ for $1\leq k\leq 4$.

Subtracting and dividing out by $10^{d_{n}}$, we have $8+1000(a-b)\equiv 0\pmod{10000}$. So $a$ and $b$ must have different parities. If $b_{n}=1$ is $3$-forceable, then $a$ is even, so $b$ is odd. Therefore $b_{n}=9$ is $3$-unforceable, as desired. The same logic holds if $b_{n}=9$ is $3$-forceable. The claim is thus proved. ∎

3 Heuristic argument for the expected sparsity of nonzero digits

We justify a heuristic that predicts the expected number of zeros between nonzero digits in our sequence.

Heuristic 1** (Expected Gap of zeros).**

Let $\mathbb{E}[x]$ be the expected value of $x$. Then

[TABLE]

Proof.

Empirical evidence (refer to Table 3) suggests that the probability that a nonzero digit is 1 or 9 is approximately $\dfrac{1}{4}$. We can also think of $1$ and $9$ being their own group of digits, with $3,5,$ and $7$ being the other three groups of digits. According to Corollaries 1 and 2, if the nonzero digit is $3,5,$ or $7$, there is only one choice of starting digit that is $2$-forceable. We have already seen that if the leading digit is $3,5,$ or $7$, we can force two zeros with a greedy algorithm, and if the leading digit is $1$ or $9$, we can force three zeros with a greedy algorithm. Then $\mathbb{E}[\text{\# of zeros between nonzero digits}]$ is equal to

[TABLE]

For convenience, let $N=\mathbb{E}[\text{Starting digit 3, 5, or 7}]$ and $M=\mathbb{E}[\text{Starting digit 1 or 9}]$. If we assume that the parity of each digit thereafter is uniformly random, then the probability that we obtain exactly $n$ zeros after the two forced zeros is $\frac{1}{2^{n+1}}$. Hence,

[TABLE]

From elementary calculus, we have

[TABLE]

Therefore, $N$ evaluates as

[TABLE]

We now turn to evaluating $M$. Intuitively, Corollary 4 tells us that $M$ should be larger than $N$. More precisely, Corollary 4 shows that we have three guaranteed zeros. Assuming that every zero thereafter is uniformly random, we have

[TABLE]

Thus, $\mathbb{E}[\text{\# of zeros between nonzero digits}]=\dfrac{1}{4}M+\dfrac{3}{4}N=\dfrac{1}{4}\cdot 4+\dfrac{3}{4}\cdot 3=\dfrac{13}{4}$, as desired. ∎

The estimation of $\frac{13}{4}=3.25$ is remarkably close to the empirical average of $3.247$ gathered for $d_{1013}$ (see Tables 1 and 2).

4 Acknowledgments

The author would like to thank Lawrence C. Washington for suggesting the problem and for his helpful guidance throughout the process.

Appendix A Data

Appendix B Code

We define our algorithm as follows, where $m=p_{k}$.