Cantor-solus and Cantor-multus Distributions

1 Cantor Distribution

Let $0 < ϑ \leq 1 / 2$ ; for instance, we could take $ϑ = 1 / 3$ as in the classical case. Let $¯ ϑ = 1 - ϑ$ . Consider a mapping [5]

F (ω_{1} ω_{2} ω_{3} \dots ω_{m}) = \frac{¯ ϑ}{ϑ} m \sum i = 1 ω_{i} ϑ^{i}

from the set $Ω$ of finite bitstrings ( $m < \infty$ ) to the nonnegative reals. The $2^{m}$ bitstrings in $Ω$ of length $m$ are assumed to be equiprobable. Consider the generating function [1]

G_{n} (z) = \sum ω \in Ω F (ω)^{n} z^{| ω |}

where $| ω |$ denotes the length of the bitstring. Clearly

G_{0} (z) = \sum ω \in Ω z^{| ω |} = \infty \sum m = 0 2^{m} z^{m} = \frac{1}{1 - 2 z} .

The quantity

\frac{[z^{m}] G_{n} (z)}{[z^{m}] G_{0} (z)} = \frac{1}{2^{m}} [z^{m}] G_{n} (z)

is the $n^{th}$ Cantor moment for strings of length $m$ ; let $μ_{n}$ denote the limit of this as $m \to \infty$ . Denote the empty string by $ε$ . From values

\begin{matrix} F (ε) = 0, & F (0 ω) = ϑ F (ω), & F (1 ω) = ¯ ϑ + ϑ F (ω) \end{matrix}

and employing the recurrence [6]

Ω = ε + {0, 1} \times Ω,

we have

	$G_{n} (z)$	$= \sum ω \in Ω ϑ^{n} F (ω)^{n} z^{1 + \| ω \|} + \sum ω \in Ω {(¯ ϑ + ϑ F (ω))}^{n} z^{1 + \| ω \|}$
		$= ϑ^{n} z G_{n} (z) + z n \sum i = 0 (\frac{n}{i}) {¯ ϑ}^{n - i} ϑ^{i} G_{i} (z)$
		$= 2 ϑ^{n} z G_{n} (z) + z n - 1 \sum i = 0 (\frac{n}{i}) {¯ ϑ}^{n - i} ϑ^{i} G_{i} (z)$

for $n \geq 1$ ; thus

G_{n} (z) = \frac{z}{1 - 2 ϑ^{n} z} n - 1 \sum i = 0 (\frac{n}{i}) {¯ ϑ}^{n - i} ϑ^{i} G_{i} (z) .

Dividing both sides by $G_{0} (z)$ , we have [5, 7, 8, 9]

μ_{n} = \frac{1}{2 (1 - ϑ^{n})} n - 1 \sum i = 0 (\frac{n}{i}) {¯ ϑ}^{n - i} ϑ^{i} μ_{i}

because

lim z \to z_{0} \frac{z}{1 - 2 ϑ^{n} z} = \frac{1}{2 (1 - ϑ^{n})}

and the singularity $z_{0} = 1 / 2$ of $G_{n} (z)$ is a simple pole. In particular, when $ϑ = 1 / 3$ ,

and, up to small periodic fluctuations [9, 10, 11],

μ_{n} \sim C n^{- ln (2) / ln (3)},

C = \frac{1}{2 ln (3)} \infty \int 0 (\infty \prod k = 2 \frac{1 + e^{- 2 x / 3^{k}}}{2}) e^{- 2 x / 3} x^{ln (2) / ln (3) - 1} d x = 0.733874...

as $n \to \infty$ .

We merely mention a problem involving order statistics. Let $ξ_{n}$ denote the expected value of the minimum of $n$

independent Cantor-distributed random variables. It is known that

[12]

ξ_{n} = \frac{1}{2^{n} - 2 ϑ} [¯ ϑ + ϑ n - 1 \sum i = 1 (\frac{n}{i}) ξ_{i}]

in general. In the special case $ϑ = 1 / 3$ , it follows that

\begin{matrix} ξ_{1} = 1 / 2, & ξ_{2} = 3 / 10, & ξ_{3} = 1 / 5, & ξ_{4} = 33 / 230, & ξ_{5} = 5 / 46 \end{matrix}

and, up to small periodic fluctuations [13],

ξ_{n} \sim c n^{- ln (3) / ln (2)},

c = \frac{2}{3 ln (2)} Γ (\frac{ln (3)}{ln (2)}) ζ (\frac{ln (3)}{ln (2)}) = 1.9967049717...

as $n \to \infty$ . If $η_{n}$ denotes the expected value of the maximum of $n$ variables, then

1 - η_{n} \sim c n^{- ln (3) / ln (2)}

by symmetry.

A final problem concerns the sum of all moments of the classical Cantor distribution [14]:

	$\infty \sum n = 0 μ_{n}$	$= - \frac{1}{3} + \frac{2}{3} \infty \sum k = 1 {(\frac{2}{3})}^{k} 2^{k} \sum j = 1 \frac{1}{j}$
		$= 3.3646507281...$

answering a question asked in [15].

2 Cantor-solus Distribution

We examine here the set $Ω$ of finite solus bitstrings ( $m < \infty$ ). Let

\begin{matrix} f_{k} = f_{k - 1} + f_{k - 2}, & f_{0} = 0, & f_{1} = 1 \end{matrix}

denote the Fibonacci numbers. The $f_{m + 2}$ bitstrings in $Ω$ of length $m$ are assumed to be equiprobable. Clearly

G_{0} (z) = \sum ω \in Ω z^{| ω |} = \infty \sum m = 0 f_{m + 2} z^{m} = \frac{1 + z}{1 - z - z^{2}} .

From additional values

\begin{matrix} F (1) = ¯ ϑ, & F (10 ω) = ¯ ϑ + ϑ^{2} F (ω) \end{matrix}

and employing the recurrence [6]

Ω = ε + 1 + {0, 10} \times Ω,

we have

	$G_{n} (z)$	$= {¯ ϑ}^{n} z + \sum ω \in Ω ϑ^{n} F (ω)^{n} z^{1 + \| ω \|} + \sum ω \in Ω {(¯ ϑ + ϑ^{2} F (ω))}^{n} z^{2 + \| ω \|}$
		$= {¯ ϑ}^{n} z + ϑ^{n} z G_{n} (z) + z^{2} \sum i + j = n (\frac{n}{i, j}) {¯ ϑ}^{i} ϑ^{2 j} G_{j} (z)$
		$= {¯ ϑ}^{n} z + ϑ^{n} z G_{n} (z) + ϑ^{2 n} z^{2} G_{n} (z) + z^{2} \sum \begin{matrix} i + j = n, j < n \end{matrix} (\frac{n}{i, j}) {¯ ϑ}^{i} ϑ^{2 j} G_{j} (z)$

for $n \geq 1$ ; thus

G_{n} (z) = \frac{1}{1 - ϑ^{n} z - ϑ^{2 n} z^{2}} ⎡ ⎢ ⎢ ⎢ ⎣ {¯ ϑ}^{n} z + z^{2} \sum \begin{matrix} i + j = n, j < n \end{matrix} (\frac{n}{i, j}) {¯ ϑ}^{i} ϑ^{2 j} G_{j} (z) ⎤ ⎥ ⎥ ⎥ ⎦ .

The purpose of using multinomial coefficients here, rather than binomial coefficients as in Section 1, is simply to establish precedent for Section 3. Let $φ = (1 + \sqrt{5}) / 2 = 1.6180339887...$ be the Golden mean. Dividing both sides by $G_{0} (z)$ , we have [1]

	$μ_{n}$	$= \frac{1}{1 - ϑ^{n} / φ - ϑ^{2 n} / φ^{2}} ⎡ ⎢ ⎢ ⎢ ⎣ 0 + \frac{1}{φ^{2}} \sum \begin{matrix} i + j = n, j < n \end{matrix} (\frac{n}{i, j}) {¯ ϑ}^{i} ϑ^{2 j} μ_{j} ⎤ ⎥ ⎥ ⎥ ⎦$
		$= \frac{1}{φ^{2} - ϑ^{n} φ - ϑ^{2 n}} \sum \begin{matrix} i + j = n, j < n \end{matrix} (\frac{n}{i, j}) {¯ ϑ}^{i} ϑ^{2 j} μ_{j}$

because

lim z \to z_{0} \frac{{¯ ϑ}^{n} z}{G_{0} (z)} = lim z \to z_{0} \frac{1 - z - z^{2}}{1 + z} {¯ ϑ}^{n} z = 0

and the singularity $z_{0} = 1 / φ$ of $G_{n} (z)$ is a simple pole. In particular, when $ϑ = 1 / 3$ ,

\begin{matrix} μ_{1} = 0.338826..., & μ_{2} = 0.203899..., & μ_{2} - μ_{1}^{2} = 0.089096... \end{matrix}

and, up to small periodic fluctuations,

μ_{n} \sim (0.616005...) n^{- ln (φ) / ln (3)} (3 / 4)^{n},

as $n \to \infty$ . An integral formula in [1] for the preceding numerical coefficient involves a generating function of exponential type:

M (x) = e^{- x / 3} \infty \sum k = 0 \frac{μ_{k}}{k!} {(\frac{4 x}{9})}^{k},

namely

\frac{1}{2 φ ln (3)} \infty \int 0 M (x) e^{- 2 x / 3} x^{ln (φ) / ln (3) - 1} d x

(we believe that the fifth decimal given in [1] is incorrect, perhaps a typo). Unlike the formula for $C$ earlier, this expression depends on the sequence $μ_{1}$ , $μ_{2}$ , $μ_{3}$ , … explicitly.

With regard to order statistics, it is known that [16]

ξ_{n} = \frac{1}{1 - ϑ φ^{- n} - ϑ^{2} φ^{- 2 n}} [¯ ϑ φ^{- 2 n} + ϑ n - 1 \sum i = 1 (\frac{n}{i}) φ^{- i} φ^{- 2 (n - i)} ξ_{i}],

η_{n} = \frac{1}{1 - ϑ φ^{- n} - ϑ^{2} φ^{- 2 n}} [¯ ϑ (1 - φ^{- n}) + ϑ^{2} n - 1 \sum j = 1 (\frac{n}{j}) φ^{- 2 j} φ^{- (n - j)} η_{j}]

in general. In the special case $ϑ = 1 / 3$ , we have, up to small periodic fluctuations,

ξ_{n} \sim (3.31661...) n^{- ln (3) / ln (φ)},

3 / 4 - η_{n} \sim (5.35114...) n^{- ln (3) / ln (φ)}

as $n \to \infty$ .

3 Cantor-multus Distribution

We examine here the set $Ω$ of finite multus bitstrings ( $m < \infty$ ). Let

\begin{matrix} f_{k} = 2 f_{k - 1} - f_{k - 2} + f_{k - 3}, & f_{0} = 0, & f_{1} = f_{2} = 1 \end{matrix}

denote the second upper Fibonacci numbers [17]. The $f_{m + 2}$ bitstrings in $Ω$ of length $m$ are assumed to be equiprobable. Clearly

G_{0} (z) = \sum ω \in Ω z^{| ω |} = \infty \sum m = 0 f_{m + 2} z^{m} = \frac{1 - z + z^{2}}{1 - 2 z + z^{2} - z^{3}} .

From additional values

F (11 ω) = ¯ ϑ + ¯ ϑ ϑ + ϑ^{2} F (ω),

F (1110 ω) = ¯ ϑ + ¯ ϑ ϑ + ¯ ϑ ϑ^{2} + ϑ^{4} F (ω)

and employing the recurrence

Ω = ε + 1 + {0, 11, 1110} \times Ω,

we have

	$G_{n} (z)$	$= {¯ ϑ}^{n} z + \sum ω \in Ω ϑ^{n} F (ω)^{n} z^{1 + \| ω \|} + \sum ω \in Ω {(¯ ϑ + ¯ ϑ ϑ + ϑ^{2} F (ω))}^{n} z^{2 + \| ω \|}$
		$+ \sum ω \in Ω {(¯ ϑ + ¯ ϑ ϑ + ¯ ϑ ϑ^{2} + ϑ^{4} F (ω))}^{n} z^{4 + \| ω \|}$
		$= {¯ ϑ}^{n} z + ϑ^{n} z G_{n} (z) + z^{2} \sum i + j + k = n (\frac{n}{i, j, k}) {¯ ϑ}^{i} {(¯ ϑ ϑ)}^{j} {(ϑ^{2})}^{k} G_{k} (z)$
		$+ z^{4} \sum i + j + k + ℓ = n (\frac{n}{i, j, k, ℓ}) {¯ ϑ}^{i} {(¯ ϑ ϑ)}^{j} {(¯ ϑ ϑ^{2})}^{k} {(ϑ^{4})}^{ℓ} G_{ℓ} (z)$
		$= {¯ ϑ}^{n} z + ϑ^{n} z G_{n} (z) + ϑ^{2 n} z^{2} G_{n} (z) + z^{2} \sum \begin{matrix} i + j + k = n, k < n \end{matrix} (\frac{n}{i, j, k}) {¯ ϑ}^{i + j} ϑ^{j + 2 k} G_{k} (z)$
		$+ ϑ^{4 n} z^{4} G_{n} (z) + z^{4} \sum \begin{matrix} i + j + k + ℓ = n, ℓ < n \end{matrix} (\frac{n}{i, j, k, ℓ}) {¯ ϑ}^{i + j + k} ϑ^{j + 2 k + 4 ℓ} G_{ℓ} (z)$

for $n \geq 1$ ; thus

	$G_{n} (z)$	$= \frac{1}{1 - ϑ^{n} z - ϑ^{2 n} z^{2} - ϑ^{4 n} z^{4}} ⎡ ⎢ ⎢ ⎣ {¯ ϑ}^{n} z + z^{2} \sum \begin{matrix} i + j + k = n, k < n \end{matrix} (\frac{n}{i, j, k}) {¯ ϑ}^{i + j} ϑ^{j + 2 k} G_{k} (z)$
		$+ z^{4} \sum \begin{matrix} i + j + k + ℓ = n, ℓ < n \end{matrix} (\frac{n}{i, j, k, ℓ}) {¯ ϑ}^{i + j + k} ϑ^{j + 2 k + 4 ℓ} G_{ℓ} (z) ⎤ ⎥ ⎥ ⎦ .$

Let

ψ = \frac{1}{3} ⎡ ⎢ ⎣ 2 + {(\frac{25 + 3 \sqrt{69}}{2})}^{1 / 3} + {(\frac{}{25 - 3 \sqrt{69}} 2)}^{1 / 3} ⎤ ⎥ ⎦ = 1.7548776662...

be the second upper Golden mean [17, 18]. Dividing both sides by $G_{0} (z)$ , we have

	$μ_{n}$	$= \frac{1}{1 - ϑ^{n} / ψ - ϑ^{2 n} / ψ^{2} - ϑ^{4 n} / ψ^{4}} ⎡ ⎢ ⎢ ⎣ 0 + \frac{1}{ψ^{2}} \sum \begin{matrix} i + j + k = n, k < n \end{matrix} (\frac{n}{i, j, k}) {¯ ϑ}^{i + j} ϑ^{j + 2 k} μ_{k}$
		$+ \frac{1}{ψ^{4}} \sum \begin{matrix} i + j + k + ℓ = n, ℓ < n \end{matrix} (\frac{n}{i, j, k, ℓ}) {¯ ϑ}^{i + j + k} ϑ^{j + 2 k + 4 ℓ} μ_{ℓ} ⎤ ⎥ ⎥ ⎦$
		$= \frac{1}{ψ^{4} - ϑ^{n} ψ^{3} - ϑ^{2 n} ψ^{2} - ϑ^{4 n}} ⎡ ⎢ ⎢ ⎣ ψ^{2} \sum \begin{matrix} i + j + k = n, k < n \end{matrix} (\frac{n}{i, j, k}) {¯ ϑ}^{i + j} ϑ^{j + 2 k} μ_{k}$
		$+ \sum \begin{matrix} i + j + k + ℓ = n, ℓ < n \end{matrix} (\frac{n}{i, j, k, ℓ}) {¯ ϑ}^{i + j + k} ϑ^{j + 2 k + 4 ℓ} μ_{ℓ} ⎤ ⎥ ⎥ ⎦$

because

lim z \to z_{0} \frac{{¯ ϑ}^{n} z}{G_{0} (z)} = lim z \to z_{0} \frac{1 - 2 z + z^{2} - z^{3}}{1 - z + z^{2}} {¯ ϑ}^{n} z = 0

and the singularity $z_{0} = 1 / ψ$ of $G_{n} (z)$ is a simple pole. In particular, when $ϑ = 1 / 3$ ,

\begin{matrix} μ_{1} = 0.504968..., & μ_{2} = 0.416013..., & μ_{2} - μ_{1}^{2} = 0.161020... \end{matrix}

but no asymptotics for $μ_{n}$ are known. Order statistics likewise remain open.

4 Bitsums

We turn to a more fundamental topic: given a set $Ω$ of finite bitstrings, what can be said about the bitsum $S_{n}$ of a random $ω \in Ω$ of length $n$ ? If $Ω$ is unconstrained, i.e., if all $2^{n}$ strings are included in the sample, then

\begin{matrix} E (S_{n}) = n / 2, & V (S_{n}) = n / 4 \end{matrix}

because a sum of $n$ independent Bernoulli( $1 / 2$ ) variables is Binomial( $n$ , $1 / 2$ ). Expressed differently, the average density of $1$ s in a random unconstrained string is $1 / 2$

, with a corresponding variance

1 / 4

Let us impose constraints. If $Ω$ consists of solus bitstrings, then the total bitsum $a_{n}$ of all $ω \in Ω$ of length $n$ has generating function [19, 20]

\infty \sum n = 0 a_{n} z^{n} = \frac{z}{{(1 - z - z^{2})}^{2}} = z + 2 z^{2} + 5 z^{3} + 10 z^{4} + 20 z^{5} + \dots

and the total bitsum squared $b_{n}$ has generating function

\infty \sum n = 0 b_{n} z^{n} = \frac{z (1 - z + z^{2})}{{(1 - z - z^{2})}^{3}} = z + 2 z^{2} + 7 z^{3} + 16 z^{4} + 38 z^{5} + \dots;

hence $c_{n} = f_{n + 2} b_{n} - a_{n}^{2}$ has generating function

\infty \sum n = 0 c_{n} z^{n} = \frac{z (1 - z)}{(1 + z)^{3} {(1 - 3 z + z^{2})}^{2}} = z + 2 z^{2} + 10 z^{3} + 28 z^{4} + 94 z^{5} + \dots

where $f_{n}$ is as in Section 2. Standard techniques [6] give asymptotics

lim n \to \infty \frac{E (S_{n})}{n} = lim n \to \infty \frac{a_{n}}{n f_{n + 2}} = \frac{5 - \sqrt{5}}{10} = 0.2763932022...,

lim n \to \infty \frac{V (S_{n})}{n} = lim n \to \infty \frac{c_{n}}{n f_{n + 2}^{2}} = \frac{1}{5 \sqrt{5}} = 0.0894427190...

for the average density of $1$ s in a random solus string and corresponding variance.

If instead $Ω$ consists of multus bitstrings, then the total bitsum $a_{n}$ of all $ω \in Ω$ of length $n$ has generating function [21]

\infty \sum n = 0 a_{n} z^{n} = \frac{z^{2} (2 - z)}{{(1 - 2 z + z^{2} - z^{3})}^{2}} = 2 z^{2} + 7 z^{3} + 16 z^{4} + 34 z^{5} + \dots

and the total bitsum squared $b_{n}$ has generating function

\infty \sum n = 0 b_{n} z^{n} = \frac{z^{2} (4 - 7 z + 4 z^{2} + 3 z^{3} - z^{4})}{{(1 - 2 z + z^{2} - z^{3})}^{3}} = 4 z^{2} + 17 z^{3} + 46 z^{4} + 116 z^{5} + \dots;

hence $c_{n} = f_{n + 2} b_{n} - a_{n}^{2}$ has generating function

\infty \sum n = 0 c_{n} z^{n} = \frac{z^{2} (4 - 9 z + 9 z^{2} - 9 z^{3} - 6 z^{4} + z^{5} - 6 z^{6} + z^{8})}{{(1 - z + 2 z^{2} - z^{3})}^{3} {(1 - 2 z - 3 z^{2} - z^{3})}^{2}} = 4 z^{2} + 19 z^{3} + 66 z^{4} + 236 z^{5} + \dots

where $f_{n}$ is as in Section 3. We obtain asymptotics

	$lim n \to \infty \frac{E (S_{n})}{n}$	$= lim n \to \infty \frac{a_{n}}{n f_{n + 2}}$
		$= \frac{1}{3} ⎡ ⎢ ⎣ 2 - {(\frac{23 + 3 \sqrt{69}}{1058})}^{1 / 3} + {(\frac{- 23 + 3 \sqrt{69}}{1058})}^{1 / 3} ⎤ ⎥ ⎦$
		$= 0.5885044113...,$

	$lim n \to \infty \frac{V (S_{n})}{n}$	$= lim n \to \infty \frac{c_{n}}{n f_{n + 2}^{2}}$
		$= \frac{1}{1587} {(\frac{69}{2})}^{1 / 3} [{(404685 + 35053 \sqrt{69})}^{1 / 3} + {(404685 - 35053 \sqrt{69})}^{1 / 3}]$
		$= 0.2810976123...$

for the average density of $1$ s in a random multus string and corresponding variance. Unsurprisingly $0.588 > 1 / 2 > 0.276$ and $0.281 > 1 / 4 > 0.089$ ; a clumping of $1$ s forces a higher density than a separating of $1$ s.

A famous example of an infinite aperiodic solus bitstring is the Fibonacci word [2, 3], which is the limit obtained recursively starting with $0$ and satisfying substitution rules $0 \mapsto 01$ , $1 \mapsto 0$ . The density of $1$ s in this word is $1 - 1 / φ \approx 0.382$ [22], which exceeds the average $0.276$ but falls well within the one-sigma upper limit $0.276 + \sqrt{0.089} = 0.575$ . We wonder if an analogously simple construction might give an infinite aperiodic multus bitstring with known density.

5 Longest Bitruns

We turn to a different topic: given a set $Ω$ of finite bitstrings, what can be said about the duration $R_{n, 1}$ of the longest run of $1$ s in a random $ω \in Ω$ of length $n$ ? If $Ω$ is unconstrained, then [6]

E (R_{n, 1}) = \frac{1}{2^{n}} [z^{n}] \infty \sum k = 1 (\frac{1}{1 - 2 z} - \frac{1 - z^{k}}{1 - 2 z + z^{k + 1}}),

the Taylor expansion of the numerator series is [23]

z + 4 z^{2} + 11 z^{3} + 27 z^{4} + 62 z^{5} + 138 z^{6} + 300 z^{7} + 643 z^{8} + 1363 z^{9} + 2866 z^{10} + \dots

and, up to small periodic fluctuations [24, 25],

E (R_{n, 1}) \sim \frac{ln (n)}{ln (2)} - (\frac{3}{2} - \frac{γ}{ln (2)})

as $n \to \infty$ . Of course, identical results hold for $R_{n, 0}$ , the duration of the longest run of $0$ s in $ω$ .

If $Ω$ consists of solus bitstrings, then it makes little sense to talk about $1$ -runs. For $0$ -runs, over all $ω \in Ω$ , we have

E (R_{n, 0}) = \frac{1}{f_{n + 2}} [z^{n}] \infty \sum k = 1 (\frac{1 + z}{1 - z - z^{2}} - \frac{1 + z - z^{k} - z^{k + 1}}{1 - z - z^{2} + z^{k + 1}})

and the Taylor expansion of the numerator series is [23]

z + 4 z^{2} + 9 z^{3} + 18 z^{4} + 34 z^{5} + 62 z^{6} + 110 z^{7} + 192 z^{8} + 331 z^{9} + 565 z^{10} + \dots

where $f_{n}$ is as in Section 2.

If instead $Ω$ consists of multus bitstrings, then we can talk both about $1$ -runs [23]:

E (R_{n, 1}) = \frac{1}{f_{n + 2}} [z^{n}] {\frac{- z}{(1 - z) (1 - z + z^{2})} + \infty \sum k = 1 (\frac{1 + z^{2}}{1 - 2 z + z^{2} - z^{3}} - \frac{1 + z^{2} - z^{k - 1} - z^{k}}{1 - 2 z + z^{2} - z^{3} + z^{k + 1}}) z},

num = 2 z^{2} + 7 z^{3} + 16 z^{4} + 32 z^{5} + 62 z^{6} + 118 z^{7} + 221 z^{8} + 409 z^{9} + 751 z^{10} + \dots

and $0$ -runs:

E (R_{n, 0}) = \frac{1}{f_{n + 2}} [z^{n}] \infty \sum k = 1 (\frac{1 + z^{2}}{1 - 2 z + z^{2} - z^{3}} - \frac{1 + z^{2} - z^{k - 1} + z^{k} - 2 z^{k + 1}}{1 - 2 z + z^{2} - z^{3} + z^{k + 2}}) z,

num = z + 2 z^{2} + 5 z^{3} + 11 z^{4} + 23 z^{5} + 45 z^{6} + 87 z^{7} + 165 z^{8} + 309 z^{9} + 573 z^{10} + \dots

where $f_{n}$ is as in Section 3. Proof: the number of multus bitstrings with no runs of $k$ $1$ s has generating function [26]

\begin{matrix} \frac{1 + z^{2} - z^{k - 1} - z^{k}}{1 - 2 z + z^{2} - z^{3} + z^{k + 1}} z & if k > 1; & \frac{z}{1 - z} & if k = 1; \end{matrix}

we conclude by use of the summation identity

\infty \sum j = 0 j \cdot h_{j} (z) = \infty \sum k = 0 (\infty \sum i = 0 h_{i} (z) - k \sum i = 0 h_{i} (z)) .

Study of runs of $k$ $0$ s proceeds analogously [27]. The solus and multus results here are new, as far as is known. Asymptotics would be good to see someday.

6 Acknowledgements

I am thankful to Alois Heinz for helpful discussions and for providing the generating function associated with $4,$ $19,$ $66,$ $236,$ $\dots$ via the Maple gfun package; R and Mathematica have been useful throughout. I am also indebted to a friend, who wishes to remain anonymous, for giving encouragement and support (in these dark days of the novel coronavirus outbreak).

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Steven Finch

MIT Sloan School of Management

Cambridge, MA, USA

steven_finch@harvard.edu

Cantor-solus and Cantor-multus Distributions

On Finding a First-Order Sentence Consistent with a Sample of Strings

On String Contact Representations in 3D

Testing Distributions of Huge Objects

Divergence Scaling for Distribution Matching

On a recolouring version of Hadwiger's conjecture

Learnable Visual Markers

A Characterization of Guesswork on Swiftly Tilting Curves

1 Cantor Distribution

2 Cantor-solus Distribution

3 Cantor-multus Distribution

4 Bitsums

5 Longest Bitruns

6 Acknowledgements

References

Cantor-solus and Cantor-multus Distributions

Related Research

On Finding a First-Order Sentence Consistent with a Sample of Strings

On String Contact Representations in 3D

Testing Distributions of Huge Objects

Divergence Scaling for Distribution Matching

On a recolouring version of Hadwiger's conjecture

Learnable Visual Markers

A Characterization of Guesswork on Swiftly Tilting Curves

1 Cantor Distribution

2 Cantor-solus Distribution

3 Cantor-multus Distribution

4 Bitsums

5 Longest Bitruns

6 Acknowledgements

References