A Counterexample to a Purported Construction of Normal Numbers

John M. Campbell Department of Mathematics and Statistics Dalhousie University Halifax, NS B3H 4R2 Canada

                        jh241966@dal.ca

Abstract

Pollack and Vandehey noted that a purported construction of normal numbers given by Szüsz and Volkmann requires an additional condition. As we demonstrate, not only is this condition required in part of the proof of Szüsz and Volkmann’s main result, but their original claim is false. We give an explicit counterexample via a decimal expansion that satisfies the conditions of Szüsz and Volkmann but is not normal. This counterexample provides a non-normal analogue of the normal number 0.(1)(4)(9)(16)(25) ⋯ considered by Besicovitch.

1 Introduction

In the field of probabilistic number theory, the study of distributions of sequences of integers and digits plays a fundamental role. Such distributions give rise to questions as in the following. For a given real number r, does a given sequence of digits appear, in some specified way, in the decimal expansion of r as often as every other sequence of the same length? Informally, a real number satisfying this property is said to be normal (in base 10). There is a long and rich history concerning normal numbers, and a notable construction of normal numbers is due to Szüsz and Volkmann [5]. A corrected version of this construction given by Pollack and Vandehey [4] implies the normality of one of the most famous constants in the study of normal numbers.

By taking the sequence (1, 2, 3, 4, …) of consecutive positive integers, and forming the real number

0.123456789101112 …  (1)

by concatenating their decimal expansions, we get Champernowne’s constant, indexed in the On-Line Encyclopedia of Integer Sequences [6] as sequence \underline{A033307}. This constant was proved to be normal by Champernowne in 1933 [2] and provides the first known and explicit construction of a normal number. Similarly, by taking the sequence (2, 3, 5, 7, …) of increasing prime numbers (given in the OEIS as \underline{A000040}), we obtain the constant

0.235711131719232 …  (2)

known as the Copeland–Erdős constant (see \underline{A033308}). This constant was proved to be normal by Copeland and Erdős in 1946 [3]. The research interest surrounding the constants in (1) and (2) raises questions as to whether or not a given sequence of nonnegative integers gives rise to a normal number using the analogous construction.

Informally, a real number α is said to be normal in base g ≥ 2 if, for a sequence of n base-g digits, this sequence appears with frequency g^{−n} in the base-g expansion of α. For a nonnegative, real-valued function f defined on ℕ, we form the base-g expansion

0.⌊f(1)⌋⌊f(2)⌋ …  (3)

by analogy with the base-10 concatenations in (1) and (2). A construction due to Szüsz and Volkmann [5] purports to ensure the base-g normality of a concatenation of the form indicated in (3) for a differentiable function f satisfying certain growth properties. Pollack and Vandehey [4] noted (as discussed in more detail below) that Szüsz and Volkmann’s method requires that

lim_{x→∞} \frac{log f(x)}{log x} ≤ 1,  (4)

but the condition in (4) was not included in Szüsz and Volkmann’s work. The corrected version of Szüsz and Volkmann’s construction provides a way of proving the normality of Champernowne’s constant, for the f(n) = n case of (3).

Since the required condition (4) was not included in Szüsz and Volkmann’s main result [5], this raises the problem of constructing an explicit counterexample. We give such a construction proving the non-normality of our constructed constant, which may be seen as a non-normal analogue of the normal number

0.149162536496481 …  (5)

related to the work of Besicovitch [1] (see also \underline{A001191}), where (5) is given by the f(n) = n^{2} case of (3). Observe that by setting f(x) = x^{2}, the required condition in (4) is not satisfied, so that Szüsz and Volkmann’s method cannot be used to prove the normality of (5) [4]. Explicitly, Pollack and Vandehey noted the following, where η(f) = β in the notation of Szüsz and Volkmann:

“Szüsz and Volkmann mistakenly claim that their result is strong enough to prove [the normality of (5)]. In Theorem 2 of their paper, they need an additional condition that β ≤ 1, because if β > 1, then the bound in line (3.11) would be M_{k} = O(1), which would cause their condition (v) to fail.” [4, p. 759]

1.1 Szüsz and Volkmann’s purported construction

Our terminology and notation concerning normal numbers are mainly based upon Szüsz and Volkmann’s work [5]. In this regard, we let g ≥ 2 denote a fixed base, we let α be a real value, we let E denote a block of digits, and we let A_{E}(α, n) denote the number of copies of E within the first n digits of α. The value α is normal of order k in base g if

lim_{n→∞} \frac{A_{E}(α, n)}{n} = \frac{1}{g^{ℓ(E)}}  (6)

for all blocks E satisfying ℓ(E) = k, letting ℓ(E) denote the number of digits in E, counting multiplicities. The value α is normal in base g if it is normal (in base g) for all possible orders k ≥ 1.

Let f denote a real-valued function on a domain containing ℕ, and let f satisfy f(n) ≥ 0 and f(n) > 0 for sufficiently large n. We then let α(f) = α_{g}(f) denote the real number such that the base-g expansion of α(f) equals 0.b_{1}b_{2}..., writing b_{n} in place of the base-g expansion of ⌊f(n)⌋. In this regard, a Champernowne function is a function f satisfying lim_{x→∞} f(x) = ∞ and such that α_{g}(f) is normal in base g for all possible integers g ≥ 2. To construct Champernowne functions, according to the purported construction given by Szüsz and Volkmann [5], this requires constraints on the growth of f, formulated in terms of the η-function given as follows. For a function h that is real-valued and positive, we set η(h) = lim_{x→∞} \frac{log h(x)}{log x}, assuming this limit exists.

The main result in Szüsz and Volkmann’s paper is given as Theorem 2 [5] and asserts that if f is differentiable and if f is monotonically increasing and positive for all sufficiently large arguments and if η(f) and η(f′) exist and η(f) > 0, then f is a Champernowne function. Since the condition such that η(f) ≤ 1 would also be required in the proof of this result [4], this leads us to the problem of constructing a function f that satisfies all of Szüsz and Volkmann’s conditions and such that (3) is not normal, i.e., with η(f) > 1.

2 Main construction

For positive integers n, we let

h_{1}(n) := \begin{cases}0, & if n < 10; \\ 10^{⌊log_{10}(n)⌋}, & otherwise. \end{cases}

Also, for positive integers n, we let

h_{2}(n) := 10^{⌊log_{10}(n)⌋} ⌊\frac{∑_{i=1}^{10^{⌊log_{10}(n)⌋}} h_{1}(i)}{10^{⌊log_{10}(n)⌋}}⌋ + ∑_{i=10^{⌊log_{10}(n)⌋}+1}^{n} h_{1}(i).  (7)

The integer sequence (h_{2}(n))_{n≥1} is given in the OEIS as \underline{A380904}.

Example 1. Initial entries in the integer sequence given by (7) are such that

(h_{2}(n))_{n≥1} = (0, 0, 0, 0, 0, 0, 0, 0, 0, 10, 20, 30, 40, 50, 60, 70, 80, 90, 100, 110, 120, …),

and the behavior of the integer sequence (h_{2}(n))_{n≥1} is further illustrated in Section 2.1 below.

2.1 Outline

To provide an explicit counterexample to Szüsz and Volkmann’s claim, we construct a function f(x) (satisfying the required conditions in Szüsz and Volkmann’s claim) on ℝ_{≥0}, as below, such that the integer part of f(n) is h_{2}(n) for positive integers n, and we prove the desired equality ⌊f(n)⌋ = h_{2}(n) in Lemma 3 below. The motivation behind the use of the integer sequence (h_{2}(n))_{n≥1} for the purposes of this paper is outlined below.

Informally, the sequence (h_{2}(n))_{n≥1} is defined in such a way so that it grows quadratically, with η(f) = 2, but in such a way so that the number of trailing zeroes in the entries of (h_{2}(n))_{n≥1} is made to be sufficiently large so as to prevent normality. Moreover, this needs to be in such a way so that f is sufficiently well-behaved, i.e., so that the limit associated with η(f′) exists. This is illustrated below.

For n ≥ 9, differences between consecutive terms in (h_{2}(n))_{n≥1} are all positive powers of 10. Informally, these powers of 10 are determined in such a way so that the number of trailing zeroes among the terms in (h_{2}(n))_{n≥1} continues to grow in a sufficiently well-behaved way. With regard to the initial values of h_{2} shown in Example 1, starting from h_{2}(10) = 10, we add the value 10 iteratively to form new terms until we reach an index of the form 10^{m} − 1, with (h_{2}(n))_{10≤n≤99} = (10, 20, 30, …, 890, 900). The next term h_{2}(100) is the next value after h_{2}(99) = 900 with at least 2 trailing zeroes, with h_{2}(100) = 1000. We then add the value 100 iteratively to form new terms until we again reach an index of the form 10^{m} − 1, with (h_{2}(n))_{100≤n≤999} = (1000, 1100, 1200, …, 90800, 90900), and, mimicking a previous step, the next term h_{2}(1000) is the next value after h_{2}(999) = 90900 with at least 3 trailing zeroes, with h_{2}(1000) = 91000. We then continue in the suggested manner. The explicit formula in (7) formalizes the construction illustrated above.

2.2 Lemmata

Let y_{1}(n) := h_{2} (10^{n−1}) and y_{2}(n) := h_{2}(10^{n} − 1). Our construction requires the following properties concerning the functions h_{1}, y_{1}, and y_{2}.

Lemma 2. For positive integers ℓ, the relation

∑_{i=1}^{10^{ℓ+1}−1} h_{1}(i) = \underbrace{9090⋯90}_{ℓ} 0  (8)

holds. For integers ℓ ≥ 2, the relations

\frac{y_{2}(ℓ) − y_{1}(ℓ)}{9 · 10^{ℓ−1} − 1} = 10^{ℓ−1} and  (9)
y_{1}(ℓ + 1) − y_{2}(ℓ) = 10^{2⌊\frac{ℓ}{2}⌋}  (10)

hold.

Proof. For positive integers ℓ, for j ∈ {1, 2, …, ℓ}, and for i ∈ {10^{j}, 10^{j+1} − 1}, since h_{1}(i) = 10^{j} is constant, we rewrite the left-hand side of (8) as

∑_{i=10^{1}}^{10^{2}−1} h_{1}(i) + ∑_{i=10^{2}}^{10^{3}−1} h_{1}(i) + ⋯ + ∑_{i=10^{ℓ}}^{10^{ℓ+1}−1} h_{1}(i) = ∑_{i=10^{1}}^{10^{2}−1} 10^{1} + ∑_{i=10^{2}}^{10^{3}−1} 10^{2} + ⋯ + ∑_{i=10^{ℓ}}^{10^{ℓ+1}−1} 10^{ℓ}

to obtain the decimal expansion on the right of (8).

We find that the relation

h_{2} (10^{ℓ}) = 10^{ℓ} ⌊\frac{∑_{i=1}^{10^{ℓ}} h_{1}(i)}{10^{ℓ}}⌋  (11)

follows in a direct way by setting n = 10^{ℓ} in (7). Setting n = 10^{ℓ} − 1 in (7), we find that

h_{2}(10^{ℓ} − 1) = 10^{ℓ−1} ⌊\frac{∑_{i=1}^{10^{ℓ−1}} h_{1}(i)}{10^{ℓ−1}}⌋ + ∑_{i=10^{ℓ−1}+1}^{10^{ℓ}−1} h_{1}(i),

and, since h_{1}(i) = 10^{ℓ−1} is constant for i ∈ {10^{ℓ−1} + 1, 10^{ℓ−1} + 2, …, 10^{ℓ} − 1}, we obtain the right-hand expression in the relation

h_{2} (10^{ℓ} − 1) = 10^{ℓ−1} ⌊\frac{∑_{i=1}^{10^{ℓ−1}} h_{1}(i)}{10^{ℓ−1}}⌋ + 10^{ℓ−1} (10^{ℓ} − 10^{ℓ−1} − 1)  (12)

for ℓ ≥ 2. Since

\frac{y_{2}(ℓ) − y_{1}(ℓ)}{9 · 10^{ℓ−1} − 1} = \frac{h_{2}(10^{ℓ} − 1) − h_{2}(10^{ℓ−1})}{9 · 10^{ℓ−1} − 1},  (13)

we may evaluate the right-hand side of (13) in closed form using (11) and (12) together for ℓ ≥ 2.

The evaluation in (8) gives us that

⌊\frac{∑_{i=1}^{10^{ℓ}} h_{1}(i)}{10^{ℓ}}⌋ = \begin{cases}10^{ℓ−1}, & if ℓ ≤ 2; \\ \underbrace{9090 ⋯ 90}_{⌊\frac{ℓ−1}{2}⌋−1} 91, & if ℓ ≠ 1 is odd; \\ \underbrace{9090 ⋯ 90}_{⌊\frac{ℓ−1}{2}⌋−1} 910, & if ℓ ≠ 2 is even. \end{cases}  (14)

holds. Consequently, from (11), we find that

y_{1}(ℓ) = \underbrace{9090 ⋯ 90}_{⌊\frac{ℓ}{2}⌋−2} 91 \underbrace{00 ⋯ 0}_{2⌊\frac{ℓ−1}{2}⌋+1}  (15)

holds for ℓ ≥ 4. Similarly, from (12) and (14) together, we find that

y_{2}(ℓ) = \underbrace{9090 ⋯ 90}_{⌊\frac{ℓ+1}{2}⌋} \underbrace{00 ⋯ 0}_{2⌊\frac{ℓ}{2}⌋−1}  (16)

holds for ℓ ≥ 2. From (15) and (16) together with the base cases for ℓ ≤ 3, we obtain that (10) holds for ℓ ≥ 2.

For 0 ≤ x ≤ \frac{100}{11}, we set

f(x) = \frac{89}{110}e^{\frac{100}{89} (11x−100)} + \frac{1}{10}.  (17)

For \frac{100}{11} ≤ x ≤ 10, we set f(x) = 10x − 90. Now, let m ≥ 2 be an integer, and set

H(m) := y_{1}(m + 1) − y_{2}(m),  (18)
a(m) := \frac{20H(m) − 11 · 10^{m}}{9} + 9 · 10^{m−1},  (19)
b(m) := \frac{(10^{m} − 1) (−20 H(m) + 11 · 10^{m})}{9} + 10^{m} − 9 · 10^{2m−1},  (20)
c(m) := −20H(m) + 9 · 10^{m−1} + 11 · 10^{m},  (21)
d(m) := 20 · 10^{m} H(m) − 11 · 10^{2m} − 9 · 10^{2m−1} + 10^{m}.  (22)

Again for m ∈ ℕ_{≥2}, we define the function h_{3}^{(m)} on the interval [10^{m} − 1, 10^{m}] so that

h_{3}^{(m)}(x) := \begin{cases}a(m)x + b(m), & if 10^{m} − 1 ≤ x ≤ 10^{m} − \frac{1}{10}; \\ c(m)x + d(m), & if 10^{m} − \frac{1}{10} ≤ x ≤ 10^{m}. \end{cases}

We then set

f(x) := \frac{x(y_{2}(m) − y_{1}(m))}{9 · 10^{m−1} − 1} − \frac{y_{1}(m) + 10^{m−1}y_{2}(m) − 10^{m}y_{1}(m)}{9 · 10^{m−1} − 1}  (23)

if 10^{m−1} ≤ x ≤ 10^{m} − 1 and we set

f(x) := h_{2}(10^{m} − 1) + ∫_{10^{m}−1}^{x} h_{3}^{(m)}(t) dt

if 10^{m} − 1 ≤ x ≤ 10^{m}, again with the understanding that m ≥ 2.

Lemma 3. The equality ⌊f(n)⌋ = h_{2}(n) holds for positive integers n. Moreover, the equality f(n) = h_{2}(n) holds for integers n ≥ 10.

Proof. The sequence (h_{2}(n))_{n≥1} may be constructed in an equivalent way as follows. Let S(0) = 0. Let n be a positive integer. If n is not of the form 10^{j} for j > 1, we set

S(n) = S(n − 1) + h_{1}(n).  (24)

Otherwise, we set S(n) = n ⌊\frac{S(n−1)+h_{1}(n)}{n}⌋. We claim that S(n) = h_{2}(n) for all natural numbers n. This leads us to show that h_{2}(n) satisfies the same recurrence as S(n), and we find that the base case holds. If n is not of the form 10^{j} for j > 1, then the desired recurrence h_{2}(n) = h_{2}(n − 1) + h_{1}(n) follows in a direct way from the definition in (7) by using the property that ⌊log_{10}(n)⌋ = ⌊log_{10}(n − 1)⌋. Now, suppose that n is of the form 10^{j} for j > 1. Then n − 1 is not of this form, so, inductively, we have that

n ⌊\frac{S(n − 1) + h_{1}(n)}{n}⌋ = 10^{j} ⌊\frac{10^{j−1} ⌊^{\frac{∑_{i=1}^{10^{j−1}} h_{1}(i)}{10^{j−1}}}⌋ + ∑_{i=10^{j−1}+1}^{10^{j}} h_{1}(i)}{10^{j}} ⌋ ,

so it remans to prove that

⌊\frac{∑_{i=1}^{10^{j}} h_{1}(i)}{10^{j}}⌋ = ⌊\frac{10^{j−1} ⌊^{\frac{∑_{i=1}^{10^{j−1}} h_{1}(i)}{10^{j−1}}}⌋ + ∑_{i=10^{j−1}+1}^{10^{j}} h_{1}(i)}{10^{j}} ⌋ .  (25)

We then find that (25) is equivalent to

⌊\frac{∑_{i=1}^{10^{j}} h_{1}(i)}{10^{j}}⌋ = ⌊\frac{1}{10} ⌊\frac{∑_{i=1}^{10^{j−1}} h_{1}(i)}{10^{j−1}}⌋ + \frac{9}{10} (1 + 10^{j−1})⌋ .  (26)

For j ≥ 3, the desired recurrence in (26) then follows in a direct way from the evaluation in (14).

From the definition for f(x) for x ≤ 10, it is immediate that ⌊f(n)⌋ = 0 = h_{2}(n) for n ≤ 10, and we may similarly routinely verify that f(n) = h_{2}(n) holds for 10 ≤ n ≤ 100.

Suppose that n satisfies 10^{m−1} < n ≤ 10^{m} − 1, again for some m ∈ ℕ_{≥2}. From the first case of the definition for f, we find that

f(n) − f(n − 1) = \frac{y_{2}(m) − y_{1}(m)}{9 · 10^{m−1} − 1}.

The relation (9) in Lemma 2 then gives us that f(n) − f(n − 1) = 10^{m−1} = h_{1}(n), i.e., so that the same recurrence in (24) holds if n is not of the form 10^{j} for j > 1. Similarly, by setting x = 10^{m−1} in (23), we obtain that f(n) = h_{2}(n).

Lemma 4. The mapping f: ℝ_{≥0} → ℝ_{≥0} is differentiable on ℝ_{>0}.

Proof. The elementary functions used to define f for x ∈ [0, \frac{100}{11}] and for x ∈ [\frac{100}{11}, 10] are such that f is differentiable on (0, \frac{100}{11}) and on (\frac{100}{11}, 10), and, moreover, we have that

lim ; x→(\frac{100}{11})^{−} f(x) = lim ; x→(\frac{100}{11})^{+} f(x) = f (\frac{100}{11}) = \frac{10}{11}

and that

lim ; x→(\frac{100}{11})^{−} f′(x) = lim ; x→(\frac{100}{11})^{+} f′(x) = 10.

We also find that the evaluation f(x) = 10x − 90 holds for x ∈ [\frac{100}{11}, 99].

Let m ∈ ℕ_{≥2}. For x ∈ (10^{m−1}, 10^{m} − 1), we have that f(x) agrees with a polynomial of degree 1, with

f′(x) = \frac{y_{2}(m) − y_{1}(m)}{9 · 10^{m−1} − 1} = 10^{m−1}

as the leading coefficient, by the relation (9) in Lemma 2. Similarly, for x ∈ (10^{m} − 1, 10^{m}), we have that f′(x) = h_{3}^{(m)}(x), and the expansions among (18)–(22) give us that lim_{x→(10^{m}−1)^{+}} h_{3}^{(m)}(x) = 10^{m−1}.

Finally, the expansions among (18)–(22) give us that lim_{x→(10^{m})^{−}} h_{3}^{(m)}(x) = 10^{m}, and this value agrees with f′(x) for x ∈ (10^{m}, 10^{m+1} −1).

Lemma 5. The mapping f: ℝ_{≥0} → ℝ_{≥0} is monotonically increasing on its domain.

Proof. The elementary function in (17) gives us that f is monotonically increasing for x ≤ \frac{100}{11}, and, as above, we have that f(x) = 10x − 90 for x ∈ [\frac{100}{11}, 99]. For m ∈ ℕ_{≥2}, and for 10^{m−1} ≤ x ≤ 10^{m} − 1, we have, as above, that f(x) agrees with a degree-1 polynomial with 10^{m−1} as its leading coefficient. For 10^{m} −1 ≤ x ≤ 10^{m}, we consider two cases. First suppose that 10^{m} − 1 ≤ x ≤ 10^{m} − \frac{1}{10}. In this case, we have that h_{3}^{(m)}(x) agrees with a line segment, and its value at 10^{m} − 1 is 10^{m−1} and its value at 10^{m} − \frac{1}{10} is

2y_{1}(m + 1) − 2y_{2}(m) − 19 · 10^{m−2}.  (27)

From (27), together with (10) in Lemma 2 and the inequality \frac{9}{10} > \frac{1}{9}(10 − 2 · 10^{1−m+2⌊ \frac{m}{2} ⌋}), we find that the expression displayed in (27) is positive. For 10^{m} − \frac{1}{10} ≤ x ≤ 10^{m}, we have that h_{3}^{(m)}(x) agrees with a line segment, and its value at 10^{m} − \frac{1}{10} again equals (27), and its value at 10^{m} is 10^{m}.

Lemma 6. The limit associated with η(f) exists, with η(f) = 2.

Proof. The monotonicity of f gives that for every x ∈ [10^{m−1}, 10^{m}), it holds that y_{1}(m) ≤ f(x) ≤ y_{1}(m + 1), and thus

\frac{9}{100}10^{2⌊log_{10}(x)⌋} ≤ f(x) ≤ 10 · 10^{2⌊log_{10}(x)⌋}.  (28)

From (28), we write

\frac{log (\frac{9}{100}10^{2⌊log_{10}(x)⌋})}{log(x)} ≤ \frac{log(f(x))}{log(x)} ≤ \frac{log (10 · 10^{2⌊log_{10}(x)⌋})}{log(x)}.  (29)

By taking the limits as x → ∞ of the upper and lower bounds in (29), we find that η(f) exists, with η(f) = 2.

Lemma 7. The limit associated with η(f′) exists, with η(f′) = 1.

Proof. For x ∈ (10^{m−1}, 10^{m} − 1), we have that f′(x) = 10^{m−1}. For x ∈ (10^{m} − 1, 10^{m}), we have that f′(x) = h_{3}^{(m)}(x). As indicated above, on the interval [10^{m} − 1, 10^{m}], the function h_{3}^{(m)}(x) agrees with a line segment from 10^{m−1} to

2 · 10^{2⌊\frac{m}{2}⌋} − 19 · 10^{m−2}  (30)

and a line segment from (30) to 10^{m}. By considering the parity of m in (30), we have, regardless of the parity of m, that

10^{m−2} ≤ f′(x) ≤ 2 · 10^{m},  (31)

letting m ∈ ℕ_{≥2}. The upper and lower bounds in (31) hold for 10^{m−1} < x < 10^{m}, giving us that

10^{⌊log_{10}(x)⌋−1} ≤ f′(x) ≤ 2 · 10^{⌊log_{10}(x)⌋+1},  (32)

for x not equal to a power of 10, and a similar argument gives us that (32) also holds for x equal to a power of 10. From (32), we write

\frac{log (10^{⌊log_{10}(x)⌋−1})}{log x} ≤ \frac{log (f′(x))}{log x} ≤ \frac{log (2 · 10^{⌊log_{10}(x)⌋+1})}{log x},  (33)

Taking the limit as x → ∞, the limits of the upper and lower bounds in (33) both reduce to 1.

2.3 Non-normality

As below, for a block B of digits, we let ℓ^{0}(B) denote the total number of zeroes, counting repetitions, in B.

Theorem 8. The concatenation 0.h_{2}(1)h_{2}(2) … is not normal in base 10.

Proof. From the lower and upper bounds in (28), we obtain that

2 ⌊log_{10}(n)⌋ − 2 < log_{10}(f(n)) ≤ 2 ⌊log_{10}(n)⌋ + 1.  (34)

The inequalities in (34) then give us that

2 ⌊log_{10}(n)⌋ − 2 ≤ # of digits in h_{2}(n) ≤ 2 ⌊log_{10}(n)⌋ + 1  (35)

for all positive integers n. So, by taking the concatenation 0.h_{2}(1)h_{2}(2) …, and by then taking the sequence of digits to the right of the decimal point given by the concatenation of the digits in the first m ∈ ℕ entries of the integer sequence (h_{2}(n) : n ∈ ℕ), the total number of digits in this sequence is ∑_{n=1}^{m}(# of digits in h_{2}(n)), and this satisfies, from (35), the inequalities such that

∑_{n=1}^{m}(2 ⌊log_{10}(n)⌋ − 2) ≤ ∑_{n=1}^{m}(# of digits in h_{2}(n)) ≤ ∑_{n=1}^{m}(2 ⌊log_{10}(n)⌋ + 1).  (36)

For the sequence h_{2}(1)h_{2}(2) … (denoted as an infinite word), we take an initial and finite segment S of this sequence. Then there is a well-defined value m = m(S) such that

∑_{n=1}^{m}(# of digits in h_{2}(n)) ≤ ℓ(S) < ∑_{n=1}^{m+1}(# of digits in h_{2}(n)).  (37)

We then write

ℓ(S) = ∑_{n=1}^{m}(# of digits in h_{2}(n)) + ℓ (\overline{S})  (38)

for a possibly empty string \overline{S} given by taking the ℓ(S) − ∑_{n=1}^{m}(# of digits in h_{2}(n)) digits to the right of S. From the well-definedness of m(S) given by (37), we find that

0 ≤ ℓ (\overline{S}) < # of digits in h_{2}(m + 1).  (39)

From (36), we obtain that

∑_{n=1}^{m}(2 ⌊log_{10}(n)⌋ − 2) + ℓ (\overline{S}) ≤ ℓ(S) ≤ ∑_{n=1}^{m}(2 ⌊log_{10}(n)⌋ + 1) + ℓ (\overline{S}) .  (40)

From (39) and (40), we find that

∑_{n=1}^{m}(2 ⌊log_{10}(n)⌋ − 2) ≤ ℓ(S) < ∑_{n=1}^{m}(2 ⌊log_{10}(n)⌋ + 1) + # of digits in h_{2}(m + 1).

This and (35) give us that

∑_{n=1}^{m}(2 ⌊log_{10}(n)⌋ − 2) ≤ ℓ(S) < ∑_{n=1}^{m+1}(2 ⌊log_{10}(n)⌋ + 1).

Using the recursion for h_{2} given in the proof of Lemma 3, we have that the following holds. For an integer n such that 10^{k} ≤ n ≤ 10^{k+1} − 1, the base-10 digit expansion of h_{2}(n) is such that its last k digits are all equal to 0, letting k be a positive integer. So, the number of zeroes appearing in the digital expansion of h_{2}(n) is at least ⌊log_{10}(n)⌋, with

⌊log_{10}(n)⌋ ≤ # of zeroes in h_{2}(n) ≤ 2⌊log_{10}(n)⌋ + 1,

and this holds for positive integers n. By analogy with (38), and for the strings S and \overline{S} as before, we find that ℓ^{0}(S) = ∑_{n=1}^{m}(# of zeroes in h_{2}(n)) + ℓ^{0} (\overline{S}). Since

∑_{n=1}^{m} ⌊log_{10}(n)⌋ + ℓ^{0} (\overline{S}) ≤ ∑_{n=1}^{m}(# of zeroes in h_{2}(n)) + ℓ^{0} (\overline{S}) ≤

∑_{n=1}^{m}(2⌊log_{10}(n)⌋ + 1) + ℓ^{0} (\overline{S}) ,

we find that ∑_{n=1}^{m} ⌊log_{10}(n)⌋ ≤ ℓ^{0}(S) < ∑_{n=1}^{m+1}(2⌊log_{10}(n)⌋ + 1), so that

\begin{matrix}∑ & {m \atop n=1} & ⌊log_{10}(n)⌋ \\ ∑ & {m+1 \atop n=1} & (2 ⌊log_{10}(n)⌋ + 1) \end{matrix} ≤ \frac{ℓ^{0}(S)}{ℓ(S)} ≤ 1.  (41)

Using an Abel summation formula, we have that ∑_{n=1}^{m}⌊log_{10}(n)⌋ ∼ \frac{m log m}{log 10} and that the partial sum in the denominator of the lower bound in (41) is asymptotic to \frac{2m log(m)}{log(10)}. So, the limit as m → ∞ of the lower bound in (41) is equal to \frac{1}{2}. So, with respect to the notation in (6), by setting α as the real value 0.⌊h_{2}(1)⌋⌊h_{2}(2)⌋ …, and by setting E as the block of digits given by the single digit 0, the limit lim_{n→∞} \frac{A_{E}(α,n)}{n} would not be equal to the desired target value of \frac{1}{10}. So, the value α is not normal in base 10.

We have shown in Section 2.2 that f satisfies all of the required conditions given in Theorem 2 from the work of Szüsz and Volkmann [5]. However, the concatenation 0.⌊f(1)⌋⌊f(2)⌋ … we have constructed is not normal in base 10, according to Lemma 3 and Theorem 8, yielding the desired counterexample. The non-normality of our new constant is illustrated in Table 1.

p Ratio of zero digits among
the digits of (h_{2}(n))_{1≤n≤10^{p}}
1  0.909091
2  0.405109
3  0.446318
4  0.476029
5  0.492968
6  0.503644
7  0.510957

Table 1: An illustration of the non-normality of 0.⌊f(1)⌋⌊f(2)⌋ . . ..

3 Acknowledgments

The author was supported by a Killam Postdoctoral Fellowship from the Killam Trusts and thanks Karl Dilcher for many useful and engaging discussions related to this paper. The author is thankful for the exceptionally helpful reviewer feedback that has been provided and that has led to many substantial improvements to this research paper.

References

[1] A. S. Besicovitch, The asymptotic distribution of the numerals in the decimal representation of the squares of the natural numbers, Math. Z. 39 (1935), 146–156.

[2] D. G. Champernowne, The construction of decimals normal in the scale of ten, J. London Math. Soc. 8 (1933), 254–260.

[3] A. H. Copeland and P. Erdős, Note on normal numbers, Bull. Amer. Math. Soc. 52 (1946), 857–860.

[4] P. Pollack and J. Vandehey, Besicovitch, bisection, and the normality of 0.(1)(4)(9) (16)(25) ⋯, Amer. Math. Monthly 122 (2015), 757–765.

[5] P. Szüsz and B. Volkmann, A combinatorial method for constructing normal numbers, Forum Math. 6 (1994), 399–414.

[6] N. J. A. Sloane et al., The On-Line Encyclopedia of Integer Sequences, 2025. Available

   at https://oeis.org.

2020 Mathematics Subject Classification: Primary 11K16. Keywords: normal number, decimal expansion, integer sequence.

(Concerned with sequences \underline{A001191}, \underline{A033307}, \underline{A033308}, and \underline{A380904}.) Received January 29 2025; revised versions received February 25 2025; January 12 2026. Published in Journal of Integer Sequences, May 20 2026. Return to Journal of Integer Sequences home page.