Normal Numbers and Pseudorandom Generators

Abstract

For an integer b ≥ 2 a real number α is b -normal if, for all m > 0, every m-long string of digits in the base-b expansion of α appears, in the limit, with frequency b ^{− m}. Although almost all reals in [0, 1] are b-normal for every b, it has been rather difficult to exhibit explicit examples. No results whatsoever are known, one way or the other, for the class of “natural” mathematical constants, such as $\pi,\,e,\,\sqrt{2}$ and log2. In this paper, we summarize some previous normality results for a certain class of explicit reals and then show that a specific member of this class, while provably 2-normal, is provably not 6-normal. We then show that a practical and reasonably effective pseudorandom number generator can be defined based on the binary digits of this constant and conclude by sketching out some directions for further research.

1.1 Appendix

Proof.

α _2, 3 is not 6-normal.

Let Q _m be the base-6 expansion of α _2, 3 immediately following position 3^m (i.e., after the “decimal” point has been shifted to the right 3^m digits). We can write

$$\displaystyle\begin{array}{rcl} Q_{m}& =& {6}^{{3}^{m} }\alpha _{2,3}\bmod 1 \\ & =& \left (\sum _{k=1}^{m}{3}^{{3}^{m}-k }{2}^{{3}^{m}-{3}^{k} }\right )\bmod 1 +\sum _{ k=m+1}^{\infty }{3}^{{3}^{m}-k }{2}^{{3}^{m}-{3}^{k} }.{}\end{array}$$

(1.22)

The first portion of this expression is zero, since all terms in the summation are integers. The small second portion is very accurately approximated by the first term of the series, namely ${(3/4)}^{{3}^{m} }/{3}^{m+1}$. In fact, for all m ≥ 1,

$$\displaystyle\begin{array}{rcl} \frac{{(3/4)}^{{3}^{m} }} {{3}^{m+1}} & <& Q_{m}\; <\; \frac{{(3/4)}^{{3}^{m} }} {{3}^{m+1}} (1 + 2 \cdot 1{0}^{-6}).{}\end{array}$$

(1.23)

Let $Z_{m} = \lfloor \log _{6}1/Q_{m}\rfloor$ be the number of zeroes in the base-6 expansion of α that immediately follow position 3^m. Then for all m ≥ 1, (1.23) can be rewritten

$$\displaystyle\begin{array}{rcl} & & {3}^{m}\log _{ 6}\left (\frac{4} {3}\right ) + (m + 1)\log _{6}3 - 2 \\ & & < Z_{m}\; <\; {3}^{m}\log _{ 6}\left (\frac{4} {3}\right ) + (m + 1)\log _{6}3.{}\end{array}$$

(1.24)

Now let F _m be the fraction of zeroes in the base-6 expansion of α up to position 3^m + Z _m (i.e., up to the end of the block of zeroes that immediately follows position 3^m). Clearly

$$\displaystyle\begin{array}{rcl} F_{m}& >& \frac{\sum _{k=1}^{m}Z_{k}} {{3}^{m} + Z_{m}},{}\end{array}$$

(1.25)

since the numerator only counts zeroes in the long stretches. The summation in the numerator satisfies, for all sufficiently large m,

$$\displaystyle\begin{array}{rcl} \sum _{k=1}^{m}Z_{ k}& >& \frac{3} {2}\left ({3}^{m} -\frac{1} {3}\right )\log _{6}\left (\frac{4} {3}\right ) + \frac{m(m + 3)} {2} \log _{6}3 - 2m \\ & >& \frac{3} {2} \cdot {3}^{m}\log _{ 6}\left (\frac{4} {3}\right ) -\frac{1} {2}\log _{6}\left (\frac{4} {3}\right ) - 2m. {}\end{array}$$

(1.26)

Now given any ε > 0, we can write, for all sufficiently large m,

$$\displaystyle\begin{array}{rcl} F_{m}& >& \frac{\frac{3} {2} \cdot {3}^{m}\log _{ 6}\left (\frac{4} {3}\right ) -\frac{1} {2}\log _{6}\left (\frac{4} {3}\right ) - 2m} {{3}^{m} + {3}^{m}\log _{6}\left (\frac{4} {3}\right ) + (m + 1)\log _{6}3} \\ & =& \frac{\frac{3} {2}\log _{6}\left (\frac{4} {3}\right ) - \frac{1} {{3}^{m}}\left (\frac{1} {2}\log _{6}\left (\frac{4} {3}\right ) + 2m\right )} {1 +\log _{6}\left (\frac{4} {3}\right ) + \frac{(m+1)\log _{6}3} {{3}^{m}} } \\ & \geq & \frac{\frac{3} {2}\log _{6}\left (\frac{4} {3}\right )-\epsilon } {1 +\log _{6}\left (\frac{4} {3}\right )+\epsilon }\; \geq \; \frac{1} {2}\log _{2}\left (\frac{4} {3}\right ) - 2\epsilon.{}\end{array}$$

(1.27)

But $\beta = \frac{1} {2}\log _{2}(4/3)$ (which has numerical value 0. 2075187496…) is clearly greater than 1 ∕ 6, since ${(4/3)}^{3} = 64/27 > 2$. This means that infinitely often (namely, whenever $n = {3}^{m} + Z_{m}$) the fraction of zeroes in the base-6 expansion of α up to position n exceeds $\frac{1} {2}(1/6+\beta ) > 1/6$. Thus α is not 6-normal. ■ 

Proof.

Given co-prime integers b ≥ 2 and c ≥ 2, the constant $\alpha _{b,c} =\sum _{k\geq 1}1/({c}^{k}{b}^{{c}^{k} })$ is not bc-normal.

Let Q _m (b, c) be the base-bc expansion of α _b, c immediately following position c ^m. Then

$$\displaystyle\begin{array}{rcl} Q_{m}(b,c)& =& {(bc)}^{{c}^{m} }\alpha _{b,c}\bmod 1 \\ & =& \left (\sum _{k=1}^{m}{c}^{{c}^{m}-k }{b}^{{c}^{m}-{c}^{k} }\right )\bmod 1 +\sum _{ k=m+1}^{\infty }{c}^{{c}^{m}-k }{b}^{{c}^{m}-{c}^{k} }.{}\end{array}$$

(1.28)

As above, the first portion of this expression is zero, since all terms in the summation are integers, and the second portion is very accurately approximated by the first term of the series, namely ${[ \frac{c} {b(c-1)}]}^{{c}^{m} }/{c}^{m+1}$. In fact, for any choice of b and c as above, and for all m ≥ 1,

$$\displaystyle\begin{array}{rcl} \frac{1} {{c}^{m+1}}{\left [ \frac{c} {b(c - 1)}\right ]}^{{c}^{m} }& <& Q_{m}(b,c)\; <\; \frac{1} {{c}^{m+1}}{\left [ \frac{c} {b(c - 1)}\right ]}^{{c}^{m} } \cdot (1 + 1/10).{}\end{array}$$

(1.29)

Let $Z_{m}(b,c) = \lfloor \log _{bc}1/Q_{m}(b,c)\rfloor$ be the number of zeroes that immediately follow position c ^m. Then for all m ≥ 1, (1.29) can be rewritten as

$$\displaystyle\begin{array}{rcl} & & {c}^{m}\log _{ bc}\left [\frac{b(c - 1)} {c} \right ] + (m + 1)\log _{bc}c - 2 \\ & & < Z_{m}(b,c)\; <\; {c}^{m}\log _{ bc}\left [\frac{b(c - 1)} {c} \right ] + (m + 1)\log _{bc}c.{}\end{array}$$

(1.30)

Now let F _m (b, c) be the fraction of zeroes up to position c ^m + Z _m (b, c). Clearly

$$\displaystyle\begin{array}{rcl} F_{m}(b,c)& >& \frac{\sum _{k=1}^{m}Z_{k}(b,c)} {{c}^{m} + Z_{m}(b,c)},{}\end{array}$$

(1.31)

since the numerator only counts zeroes in the long stretches. The summation in the numerator of F _m (b, c) satisfies

$$\displaystyle\begin{array}{rcl} \sum _{k=1}^{m}Z_{ k}(b,c)& >& \frac{c} {c - 1}\left ({c}^{m} -\frac{1} {c}\right )\log _{bc}\left [\frac{b(c - 1)} {c} \right ] + \frac{m(m + 3)} {2} \log _{bc}c - 2m \\ & >& \frac{{c}^{m+1}} {c - 1}\log _{bc}\left [\frac{b(c - 1)} {c} \right ] - \frac{1} {c - 1}\log _{bc}\left [\frac{b(c - 1)} {c} \right ] - 2m. {}\end{array}$$

(1.32)

Thus given any ε > 0, we can write, for all sufficiently large m,

$$\displaystyle\begin{array}{rcl} F_{m}(b,c)& >& \frac{\frac{{c}^{m+1}} {c-1} \log _{bc}\left [\frac{b(c-1)} {c} \right ] - \frac{1} {c-1}\log _{bc}\left [\frac{b(c-1)} {c} \right ] - 2m} {{c}^{m} + {c}^{m}\log _{bc}\left (\frac{b(c-1)} {c} \right ) + (m + 1)\log _{bc}c} \\ & =& \frac{ \frac{c} {c-1}\log _{bc}\left [\frac{b(c-1)} {c} \right ] - \frac{1} {{c}^{m}}\left ( \frac{1} {c-1}\log _{bc}\left [\frac{b(c-1)} {c} \right ] + 2m\right )} {1 +\log _{bc}\left [\frac{b(c-1)} {c} \right ] + \frac{(m+1)\log _{bc}c} {{c}^{m}} } \\ & \geq & \frac{ \frac{c} {c-1}\log _{bc}\left [\frac{b(c-1)} {c} \right ]-\epsilon } {1 +\log _{bc}\left [\frac{b(c-1)} {c} \right ]+\epsilon } \\ & \geq & \frac{c} {c - 1} \cdot \frac{\log _{bc}\left [\frac{b(c-1)} {c} \right ]} {1 +\log _{bc}\left [\frac{b(c-1)} {c} \right ]} - 2\epsilon {}\end{array}$$

(1.33)

$$\displaystyle\begin{array}{rcl} & =& T(b,c) - 2\epsilon,{}\end{array}$$

(1.34)

where

$$\displaystyle\begin{array}{rcl} T(b,c)& =& \frac{c} {c - 1} \cdot \frac{\log _{bc}\left [\frac{b(c-1)} {c} \right ]} {1 +\log _{bc}\left [\frac{b(c-1)} {c} \right ]}.{}\end{array}$$

(1.35)

To establish the desired result that T(b, c) > 1 ∕ (bc), first note that

$$\displaystyle\begin{array}{rcl} T(b,c)& >& \frac{1} {2}\log _{bc}\left [\frac{b(c - 1)} {c} \right ]\; \geq \; \frac{1} {2}\log _{bc}\left (\frac{b} {2}\right ).{}\end{array}$$

(1.36)

Raise bc to the power of the right-hand side and also to the power 1 ∕ (bc). Then it suffices to demonstrate that

$$\displaystyle\begin{array}{rcl} \frac{b} {2}& >&{ \left [{(bc)}^{1/(bc)}\right ]}^{2}.{}\end{array}$$

(1.37)

The right-hand side is bounded above by ${({e}^{1/e})}^{2} = 2.0870652286\ldots$. Thus this inequality is clearly satisfied whenever b ≥ 5.

If we also presume that c ≥ 5, then by examining the middle of (1.36), it suffices to demonstrate that

$$\displaystyle\begin{array}{rcl} \frac{1} {2}\log _{bc}\frac{4b} {5} & >& \frac{1} {bc}{}\end{array}$$

(1.38)

or

$$\displaystyle\begin{array}{rcl} \frac{4b} {5} & >&{ \left ({e}^{1/e}\right )}^{2}.{}\end{array}$$

(1.39)

But this is clearly satisfied whenever b ≥ 3. For the case b = 2 and c ≥ 5, we can write

$$\displaystyle\begin{array}{rcl} T(b,c)& =& \frac{c} {c - 1} \cdot \frac{\log _{2c}\left [\frac{2(c-1)} {c} \right ]} {1 +\log _{2c}\left [\frac{2(c-1)} {c} \right ]}\; \geq \; \frac{\log _{2c}\left [\frac{2(c-1)} {c} \right ]} {1 +\log _{10}2},{}\end{array}$$

(1.40)

so by similar reasoning it suffices to demonstrate that

$$\displaystyle\begin{array}{rcl} \frac{2(c - 1)} {c} & >&{ \left ({e}^{1/e}\right )}^{1+\log _{10}2}\; =\; 1.61384928833\ldots.{}\end{array}$$

(1.41)

But this is clearly satisfied whenever c ≥ 6.

The five remaining cases, namely (2, 3), (2, 5), (3, 2), (3, 4), (4, 3), are easily verified by explicitly computing numerical values of T(b, c) using (1.35). As it turns out, the simple case that we worked out in detail above, namely b = 2 and c = 3, is the worst case, in the sense that for all other (b, c), the fraction T(b, c) exceeds the natural frequency 1 ∕ (bc) by greater margins. ■