# Filtered rays over iterated absolute differences on layers of integers

Raghavendra N. Bhat  Department of Mathematics, University of Illinois, 1409
West Green Street, Urbana, IL 61801, USA<EMAIL_ADDRESS>,  Cristian
Cobeli  ”Simion Stoilow” Institute of Mathematics of the Romanian Academy, 21
Calea Grivitei Street, P. O. Box 1-764, Bucharest 014700, Romania
<EMAIL_ADDRESS>and  Alexandru Zaharescu  Department of Mathematics,
University of Illinois, 1409 West Green Street, Urbana, IL 61801, USA, and
”Simion Stoilow” Institute of Mathematics of the Romanian Academy, 21 Calea
Grivitei Street, P. O. Box 1-764, Bucharest 014700, Romania
<EMAIL_ADDRESS>
###### Abstract.

The dynamical system generated by the iterated calculation of the high order
gaps between neighboring terms of a sequence of natural numbers is remarkable
and only incidentally characterized at the boundary by the notable Proth-
Glibreath Conjecture for prime numbers.

We introduce a natural extension of the original triangular arrangement,
obtaining a growing hexagonal covering of the plane. This is just the base
level of what further becomes an endless discrete helicoidal surface. Although
the repeated calculation of higher-order gaps causes the numbers that generate
the helicoidal surface to decrease, there is no guarantee, and most often it
does not even happen, that the levels of the helicoid have any regularity, at
least at the bottom levels.

However, we prove that there exists a large and nontrivial class of sequences
with the property that their helicoids have all levels coinciding with their
base levels. This class includes in particular many ultimately binary
sequences with a special header. For almost all of these sequences, we
additionally show that although the patterns generated by them seem to fall
somewhere between ordered and disordered, exhibiting fractal-like and random
qualities at the same time, the distribution of zero and non-zero numbers at
the base level has uniformity characteristics. Thus, we prove that a multitude
of straight lines that traverse the patterns encounter zero and non-zero
numbers in almost equal proportions.

###### 2020 Mathematics Subject Classification:

Primary 11B37; Secondary 11B39, 11B50.

Key words and phrases: Proth-Gilbreath Conjecture, formal power series, SP-
numbers, gaps between primes.

##  1\. Introduction

Let $\mathfrak{u}=\\{a_{k}\\}_{k\geq 0}$ be a sequence of non-negative
integers. We place the sequence $\mathfrak{u}$ on the top row of a triangle
whose subsequent rows are recursively obtained as sequences of numbers given
by the absolute values of the differences between neighboring terms on the
previous line. The infinite equilateral triangle obtained in this way is
defined by

$a_{0}$ |  | $a_{1}$ |  | $a_{2}$ |  | $a_{3}$ |  | $a_{4}$ |  | $a_{5}$ |  | $a_{6}$ |  | $\dots$ |
---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---
| $d_{0}^{(1)}$ |  | $d_{1}^{(1)}$ |  | $d_{2}^{(1)}$ |  | $d_{3}^{(1)}$ |  | $d_{4}^{(1)}$ |  | $d_{5}^{(1)}$ |  | $\dots$ |  |
|  | $d_{0}^{(2)}$ |  | $d_{1}^{(2)}$ |  | $d_{2}^{(2)}$ |  | $d_{3}^{(2)}$ |  | $d_{4}^{(2)}$ |  | $\dots$ |  |  |
|  |  | $d_{0}^{(3)}$ |  | $d_{1}^{(3)}$ |  | $d_{2}^{(3)}$ |  | $d_{3}^{(3)}$ |  | $\dots$ |  |  |  |
|  |  |  | $\dots$ |  | $\dots$ |  | $\dots$ |  | $\dots$ |  |  |  |  |
(P-G)

where

$d_{k}^{(j+1)}:=\big{|}d_{k+1}^{(j)}-d_{k}^{(j)}\big{|}\quad\text{ and }\quad
d_{k}^{(0)}:=a_{k}\quad\text{ for $j,k\geq 0$.}\\\ $ (1)

Let $\mathfrak{w}=\\{b_{j}\\}_{j\geq 0}$ be the sequence of numbers on the
left edge of this triangle, that is, $b_{0}=a_{0}$ and $b_{j}=d_{0}^{(j)}$ for
$j\geq 1$. We also denote by
$\mathfrak{w}_{k}=\big{\\{}d_{k}^{(j)}\big{\\}}_{j\geq 0}$, for $k\geq 0$, the
column or the ray that passes through the triangle parallel to the edge on the
left, with the first component being $a_{k}$, so that
$\mathfrak{w}=\mathfrak{w}_{0}$.

Taking successively the absolute values of higher-order differences, the
resulting numbers on the lower rows become smaller and smaller. It is then
natural to observe this comprehensive phenomenon on the left edge. However,
the components of $\mathfrak{w}$ may not necessarily become all equal to $0$,
but we should expect that from a certain point onwards they will take at most
two values, one being zero and the other an integer different from zero,
provided $\mathfrak{u}$ does not grow too fast. Considering, for example, the
simpler revealing case where the components of $\mathfrak{u}$ only take the
values $0$ or $a$, where $a$ is a positive integer, then the numbers on the
left edge are also only $0$’s or $a$’s.

The famous case of this type of construction, where the top row is the
sequence of prime numbers, is the subject of the Proth-Gilbreath conjecture
(see Proth [20], Gilbreath [13], Killgrove and Ralston [16], Odlyzko [18] and
the problem sets of Guy [14, Example 12], [15, Problem A10] and Montgomery
[17, Appendix Problem 68]). In accordance with common sense and extensive
numerical observations (see [16, 18]), the conjecture states that all the
entries of $\mathfrak{w}$ except $b_{0}$ are equal to $1$.

######  Conjecture 1 (Proth-Gilbreath).

All the differences on the western edge of the (P-G) triangle generated by the
sequence of all primes are equal to $1$.

However, to this day, in the sequence that is believed, as stated in
Conjecture 1, to be constantly equal to $1$, it is not even known if there are
an infinite number of $1$’s.

In fact, the reality generated by the primes is even more intriguing beyond
the left edge $\mathfrak{w}_{0}$. Inside the (P-G) triangle, the phenomenon is
exactly at the opposite end. First, let us observe that on $\mathfrak{w}_{1}$,
the immediately adjacent parallel line to the left edge, there are only $0$’s
or $2$’s, according to Conjecture 1. Furthermore, numerical evidence shows
that these values are present on this line in approximately equal proportions.
And likewise, when moving inside towards the right, on the following parallel
rays $\mathfrak{w}_{j}$, $j\geq 2$, it is also very likely that the same fact
holds true (see Table 1).

######  Conjecture 2.

Let $\mathfrak{w}_{j}$, $j\geq 1$, be any line parallel to the left edge of
the (P-G) triangle generated by the sequence of all primes. Denote by
$\nu_{d}(n)$ the number of $d$’s among the first $n$ elements of
$\mathfrak{w}_{j}$. Then, for $d\in\\{0,2\\}$, there exists a constant $c>0$
and an integer $n_{j}$, such that

$\Big{|}\nu_{d}(n)-\mbox{\small$\displaystyle\frac{n}{2}$}\Big{|}<c\sqrt{n},\quad\text{
for $n\geq n_{j}$.}$

We expect the same type of uniform distribution to occur on other rays that
cross the infinite (P-G), for example, those that pick the differences at
equally spaced intervals in different directions. Similarly, in convex
domains, that are sufficiently large and are situated far enough from the
starting row, the number of $0$’s and the number of $2$’s should be
approximately the same with probability one.

The structure of the discrete dynamical system generated by iteratively
calculating neighbor gaps recorded in the (P-G) triangle can be better
understood when viewed in a broader context (see [6]) that is related to
phenomena occurring in Pascal’s triangle (see [3, 9] and Prunescu et al. [8,
21]) or the patterns of numbers generated in Ducci type games (see Caragiu,
Zaki et al. [12, 11, 10, 5, 7]).

Figure 1. The triangle generated by $\mathfrak{u}$ that contains the first
$30$ square-prime numbers. (See Section 3 for the formal definition and some
appealing properties that the sequence of square-primes has.) Then, with
$\Upsilon^{(n)}(\mathfrak{u})$ as initial rows, five more triangles are
generated. The six figures represent the intermediate steps in forming the
first layer of the helicoid.

Let $\mathcal{L}$ denote the set of sequences of non-negative integers, and
let $\mathcal{L}_{0}\subset\mathcal{L}$ be the set of sequences with terms
equal to $0$ or $1$, only. Similarly, for any integer $N\geq 0$, let
$\mathcal{L}(N)$ and $\mathcal{L}_{0}(N)$ be the sets of finite sequences with
$N$ non-negative integer elements, and sequences consisting only of $0$’s or
$1$’s, respectively. If $\mathfrak{s}$ is a sequence and $N\geq 0$, we denote
by $\mathfrak{s}(N)$ the partial finite sequence formed by the first $N$
elements of $\mathfrak{s}$.

In this paper, we examine the overlying operator $\Upsilon$ that transforms
the top sequence $\mathfrak{u}$ into the one on the left edge $\mathfrak{w}$
in the (P-G) triangle. Then $\Upsilon$ is defined by

$\begin{split}\Upsilon:\mathcal{L}\to\mathcal{L}\text{ and
}\Upsilon(\mathfrak{u}):=\mathfrak{w}.\end{split}$

Let $\Psi:\mathcal{L}\to\mathcal{L}$ be the operator that transforms a
horizontal row of the triangle into the immediately following row. Note that
the entire triangle (P-G) is composed of the sequence of successive horizontal
rows $\Psi^{(j)}(\mathfrak{u})$, for $j\geq 0$, where
$\Psi^{(0)}(\mathfrak{u})=\mathfrak{u}$ is the top generating row. Also note
that the restrictions of $\Psi$ and $\Upsilon$ to $\mathcal{L}_{0}$ have in
their image only sequences in $\mathcal{L}_{0}$. The same type of property
occurs with the action of $\Psi$ and $\Upsilon$ on sequences $\mathfrak{s}$ of
$0$’s and $1$’s, except for a finite number of terms. For these sequences,
$\Psi(\mathfrak{s})$ is also ultimately composed only of $0$’s and $1$’s, but
this is not generally true for $\Upsilon(\mathfrak{s})$.

The geometric correspondent of each iteration of $\Upsilon$ applied on
$\mathfrak{u}$ results in the construction of a new equilateral triangle,
rotated clockwise around the first component $a_{0}$ of $\mathfrak{u}$ by an
angle of $60$ degrees. After six iterations of $\Upsilon$, the initial
sequence $\mathfrak{u}$ is geometrically reached again. This results in the
completion of the first layer or level of what can further be seen as a
helicoidal discrete surface, since in general
$\Upsilon^{(6)}(\mathfrak{u})\neq\mathfrak{u}$ (see Figure 1 for an example of
these iterations with a finite sequence of integers). Continuing the
iterations produces a discrete helicoid denoted
$\mathcal{H}=\mathcal{H}(\mathfrak{u})$, a “Riemann surface”-like structure of
non-negative integers. Let $\mathcal{H}_{n}=\mathcal{H}_{n}(\mathfrak{u})$,
$n\geq 1$, denote the $n$-th levels of the helicoid, so that

$\mathcal{H}=\bigcup\limits_{n\geq 1}\mathcal{H}_{n}.$ (2)

The first level $\mathcal{H}_{1}$, called also the base layer, is the union of
six equilateral triangles with a vertex in $a_{0}$ and edges
$\Upsilon^{(k)}(\mathfrak{u})$ and $\Upsilon^{(k+1)}(\mathfrak{u})$ for
$k=0,\dots,5$. The subsequent levels are each generated similarly by their
initial sequence $\Upsilon^{(6k)}(\mathfrak{u})$ for $k\geq 0$. Note that all
layers are congruent geometrically, each of them covering the entire plane
with numbers, if $\mathfrak{u}$ is infinite, or having the shape of a regular
hexagon with $N+1$ numbers on each side if $\mathfrak{u}=(a_{0},\dots,a_{N})$
(see Figure 2).

Figure 2.  The first seven layers of the helicoid generated by the sequence
$(100000,59049,32768,16807,7776,3125,1024,243,32,1,0,1,0,0,0,0,0,1,0,0)$,
where the first positive integers are the first ten $5$th powers in decreasing
order. The four captures are taken in order from bottom, sides, and top.
Distinct integers are shown in different colors. The helicoid has seven
distinct layers, and starting from the $8$th, all layers coincide with the
$7$th. The vertical strip indicates the places where the last outcome row on
one layer transcends, becoming the first generating row on the upper layer.

The result of this process leads to a simple pattern when $\mathfrak{u}$ is
the sequence of prime numbers, which, at least under the assumption of the
Proth-Gilbreath Conjecture, is well understood. In this case
$\Upsilon^{(n}(\mathfrak{u})=(2,1,1,1,\dots)$ for $n\geq 1$, so that the
interior of the corresponding generated equilateral triangles are then bounded
by sequences of $1$’s and have only $0$’s inside.

Figure 3.  Left: The base layer of the helicoid generated by the prime-number-
indicator function for integers in the interval $[0,50)$. Right: The base
layer of the helicoid generated by the sequence:
$3072,1536,768,384,192,96,48,24,12,6,3$ (the first $10$ integers that are $3$
times a power of $2$ in decreasing order), followed by a sequence or ‘random
bits’ defined by $\operatorname*{ind}_{[0,1/2)}\big{(}\\{k\sqrt{2}\\}\big{)}$
for $k=1,2,\dots,40$. (Note that by Theorem 4, in both cases, the helicoids
have identical layers at each level.)

However, there is a much more interesting perspective if we mark in the
initial sequence only the positions where prime numbers appear, their size
being given by the rank of the terms in the sequence. To be precise, let
$\mathfrak{i}=\\{\operatorname*{ind}{}_{\mathcal{P}}(j)\\}_{j\geq 0}$ be the
indicator sequence of prime numbers, where

$\operatorname*{ind}{}_{\mathcal{P}}(j)=\begin{cases}1&\text{ if $j$ is
prime,}\\\ 0&\text{ else.}\end{cases}$

Among our numerical experiments, we observed with surprise that the $6$th
iteration of $\Upsilon$ brings $\mathfrak{i}$ back to its starting point.
Indeed, the helicoid generated by $\mathfrak{i}$ with its leaves composed of
six equilateral triangles each, actually has only one leaf, all the subsequent
that follow being identical to the first modulo a rotation or a reflection.

We will show that this phenomenon is not unique and that there is a large
class of generating sequences $\mathfrak{u}$ that have the same property
$\Upsilon^{(6)}(\mathfrak{u})=\mathfrak{u}$. In Theorem 4 below we consider
these sequences and note, according to Theorem 5, that their class includes a
multitude of sequences that are expected to exhibit a random behavior, like
the indicator function of primes has.

##  2\. The main results

The problem of demonstrating the uniform distribution of zeros and ones on the
sides or on certain rays that cross the (P-G) triangle generated by some
generic initial sequence is not within our reach. However, beyond the specific
results that can be obtained in particular cases, we can prove the following
existence result with square-prime numbers.

Figure 4.  Two cut-offs of the (P-G) triangles with integers taken modulo $4$
(left) and modulo $2$ (right). In the image on the left, the triangle is
generated by the first $78$ primes less than $400$, and in the image on the
right, the top row contains the $75$ square-prime numbers less than $400$.

######  Theorem 1.

There exists an infinite subsequence of square-primes $A_{1}<A_{2}<\ldots$
such that the (P-G) triangle generated by $A_{1},A_{2},\ldots$ on the first
row has $1$ as every other element on the left edge.

Note that Theorem 1 implies that there are (P-G) triangles generated by
sequences of square-prime numbers for which at least half of the elements on
the left edge are $1$.

Table 1.  The frequencies of the absolute values of the differences on the rays $\mathfrak{w}_{0},\dots,\mathfrak{w}_{9}$ that cross a cut-off of the (P-G) triangle passing parallel to its left edge. The generating row contains the prime numbers less than one million. Counting is done based on the values of the higher-order differences taken modulo $4$. The notations are as follows: $r$ is the rank of the ray, starting with the left edge $\mathfrak{w}$ that has rank $r=0$; $N$ is the number of differences on the ray (note that there are no differences on the first row of (P-G)); $z$ is the number of $0$’s, $t$ is the number of $2$’s, and $h$ is the number of values that are not equal to $0$ or $2$.  $r$ | $N$ | $z$ | $t$ | $z-t$ | $h$ | $(z-t)/N$
---|---|---|---|---|---|---
0 | 78497 | 0 | 0 | 0 | 78497 | 0.00000
1 | 78496 | 39061 | 39435 | -374 | 0 | -0.00476
2 | 78495 | 39272 | 39223 | 49 | 0 | 0.00062
3 | 78494 | 39218 | 39275 | -57 | 1 | -0.00073
4 | 78493 | 39405 | 39088 | 317 | 0 | 0.00404
5 | 78492 | 39311 | 39180 | 131 | 1 | 0.00167
6 | 78491 | 39030 | 39461 | -431 | 0 | -0.00549
7 | 78490 | 39307 | 39182 | 125 | 1 | 0.00159
8 | 78489 | 39276 | 39211 | 65 | 2 | 0.00083
9 | 78488 | 39231 | 39256 | -25 | 1 | -0.00032

What is actually specific to the (P-G) constructions is not the singular fact
that the edge on the left only has a special form, as it happens when the
generator is the sequence of prime numbers and the constant shape of the edge
on the left is given by almost an arithmetic accident. An impression of what
happens most often can be seen in Figure 4, where the triangles generated by
prime and square-prime numbers are placed side by side for comparison.

Then what is truly remarkable is that the rays $\mathfrak{w}_{j}$ parallel to
the left edge tend to have a random sequence appearance. In particular, the
sequences $\mathfrak{w}_{j}$ seem to become binary, and the statistics that
count the number of the two types of elements indicate this. Thus, Table 1
shows the closeness between the number of $0$’s and the number of $2$’s on
rays $\mathfrak{w}_{1},\dots,\mathfrak{w}_{9}$ for the partial (P-G) triangle
generated by prime numbers less than one million and the high order
differences taken modulo $4$.

To compare, the same property is observed in Table 2 when the first row begins
with the square-prime numbers less than one million and differences taken
modulo $2$. In this case, on each of the columns
$\mathfrak{w}_{0},\dots,\mathfrak{w}_{9}$, the distribution of the number of
$0$’s and the number of $1$’s is also almost evenly split in half. And this is
not only happening on the western edge, the same phenomenon occurs inside the
(P-G) triangle as well. To quantify the distribution let us measure the rate
of change on rays.

Table 2.  The frequencies of the absolute values of the differences on the rays $\mathfrak{w}_{0},\dots,\mathfrak{w}_{9}$ that cross a cut-off of the (P-G) triangle passing parallel to its left edge. The generating row contains the $69179$ square-primes less than one million. Counting is done based on the values of the higher-order differences taken modulo $2$. The notations are as follows: $r$ is the rank of the ray, starting with the left edge $r_{0}=\mathfrak{w}$; $N$ is the number of differences on the ray (note that there are no differences on the first row of (P-G)); $z$ is the number of $0$’s, $o$ is the number of $1$’s, and $h$ is the number of values that are not equal to $0$ or $1$.  $r$ | $N$ | $z$ | $o$ | $z-o$ | $h$ | $(z-o)/N$
---|---|---|---|---|---|---
0 | 69178 | 34616 | 34559 | 57 | 3 | 0.00082
1 | 69177 | 34684 | 34485 | 199 | 8 | 0.00288
2 | 69176 | 34614 | 34556 | 58 | 6 | 0.00084
3 | 69175 | 34439 | 34727 | -288 | 9 | -0.00416
4 | 69174 | 34485 | 34681 | -196 | 8 | -0.00283
5 | 69173 | 34808 | 34357 | 451 | 8 | 0.00652
6 | 69172 | 34707 | 34458 | 249 | 7 | 0.00360
7 | 69171 | 34471 | 34694 | -223 | 6 | -0.00322
8 | 69170 | 34644 | 34522 | 122 | 4 | 0.00176
9 | 69169 | 34689 | 34472 | 217 | 8 | 0.00314

The next theorem demonstrates the existence of an underlying link between the
horizontal and the vertical/diagonal rows of the (P-G) triangle.

######  Theorem 2.

Let $\mathfrak{u}=(a_{0},a_{1},\dots)\in\mathcal{L}_{0}$ be the first row of
the (P-G) triangle and let $\mathfrak{w}=(b_{0},b_{1},\dots)$ be the sequence
on its left-edge. Let $f,g\in\mathbb{F}_{2}[[X]]$ be the formal power series
with coefficients in $\mathfrak{u}$ and $\mathfrak{w}$, respectively. Let
$T:\mathbb{F}_{2}[[X]]\to\mathbb{F}_{2}[[X]]$ be the operator defined by
$T(f)=g$. Then:

1\. The operator $T$ satisfies the following formula

$\begin{split}\big{(}T(f)\big{)}(X)=f\Big{(}\mbox{\small$\displaystyle\frac{X}{1+X}$}\Big{)}\cdot\mbox{\small$\displaystyle\frac{1}{1+X}$}\,.\end{split}$
(3)

2\. The operator $T$ has the property $T^{(2)}(f)=f$ for any
$f\in\mathbb{F}_{2}[[X]]$, so that $T$ is invertible and bijective.

The phenomenon of involution between $\mathfrak{u}$ and $\mathfrak{w}$ shown
in Theorem 2 occurs even more exquisitely for all (P-G) triangles of bounded
size. Consequently, an analogous version of Theorem 2 holds true for such
triangles of finite size. Here, polynomials play the role of the formal power
series because they can be thought of as formal power series with a finite
number of nonzero coefficients.

Let $R[X]^{[n]}$ denote the set of polynomials with coefficients in $R$ and
degree at most $n$.

######  Theorem 3.

Let $N\geq 1$ be an integer and let
$\mathfrak{u}=(a_{0},a_{1},\dots,a_{N-1})\in\mathcal{L}_{0}(N)$ be the top row
and let $\mathfrak{w}=(b_{0},b_{1},\dots,b_{N-1})$ be the sequence on the
left-edge of the (P-G) triangle of side $N$. Suppose $a_{0}=b_{0}$ and let
$f,g\in\mathbb{F}_{2}[X]^{[N-2]}$, be the polynomials whose coefficients are
the components of $\mathfrak{u}$ and $\mathfrak{w},$ respectively. Let
$T_{N}:\mathbb{F}_{2}[X]^{[N-2]}\to\mathbb{F}_{2}[X]^{[N-2]}$ defined by
$T_{N}(f)=g$. Then:

1\. The operator $T_{N}$ satisfies the following formula

$\begin{split}\big{(}T_{N}(f)\big{)}(X)\equiv
f\Big{(}\mbox{\small$\displaystyle\frac{X}{1+X}$}\Big{)}\cdot\mbox{\small$\displaystyle\frac{1}{1+X}$}\left(\mathrm{mod\;}X^{N}\right)\,.\end{split}$
(4)

2\. The operator $T_{N}$ has the property $T_{N}^{(2)}(f)=f$ for any
$f\in\mathbb{F}_{2}[X]^{[N-2]}$, so that $T_{N}$ is invertible and bijective.

As a follow-up of Theorems 2 and 3, we find that
$\Upsilon^{(6)}(\mathfrak{u})=\mathfrak{u}$ for all binary finite or infinite
sequences $\mathfrak{u}$, which in particular proves to be true the inceptive
observation discussed at the end of the Introduction for the indicator
function of primes. The above theorems also imply that the helicoids generated
by binary sequences have only one distinct layer, which is a three-petal
hexagon. The next results also shows the necessity of an additional condition
that must be fulfilled by the more general sequences in $\mathcal{L}$ so that
they also generate helicoids with just a single distinct layer. The general
statement includes both the case of infinite sequences and the case of finite
sequences, as we consider the ring of polynomials embedded in the ring of
formal power series, where polynomials have only a finite number of non-zero
coefficients.

First, we define the concept of a champion in a sequence. We say that the term
of rank $n\geq 0$ in the sequence $\mathfrak{s}=\\{a_{k}\\}_{k\geq 0}$ of non-
negative integers is a champion of $\mathfrak{s}$, or shortly a champion, if
$a_{n}>0$ and $a_{j}<a_{n}$ for $0\leq j<n$. Note that an unbounded sequence
has infinitely many champions, and in a strictly increasing sequence all its
terms are champions. However, our point is at the other end, at sequences with
at most one champion.

######  Theorem 4.

Let $\mathfrak{u}\in\mathcal{L}$ and let $\Upsilon:\mathcal{L}\to\mathcal{L}$
be the operator defined by $\Upsilon(\mathfrak{u})=\mathfrak{w}$, where
$\mathfrak{w}$ is the sequence of numbers obtained as the left edge of the
(P-G) triangle generated by $\mathfrak{u}$. Then:

  1. (1)

Any discrete helicoid, as defined by relation (2), which is generated by a
finite or infinite binary sequences $\mathfrak{u}\in\mathcal{L}_{0}(N)$ or
$\mathfrak{u}\in\mathcal{L}_{0}$ has all levels equal, that is
$\Upsilon^{(6)}(\mathfrak{u})=\mathfrak{u}$.

  2. (2)

The base level of the helicoids generated by binary finite or infinite
sequences are composed by three equal diamonds rotated around the origin, and
each of these diamonds is the union of two equilateral triangles that mirror
each other with respect to the diagonal that joins the corners with obtuse
vertices of the diamond.

  3. (3)

Let $\mathfrak{u}\in\mathcal{L}$ and suppose that
$\Upsilon^{(6)}(\mathfrak{u})=\mathfrak{u}$. Then the sequence $\mathfrak{u}$
has at most one champion.

If $\mathfrak{u}$ is a finite sequence, since the numbers in the helicoid
$\mathcal{H}(\mathfrak{u})$ have a general tendency to decrease, in any case,
all of them being at most equal to the largest component of $\mathfrak{u}$, it
follows that the sequence $\\{\mathcal{H}_{n}(\mathfrak{u})\\}_{n\geq 1}$ of
the hexagonal layers of numbers that compose $\mathcal{H}(\mathfrak{u})$ is
finally periodic. The sequence of these layers should then be composed of a
precycle followed by a cycle, both of finite length. Note that the length of a
precycle can be zero, but the length of a cycle must always be positive.
According to the first part of Theorem 4, the length of the precycle is zero
and the length of the cycle is one, for all binary generators
$\mathfrak{u}\in\mathcal{L}_{0}(N)$, where $N\geq 1$, as well as for infinite
sequences in $\mathcal{L}$. Further investigation is needed to classify the
generating sequences based on the length of the precycle or the cycle of the
sequences of layers that their helicoids have. The following is an example of
such a comprehensive problem.

######  Problem 1.

Let $P\geq 0$ and $C\geq 1$ be integers. Find $\mathfrak{u}\in\mathcal{L}(N)$,
for some $N\geq 1$, such that the sequence
$\\{\mathcal{H}_{n}(\mathfrak{u})\\}_{n\geq 1}$ of layers of the helicoid
$\mathcal{H}(\mathfrak{u})$ has exactly $P+C$ distinct layers, grouped in a
precycle of length $P$, and followed by the endless repetition of a cycle of
length $C$.

For finite sequences with a single champion, we numerically tested various
decreasing sequences of integers followed by a sequence of $0$’s or $1$’s. We
found that many such sequences produce helicoids with a sequence of layers
composed by a precycle (possibly empty) and a cycle of length one. Yet this is
not generally observed, as for instance, the helicoid generated by the first
positive $77$ seventh powers arranged in decreasing order and followed by the
sequence of ten bits: $0,1,0,0,0,0,0,1,0,0$ generates a helicoid with $17$
distinct layers, of which $9$ are in a precycle, and $8$ are in the subsequent
infinitely repeated cycle.

In Figure 3 there are shown the base layer of two helicoids whose upper layers
coincide with their initial ones. In both cases the generating sequences have
exactly one champion. In Section 5 we present several other examples of
sequences with just one champion, whose helicoids have only one distinct
level. Additionally, we show that the property of having a single champion is
not sufficient to characterize the one-distinct-level helicoids. Indeed, there
are sequences with just one champion whose helicoids have more distinct
levels, such as those in Figure 8 (left) and Figure 6 (right), which have four
and nine distinct levels, respectively.

For any ray $\mathfrak{w}_{k}$, for $k\geq 0$, that is parallel to the western
edge $\mathfrak{w}_{0}$, let $R_{\mathfrak{w}}(k)$ denote the proportion of
zeros among the first $N$ components of the ray, that is,

$R_{\mathfrak{w}}(k):=\mbox{\small$\displaystyle\frac{1}{N-k}$}\\#\Big{\\{}j:d_{k}^{(j)}=1,\
0\leq j<N\Big{\\}},\ \ \text{ for $0\leq k\leq N$}.$ (5)

Symmetrical with respect to vertical axes, let us consider the eastern edges
of the cut-off (P-G). For any fixed integer $N\geq 0$, denote by
$\mathfrak{e}_{0},\mathfrak{e}_{1},\dots$ these edges, seen this time
geometrically, in order from right to left. Precisely,
$\mathfrak{e}_{k}=\mathfrak{e}_{k}(N)$, $0\leq k\leq N$, is defined by

$\begin{split}\mathfrak{e}_{k}:=\big{\\{}d_{N-k}^{(j)}:0\leq j\leq
N,\,j+k=N\big{\\}}.\end{split}$

Then, just like for the western edges, let us denote by $R_{\mathfrak{e}}(k)$
the proportion of zeros on $\mathfrak{e}_{k}$:

$R_{\mathfrak{e}}(k):=\mbox{\small$\displaystyle\frac{1}{N-k}$}\\#\Big{\\{}j:d_{N-k}^{(j)}=1,\
0\leq j<N,\,j+k=N\Big{\\}},\ \ \text{ for $0\leq k\leq N$}.$ (6)

The following theorem shows that almost all (P-G) triangles generated by
sequences of $0$’s and $1$’s have nearly equal proportions of $0$’s and $1$’s
on the rays $\mathfrak{w}_{0},\mathfrak{w}_{1},\dots$ and
$\mathfrak{e}_{0},\mathfrak{e}_{1},\dots$ The more precise result is as
follows.

######  Theorem 5.

For any $\varepsilon\in(0,1/2)$, there exits $\delta=\delta_{\varepsilon}>0$
and an integer $N_{\varepsilon,\delta}$ such that, for any integer $N\geq
N_{\varepsilon,\delta}$, there exists an exceptional subset
$\mathcal{E}(N)\subset\mathcal{L}_{0}(N)$ (possibly empty) having at most
$\varepsilon\cdot\\#\mathcal{L}_{0}(N)$ elements, such that, for any sequence
$\mathfrak{u}(N)\in\mathcal{L}_{0}(N)\setminus\mathcal{E}(N)$, all ratios,
defined by (5) and (6), of the number of zeros on the rays
$\mathfrak{w}_{0},\mathfrak{w}_{1},\dots,\mathfrak{w}_{\lfloor\delta
N\rfloor}$ and
$\mathfrak{e}_{0},\mathfrak{e}_{1},\dots,\mathfrak{e}_{\lfloor\delta
N\rfloor}$ in the corresponding triangle (P-G) satisfy

$R_{\mathfrak{w}}(0),R_{\mathfrak{w}}(2),\dots,R_{\mathfrak{w}}\big{(}{\lfloor\delta
N\rfloor}\big{)};R_{\mathfrak{e}}(0),R_{\mathfrak{e}}(2),\dots,R_{\mathfrak{e}}\big{(}{\lfloor\delta
N\rfloor}\big{)}\in[1/2-\epsilon,1/2+\epsilon].$ (7)

##  3\. Differences with square-primes

###  3.1. Preliminary notes on SP-numbers

Merging together squares larger than one and primes into the recently
introduced sequence of square-primes [4, 1, 2, 3] proves to be a bright
combination. Formally, the sequence is defined as the ordered union:

$\begin{split}S\mathcal{P}:=&\bigsqcup_{k=2}^{\infty}\\{k^{2}p\mid p\text{
prime}\\}\\\
=&\big{\\{}8,12,18,20,27,28,32,44,45,48,50,52,63,68,72,75,76,80,92,98,99,\dots\big{\\}}.\end{split}$

Note that there are no primes nor squares in the set $S\mathcal{P}$. Due to
the uniform growth of the gaps between squares, this new sequence, also called
SP-numbers, has a type of distribution that reverberates from a distance that
of the prime numbers (for the higher order differences of primes and square-
primes, in Figure 4 there are side-by-side two triangles generated by them for
comparison). However, the change in the arithmetic nature from primes to
composite numbers, makes proving some remarkable properties that the prime
number sequence has transferred to SP-numbers not as difficult if we employ
what we already know about prime numbers.

Let $s_{n}$, $n\geq 1$, denote the $n$th square-prime. A few such numbers are
$s_{1}=8$, $s_{21}=99$, $s_{76}=404$ and $s_{1000}=7900$. An asymptotic
estimate (see [1, Theorem 4.1]) shows that $s_{n}$ and $p_{n}$ have a similar
order of magnitude:

$\begin{split}s_{n}=\big{(}\zeta(2)-1\big{)}\cdot\mbox{\small$\displaystyle\frac{n}{\log
n}$}+O\left(\mbox{\small$\displaystyle\frac{n}{\log
n^{2}}$}\right),\end{split}$

and $s_{n}<p_{n}$ for large $n$ because $\zeta(2)-1\approx 0.64493<1$. An
analogous result to Dirichlet’s Theorem [1, Theorem 6.1] on primes in
arithmetic progressions holds with a different constant depending on the
progression for square-primes, also.

In the increasing sequence of square-primes, twins are pairs of neighbor
numbers that are $1$ distance apart (such as $27$ and $28$ or $44$ and $45$),
and we know that there are infinitely many such pairs [1, Theorem 4.3].
Closely related to this is the following lemma that we need in the proof of
Theorem 1. The proof of the lemma appears in [4] and for the sake of
completeness, we include here, also.

######  Lemma 3.1 ([4, Theorem 2.1]).

For any positive integer $x$, there exist two square-prime numbers $a,b$ such
that $x=a-b$.

###### Proof.

We partition the set of all positive integers into the following five subsets:

  1. (i)

$S_{1}=\\{1\\}$.

  2. (ii)

$S_{2}=\mathcal{P}$, the set of all primes.

  3. (iii)

$S_{3}=\\{x:x\not\in\mathcal{P},\,2\nmid x\\}$, the set of all odd composite
numbers.

  4. (iv)

$S_{4}=\\{x:x=2p_{1}\cdots p_{k},\,k\geq 1,\text{ for some distinct odd
primes}\\}$, the set of all even composite square-free numbers.

  5. (v)

$S_{5}=\\{x:x=p^{2}d,p\in\mathcal{P},d\geq 1\\}$, the set of non-square-free
numbers.

Note that $\mathbb{N}\setminus\\{0\\}=S_{1}\cup S_{2}\cup S_{3}\cup S_{4}\cup
S_{5}$, hence it suffices to prove the existence of a pair of $SP$-numbers
with difference $x$, separately, for $x\in S_{j}$, $1\leq j\leq 5$.

(i) We have only one candidate, $x=1$, in this case. There exist infinitely
many neighbor SP-numbers [1, Theorem 4.3], but it is enough to consider the
first example $(a,b)=(28,27)$, for which $a-b=1$, which proves the case.

(ii) Let $x\in\mathcal{P}$ be fixed, and let also $p\in\mathcal{P}$ be a
different prime number. Consider the following Pell equation in variables
$m,n\in\mathbb{Z}$:

$\begin{split}m^{2}-pxn^{2}=1.\end{split}$ (8)

Since $px$ is not a square, we know after Pell that (8) has at least a
solution with $m,n>1$. Let $(M,N)$ be such a solution. Then,
$M^{2}-pxN^{2}=1$, and multiplying this equality by $x$, we find that

$xM^{2}-p(xN)^{2}=x.$ (9)

Observing that $a:=xM^{2}$ and $b:=p(xN)^{2}$ are both SP-numbers, the
equality (9), which becomes $a-b=x$, proves the lemma in this case.

(iii) Suppose now that $x$ is an odd composite square-free number. Then,
$x=py$, where $p\geq 3$ is prime and $y\geq 5$ is prime or a product of
distinct primes $\geq 5$. Thus $y$ is also necessarily odd. Let $y=2K+1$ for
some integer $K\geq 2$. It then follows that $y$ is a difference of two
squares: $y=(K+1)^{2}-K^{2}$, which implies

$x=p\big{(}(K+1)^{2}-K^{2}\big{)}=p(K+1)^{2}-pK^{2}.$ (10)

Let $a:=p(K+1)^{2}$ and $b:=pK^{2}$. Since $K>1$, both $a$ and $b$ are square-
prime numbers and (10) shows that $x=a-b$, which concludes the proof in this
case as well.

(iv) Now let us assume that $x$ is an even composite square-free number. Then
$x=2y$, where $y\geq 3$ is prime or a product of distinct primes $\geq 3$.
Reasoning as in the previous case, we find that $y=2K+1$, and the analogue of
(10) is

$x=2\big{(}(K+1)^{2}-K^{2}\big{)}=2(K+1)^{2}-2K^{2},$ (11)

where $K$ is a positive integer that can also be equal to $1$ this time.

Let $a:=2(K+1)^{2}$ and $b:=2K^{2}$. Note that if $K>1$, then both $a$ and $b$
are SP-numbers and (11) shows that $x=a-b$.

In the remaining possibility when $K=1$, we have $y=3$, so $x=6$, which can
also be written as a difference of square-primes: $6=2\cdot 3^{2}-3\cdot
2^{2}$, concluding the argument in case (iv).

(v) Let us assume now that $x>1$ is not square-free, that is, $x=c^{2}y$ for
some integers $c>1$, $y\geq 1$, and $y$ is square-free. Then, from the proved
cases (i)-(iv), we know that there exist two square-prime numbers $a^{\prime}$
and $b^{\prime}$ such that $y=a^{\prime}-b^{\prime}$. Let us say that
$a^{\prime}=p^{\prime}s^{2}$ and $b^{\prime}=p^{\prime\prime}t^{2}$, where
$p^{\prime},p^{\prime\prime}$ are prime numbers and $s,t>1$ are integers.
These yield:

$x=c^{2}y=c^{2}(a^{\prime}-b^{\prime})=p^{\prime}s^{2}c^{2}-p^{\prime\prime}t^{2}c^{2}.$

Let $a:=p^{\prime}(sc)^{2}$ and $b:=p^{\prime\prime}(tc)^{2}$. Since
$p^{\prime},p^{\prime\prime}$ are primes and $sc,tc>1$, both $a$ and $b$ are
SP-numbers, and the above shows that $x=a-b$. This concludes the proof for
case (v) and also the entire proof of the lemma. ∎

On combining Lemma 3.1 with the fact that any distance between two square-
primes is replicated infinitely often as the difference between other square-
primes [1, Theorem 4.3], we find that all positive integers appear infinitely
often as differences between square-primes.

######  Lemma 3.2.

Any positive integer appears infinitely often as a difference between square-
primes.

###  3.2. Proof of Theorem 1

We begin by proving a related result which shows that a (P-G) triangle with
‘controlled size’ elements on the eastern edge can be enlarged by padding it
in such a way that the new southern vertex has a preset number $Z$.

######  Proposition 3.1.

Consider a (P-G) triangle with integers $0\leq B_{1}\leq B_{2}\leq\cdots\leq
B_{m}$ on the eastern edge $\mathfrak{e}_{1}$. Then, there exists an integer
$C_{1}\geq B_{1}$, such that the triangle bordered by a new eastern edge
$\mathfrak{e}_{0}$ obtained by calculating the differences generated by the
addition of $C_{1}$ at the end of the generating row has components
$C_{1},C_{2},\dots,C_{m},C_{m+1}$ with $C_{j}\geq B_{j}$ for $1\leq j\leq m$.
Moreover, given an integer $Z\geq 0$, we can choose $C_{1}$ such that
$C_{m+1}=Z$.

###### Proof.

With some arbitrary integers $A_{1},\dots,A_{m-1}$, the triangle in the
statement of the proposition is as follows:

$A_{1}$ |  | $A_{2}$ |  | $\dots$ |  | $A_{m-1}$ |  | $B_{1}$ |  | $C_{1}$ |
---|---|---|---|---|---|---|---|---|---|---|---
| $\dots$ |  | $\dots$ |  | $\dots$ |  | $B_{2}$ |  | $C_{2}$ |  |
|  | $\dots$ |  | $\dots$ |  | $B_{3}$ |  | $C_{3}$ |  |  |
|  |  | $\dots$ |  | $\dots$ |  | $\dots$ |  |  |  |
|  |  |  | $B_{m}$ |  | $C_{m}$ |  |  |  |  |
|  |  |  |  | $Z$ |  |  |  |  |  |

We proceed backwards, from bottom to top. Let us assume that $Z\geq 0$ is
given and it takes the position of $C_{m+1}$. Then, according to the
definition, we may take $C_{m}$ such that $C_{m}-B_{m}=Z=C_{m+1}$, that is,
$C_{m}=C_{m+1}+B_{m}\geq B_{m}$.

Next, take $C_{m-1}$ such that $C_{m-1}-B_{m-1}=C_{m}$, so that
$C_{m-1}=C_{m}+B_{m-1}\geq B_{m-1}$.

Likewise, inductively, it follows that we may take $C_{1}$ such that
$C_{1}-B_{1}=C_{2}$, so that $C_{1}=C_{2}+B_{1}\geq B_{1}$.

In conclusion, we obtained $C_{1}$ and the sequence $C_{1},\dots,C_{m+1}$,
which satisfies the inequalities $C_{1}\geq B_{1},\dots,C_{m}\geq B_{m}$, and
additionally $C_{m+1}=Z$, thus proving the lemma. ∎

######  Remark 3.1.

Let us note that the proof of Proposition 3.1 also allows the assumption of a
different preset order between $C_{j}$ and $B_{j}$ for $1\leq j\leq m$.
Indeed, starting in the same way from $Z$ and recursively calculating in
reverse order the elements $C_{j}$ from the new equalities given by the preset
order, we obtain $C_{1}$, the new element of the first row, which ensures that
the southern vertex of the (P-G) triangle is $Z$.

Numerical experiments show that square-prime numbers are very handy for
generating (P-G) triangles that have various properties. For example, one that
has alternately $1$ on the western edge is:

27 |  | 28 |  | 44 |  | 76 |  | 98 |  | 112 |  | 153 |  | 171 |  | 180 |  | 188 |  | 292 |  | 316 |
---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---
| 1 |  | 16 |  | 32 |  | 22 |  | 14 |  | 41 |  | 18 |  | 9 |  | 8 |  | 104 |  | 24 |  |
|  | 15 |  | 16 |  | 10 |  | 8 |  | 27 |  | 23 |  | 9 |  | 1 |  | 96 |  | 80 |  |  |
|  |  | 1 |  | 6 |  | 2 |  | 19 |  | 4 |  | 14 |  | 8 |  | 95 |  | 16 |  |  |  |
|  |  |  | 5 |  | 4 |  | 17 |  | 15 |  | 10 |  | 6 |  | 87 |  | 79 |  |  |  |  |
|  |  |  |  | 1 |  | 13 |  | 2 |  | 5 |  | 4 |  | 81 |  | 8 |  |  |  |  |  |
|  |  |  |  |  | 12 |  | 11 |  | 3 |  | 1 |  | 77 |  | 73 |  |  |  |  |  |  |
|  |  |  |  |  |  | 1 |  | 8 |  | 2 |  | 76 |  | 4 |  |  |  |  |  |  |  |
|  |  |  |  |  |  |  | 7 |  | 6 |  | 74 |  | 72 |  |  |  |  |  |  |  |  |
|  |  |  |  |  |  |  |  | 1 |  | 68 |  | 2 |  |  |  |  |  |  |  |  |  |
|  |  |  |  |  |  |  |  |  | 67 |  | 66 |  |  |  |  |  |  |  |  |  |  |
|  |  |  |  |  |  |  |  |  |  | 1 |  |  |  |  |  |  |  |  |  |  |  |
(12)

Turning now to the proof of Theorem 1, let us suppose that $m\geq 2$ is even
and the (P-G) triangle generated by $\mathfrak{u}=(A_{1}<A_{2}<\cdots<A_{m})$
is given such that it satisfies the hypothesis of the theorem. For the case
where there are only square-prime numbers on the first row and the elements on
the western edge have $1$ as every other element, there are many small
triangles satisfying these requirements (see the numerical triangle (12) that
has inserted into it a few such examples). Our objective is to border the
triangle of size $m$ with two additional edges to the east in such a way that
the larger triangle, with a side length of $m+2$, also satisfies the
requirements of Theorem 1. Then, by induction, we will conclude that the
result holds for any triangle of even size $m\geq 2$.

Denote by $\mathfrak{e}^{\prime\prime}=(A_{m},D_{1},\dots,D_{m-1})$ the
eastern edge of the given triangle of size $m$. Let $X$ and $Y$, be the two
new numbers that will continue $\mathfrak{u}$, and let us denote the bordering
edges they generate by $\mathfrak{e}^{\prime}=(X,E_{1},\dots,E_{m}-1,E_{m})$
and $\mathfrak{e}=(Y,F_{1},\dots,F_{m-1},F_{m},F_{m+1})$, respectively. Let
$Z$ be the integer on the southern vertex, that is, in the previous notation,
$F_{m+1}=Z$, and in the hypothesis of the theorem $Z=1$. All these notations
can be seen at a glance in the following display:

$A_{1}$ |  | $A_{2}$ |  | $\dots$ |  | $A_{m}$ |  | $X$ |  | $Y$ |
---|---|---|---|---|---|---|---|---|---|---|---
| $A_{2}-A_{1}$ |  | $\dots$ |  | $D_{1}$ |  | $E_{1}$ |  | $F_{1}$ |  |
|  | $\dots$ |  | $D_{2}$ |  | $E_{2}$ |  | $F_{2}$ |  |  |
|  |  | $D_{m-1}$ |  | $E_{m-1}$ |  | $F_{m-1}$ |  |  |  |
|  |  |  | $E_{m}$ |  | $F_{m}$ |  |  |  |  |
|  |  |  |  | $F_{m+1}=Z$ |  |  |  |  |  |

Let $\Delta$ denote the difference $\Delta=Y-X$. Proceeding backwards as in
the proof of Proposition 3.1, we find out the conditions that $X$ and $\Delta$
must meet. This time, we start with the necessary conditions and verify that
they actually fulfill their role. The two conditions are:

$X\geq A_{m}+Z+D_{1}+D_{2}+\cdots+D_{m-1}\phantom{Y-Z}$ (13)

and

$\Delta=Y-X=Z+\big{\llbracket}D_{1}+D_{3}+\cdots+D_{m-1}\big{\rrbracket},$
(14)

where, the double square brackets indicate that only the $D_{j}$’s with odd
indices are added.

Let us first note that both conditions (13) and (14) depend only on the
eastern edge $\mathfrak{e}^{\prime\prime}$ of the initial triangle that we
border and the integer $Z$ that we want as the southern vertex of the bordered
triangle. Then we know from Lemma 3.2 that there are pairs of square-primes,
no matter how big and how many, that fulfill them.

Taking into account condition (13), the numbers on the first added border
layer $\mathfrak{e}^{\prime}$ are:

$\begin{split}E_{1}&=X-A_{m},\\\ E_{2}&=E_{1}-D_{1}=X-A_{m}-D_{1},\\\
E_{3}&=E_{2}-D_{2}=X-A_{m}-D_{1}-D_{2},\\\ &\cdots\\\
E_{m-1}&=E_{m-2}-D_{m-2}=X-A_{m}-D_{1}-D_{2}-\cdots-D_{m-2},\\\
E_{m}&=E_{m-1}-D_{m-1}=X-A_{m}-D_{1}-D_{2}-\cdots-D_{m-2}-D_{m-1}.\end{split}$

Further, employing condition (14) as well, these show that the differences on
$\mathfrak{e}$, the outer layer of the border, are

$\begin{split}F_{1}&=Y-X=Z+\big{\llbracket}D_{1}+D_{3}+\cdots+D_{m-1}\big{\rrbracket},\\\
F_{2}&=E_{1}-F_{1}=X-A_{m}-Z-\big{\llbracket}D_{1}+D_{3}+\cdots+D_{m-1}\big{\rrbracket},\\\
F_{3}&=E_{2}-F_{2}=Z+\big{\llbracket}D_{3}+\cdots+D_{m-1}\big{\rrbracket},\\\
F_{4}&=E_{3}-F_{3}=X-A_{m}-Z-D_{1}-D_{2}-\big{\llbracket}D_{3}+\cdots+D_{m-1}\big{\rrbracket},\\\
F_{5}&=E_{4}-F_{4}=Z+\big{\llbracket}D_{5}+\cdots+D_{m-1}\big{\rrbracket},\\\
F_{6}&=E_{5}-F_{5}=X-A_{m}-Z-D_{1}-D_{2}-D_{3}-D_{4}-\big{\llbracket}D_{5}+\cdots+D_{m-1}\big{\rrbracket},\\\
&\cdots\\\
F_{m-1}&=E_{m-2}-F_{m-2}=Z+\big{\llbracket}D_{m-1}\big{\rrbracket},\\\
F_{m}&=E_{m-1}-F_{m-1}=X-A_{m}-Z-D_{1}-D_{2}-\cdots-
D_{m-2}-\big{\llbracket}D_{m-1}\big{\rrbracket},\\\
F_{m+1}&=E_{m}-F_{m}=Z.\end{split}$

In conclusion, by extending the initial sequence $\mathfrak{u}$ with the two
square-prime numbers $X$ and $Y$, we obtained a triangle whose southern vertex
is $Z=1$, as desired, which concludes the proof of Theorem 1.

It is worth noting that in the above proof, the actual value of the number $Z$
did not play any special role, as $Z$ could have taken any value $Z\geq 0$.
Thus, the following more general result holds. Given a sequence of non-
negative integers $\\{w_{j}\\}_{j\geq 1}$, there exists a (P-G) triangle
generated by an increasing sequence of square-primes whose western edge is
$(*,w_{1},*,w_{2},*,w_{3},*,\dots)$, where the stars are some unspecified non-
negative integers.

######  Theorem 6.

Let $\bm{w}=\\{w_{j}\\}_{j\geq 1}$ be a sequence of non-negative integers.
Then, there exists an increasing sequence of square-primes such that the (P-G)
triangle they generate has on the western edge a sequence whose even-indexed
elements are the elements of $\bm{w}$.

##  4\. (P-G) triangles in the mirror-rays

Suppose all entries on the top row of the (P-G) are from $\mathcal{L}_{0}$.
Then the entire triangle has the same property, also.

Next we show that there is a close, analytically expressible link between the
top row $\mathfrak{u}$ and the left edge $\mathfrak{w}$ of the triangle. This
link actually extends across the entire triangle, because if one cuts a few
rows from the top, then the link shall be maintained between the remaining
new-first row and the remaining elements on the new-left edge. On the other
hand, cutting also a few vertical (geometrically rather oblique) columns on
the left side, all parallel to edge $\mathfrak{w}$, then the link becomes one
between the first remaining row and the first remaining leftmost edge.

Denote the components of the top sequence by
$\mathfrak{u}=(a_{0},a_{1},\dots)\in\mathcal{L}_{0}$, and the componets of the
left-edge sequence by $\mathfrak{w}=(b_{0},b_{1},\dots)\in\mathcal{L}_{0}$.
Their corresponding formal power series are $f=f(X)\in\mathbb{F}_{2}[[X]]$ and
$g=g(X)\in\mathbb{F}_{2}[[X]]$, respectively, where

$f(X)=\sum_{n\geq 0}a_{n}X^{n}\ \ \text{ and }\ \ g(X)=\sum_{n\geq
0}b_{n}X^{n}.$ (15)

Suppose $a_{0}=b_{0}$ is the element in the upper left corner of (P-G), and
that it is also the constant term of both series $f$ and $g$.

###  4.1. Lemmas

The first lemma shows that starting somewhere on the right side of the Pascal
triangle and adding the numbers placed on the following rows diagonally to the
left, the partial sums that we obtain are equal to the first number on the
right of the next row.

######  Lemma 4.1.

For any integers $K,n\geq 0$, we have

$\scalebox{0.8}{$\dbinom{K}{K}$}+\scalebox{0.8}{$\dbinom{K+1}{K}$}+\cdots+\scalebox{0.8}{$\dbinom{K+n}{K}$}=\scalebox{0.8}{$\dbinom{K+n+1}{K+1}$}.$
(16)

###### Proof.

The proof follows by induction over $n$ using the recursive formula that
generates Pascal’s triangle.

If $n=1$, we have
$\scalebox{0.7}{$\dbinom{K}{K}$}+\scalebox{0.7}{$\dbinom{K+1}{K}$}=1+(K+1)=K+2=\scalebox{0.7}{$\dbinom{K+2}{K+1}$}$.

Suppose (16) holds for some $n\geq 1$. Then, since

$\begin{split}\scalebox{0.8}{$\dbinom{K+n+1}{K+1}$}+\scalebox{0.8}{$\dbinom{K+n+1}{K}$}=\scalebox{0.8}{$\dbinom{K+n+2}{K+1}$}\end{split}$

it follows that (16) also holds for $n+1$, which concludes the proof of the
lemma. ∎

The next lemma gives the formal power series expression and the rational
representation of a power of the fundamental series $F(X)=1/(1+X)$, whose
coefficients are all equal to $1$ in $\mathbb{F}_{2}$.

######  Lemma 4.2.

Let $F\in\mathbb{F}_{2}[[X]]$, $F(X)=1+X+X^{2}+\cdots$. Then, for any integer
$N\geq 0$, we have

$\big{(}F(X)\big{)}^{N+1}=\sum_{n\geq
0}\scalebox{0.8}{$\dbinom{N+n}{N}$}X^{n}=\scalebox{1.3}{$\frac{1}{(1+X)^{N+1}}$}.$
(17)

###### Proof.

The proof is by induction. If $N=1$, relation (17) holds because $F(X)\cdot
F(X)=\sum_{n\geq 0}(n+1)X^{n}$, the coefficients being given by the equality

$\sum_{r\geq 0}\sum_{\begin{subarray}{c}s\geq 0\\\ r+s=n\end{subarray}}1\cdot
1=n+1.$

Let $K\geq 1$ and suppose that the coefficient of $X^{s}$ in the power series
of $\big{(}F(X)\big{)}^{K}$ is  $\dbinom{K-1+s}{K-1}$ for all $s\geq 0$. Then,
the coefficient of $X^{n}$ in the product $F(X)\cdot F^{K}(X)$ is

$\begin{split}\sum_{r\geq 0}\sum_{\begin{subarray}{c}s\geq 0\\\
r+s=n\end{subarray}}1\cdot\scalebox{0.8}{$\dbinom{K-1+s}{K-1}$}=\scalebox{0.8}{$\dbinom{K-1+0}{K-1}$}+\scalebox{0.8}{$\dbinom{K-1+1}{K-1}$}+\cdots+\scalebox{0.8}{$\dbinom{K-1+n}{K-1}$}=\scalebox{0.8}{$\dbinom{K+n}{K}$},\end{split}$

where the last equality follows from Lemma 4.1. Since these are exactly the
coefficients of $\big{(}F(X)\big{)}^{K+1}$, this concludes the proof of the
lemma. ∎

###  4.2. Proof of Theorem 2

1\. Let $\mathfrak{w}=(b_{0},b_{1},b_{2},\dots)$ be the left edge of (P-G)
generated by $\mathfrak{u}$. Then $b_{0}=a_{0}$, $b_{1}=a_{0}+a_{1}$,
$b_{2}=a_{0}+2a_{1}+a_{2}$ and so on. Then, by induction, one finds that the
general formula for $b_{n}$ is

$b_{n}=\binom{n}{0}a_{0}+\binom{n}{1}a_{1}+\cdots+\binom{n}{n}a_{n}\,.$ (18)

The formal power series of $\big{(}T(f)\big{)}(X)$ is
$b_{0}+b_{1}X+b_{2}X^{2}+\cdots$, and to deduce a functional expression for
it, we rearrange the terms using formula (18). Collecting together similar
terms with the same coefficient $a_{n}$, we see that

$\begin{split}\big{(}T(f)\big{)}(X)=&a_{0}\big{(}1+X+X^{2}+X^{3}+\cdots\big{)}\\\
&+a_{1}X\bigg{(}\mbox{\small$\displaystyle\binom{1+0}{1}$}+\mbox{\small$\displaystyle\binom{1+1}{1}$}X+\mbox{\small$\displaystyle\binom{1+2}{1}$}X^{2}+\mbox{\small$\displaystyle\binom{1+3}{1}$}X^{3}+\cdots\bigg{)}\\\
&+a_{2}X^{2}\bigg{(}\mbox{\small$\displaystyle\binom{2+0}{2}$}+\mbox{\small$\displaystyle\binom{2+1}{2}$}X+\mbox{\small$\displaystyle\binom{2+2}{2}$}X^{2}+\mbox{\small$\displaystyle\binom{2+3}{2}$}X^{3}+\cdots\bigg{)}\\\
&+\cdots\\\
&+a_{n}X^{n}\bigg{(}\mbox{\small$\displaystyle\binom{n+0}{n}$}+\mbox{\small$\displaystyle\binom{n+1}{n}$}X+\mbox{\small$\displaystyle\binom{n+2}{n}$}X^{2}+\mbox{\small$\displaystyle\binom{n+3}{n}$}X^{3}+\cdots\bigg{)}\\\
&+\cdots\end{split}$

Using Lemma 4.2 on each of the lines of the relation above we obtain

$\begin{split}\big{(}T(f)\big{)}(X)&=\mbox{\small$\displaystyle\frac{a_{0}}{1+X}$}+\mbox{\small$\displaystyle\frac{a_{1}X}{(1+X)^{2}}$}+\mbox{\small$\displaystyle\frac{a_{2}X^{2}}{(1+X)^{3}}$}+\cdots+\mbox{\small$\displaystyle\frac{a_{n}X^{n}}{(1+X)^{n+1}}$}+\cdots\\\
&=\mbox{\small$\displaystyle\frac{1}{1+X}$}\Big{(}a_{0}+a_{1}\mbox{\small$\displaystyle\frac{X}{1+X}$}+a_{2}\bigg{(}\mbox{\small$\displaystyle\frac{X}{1+X}$}\bigg{)}^{2}+\cdots+a_{n}\bigg{(}\mbox{\small$\displaystyle\frac{X}{1+X}$}\bigg{)}^{n}+\cdots\Big{)}\\\
&=\mbox{\small$\displaystyle\frac{1}{1+X}$}f\Big{(}\mbox{\small$\displaystyle\frac{X}{1+X}$}\Big{)}\,,\end{split}$

which proves the first point of the theorem.

2\. We apply formula (3) twice. First we obtain

$\begin{split}T^{(2)}\big{(}f(X)\big{)}=T\Big{(}T\big{(}f(X)\big{)}\Big{)}=\mbox{\small$\displaystyle\frac{1}{1+X}$}T(f)\Big{(}\mbox{\small$\displaystyle\frac{X}{1+X}$}\Big{)},\end{split}$

and then, continuing, on the second application, after reducing the terms in
the rational factions of $\mathbb{F}_{2}(X)$ and making the necessary
cancellations, we obtain:

$\begin{split}T^{(2)}\big{(}f(X)\big{)}=\mbox{\small$\displaystyle\frac{1}{1+X}$}\cdot\mbox{\small$\displaystyle\frac{1}{1+\frac{X}{1+X}}$}f\bigg{(}\mbox{\small$\displaystyle\frac{\frac{X}{1+X}}{1+\frac{X}{1+X}}$}\bigg{)}=f(X).\end{split}$

It then follows that $T$ is invertible, so that is bijective and $T^{-1}=T$.
This concludes the proof of the theorem.

###  4.3. Proof of Theorem 3

For any integer $N\geq 0$, the set of finite sequences $\mathcal{L}_{0}(N)$ is
in one-to-one correspondence with the set of polynomials $\mathbb{F}_{2}[X]$,
which in turn is embedded in $\mathbb{F}_{2}[[X]]$, viewing the polynomials as
formal power series with only finitely many non-zero coefficients. In
accordance with this, the restriction of the application $T$ to
$\mathcal{L}_{0}(N)[X]$, and also that of $\Upsilon$, on the sequences side,
to finite binary sequences $\mathcal{L}_{0}(N)$, are well-defined.
Furthermore, since as we have seen in the induction process during the proof
of Theorem 2, the transformations through $\Upsilon$ that occur between the
top $\mathfrak{u}$ and the western edge $\mathfrak{w}$ actually occur in an
ordered manner, the first $N$ components of one only affect the first $N$
components of the other, for any $N\geq 0$. Therefore, Theorem 3, the finite
analogue version of Theorem 2, also holds true for (P-G) triangles of bounded
size and polynomials instead of formal power series. That is, the restriction
of $T$ to $\mathcal{L}_{0}(N)[X]$ is still an involution, just like the
corresponding restriction of $\Upsilon$ to $\mathcal{L}_{0}(N)$ is.

##  5\. Binary generators and generators with only one champion

###  5.1. Proof of Theorem 4

A consequence of the fact that $T$ is an involution, as proved in Theorems 2
and 3, on the side of the coefficients of the formal power series or of the
polynomials, is the fact that the restriction of $\Upsilon$ to binary
sequences is also an involution. This means that, on one hand,
$\Upsilon^{(2)}(\mathfrak{u})=\mathfrak{u}$ for binary sequences
$\mathfrak{u}$ and for any finite initial fragments of these sequences, as
well. It follows then that $\Upsilon^{(6k)}(\mathfrak{u})=\mathfrak{u}$ for
$k\geq 1$, meaning that the helicoid generated by
$\mathfrak{u}\in\mathcal{L}_{0}$ has all layers identical.

On the other hand, it also follows that $\Upsilon$ is invertible and its
inverse satisfies $\Upsilon^{(-1)}=\Upsilon$. Therefore, if
$\Upsilon(\mathfrak{u})=\mathfrak{w}$, it follows that
$\Upsilon(\mathfrak{w})=\Upsilon^{(2)}(\mathfrak{u})=\mathfrak{u}$ for all
finite or infinite binary sequences $\mathfrak{u},\mathfrak{w}$.
Geometrically, this means that the single layer of a helicoid generated by a
binary sequence is composed of three identical diamond petals. Moreover, the
diamond is the union of two equilateral triangles, positioned symmetrically
across the short diagonal $\mathfrak{w}$. This proves the first two parts of
Theorem 4.

Figure 5. Two helicoids with identical layers on all levels. They are
generated by $\mathfrak{u}$ given by the first thirty elements of the
Fibonacci sequence $F_{n}$ (left) and the Bisection of Fibonacci sequence
$F_{2n}$ [19, A001906] (right) in decreasing order, followed by the sequence
of ten bits: $0,1,0,0,0,0,0,1,0,0$. Distinct integers are shown in different
colors. Under the helicoids, the corresponding generating sequences
$\Upsilon^{(0)}(\mathfrak{u})$,
$\Upsilon^{(1)}(\mathfrak{u}),\dots,\Upsilon^{(6)}(\mathfrak{u})$ of the
intermediate triangles are shown. In both helicoids, the initial
$\mathfrak{u}=\Upsilon^{(0)}(\mathfrak{u})$ is covered by
$\Upsilon^{(6)}(\mathfrak{u})$, but they can be seen for comparison on the
accompanying maps.

Figure 6. The first level of the helicoid generated by $\mathfrak{u}^{\prime}$
given by the first twenty $4$th positive powers (left) and the fourth level of
the helicoid (out of the nine distinct it has) generated by
$\mathfrak{u}^{\prime\prime}$ given by the first twenty $5$th positive powers
(right), both in decreasing order, and then followed each by the same sequence
of ten bits: $0,1,0,0,0,0,0,1,0,0$. Distinct integers are shown in different
colors. Under the represented levels, the corresponding generating sequences
of their intermediate triangles are shown:
$\Upsilon^{(0)}(\mathfrak{u}^{\prime})$,
$\Upsilon^{(1)}(\mathfrak{u}^{\prime}),\dots,\Upsilon^{(6)}(\mathfrak{u}^{\prime})$
(left), and $\Upsilon^{(18)}(\mathfrak{u}^{\prime\prime})$,
$\Upsilon^{(19)}(\mathfrak{u}^{\prime\prime}),\dots,\Upsilon^{(24)}(\mathfrak{u}^{\prime\prime})$
(right, numbered also from $0$ to $6$). In the representation on the left,
$\mathfrak{u}^{\prime}=\Upsilon^{(0)}(\mathfrak{u}^{\prime})$ is covered by
$\Upsilon^{(6)}(\mathfrak{u}^{\prime})$ and in the representation on the right
$\Upsilon^{(18)}(\mathfrak{u}^{\prime\prime})$ is covered by
$\Upsilon^{(24)}(\mathfrak{u}^{\prime\prime})$, but their elements can be seen
for comparison on the last two rows of the maps underneath.

Figure 7. The basic levels of two helicoids that have identical layers from
the second level on. They are generated by $\mathfrak{u}$ given by the first
twenty primes (left) and the first twenty square-primes (right) in decreasing
order, followed by the sequence of ten bits: $0,1,0,0,0,0,0,1,0,0$. Distinct
integers are shown in different colors. Under the layers, the corresponding
generating sequences $\Upsilon^{(0)}(\mathfrak{u})$,
$\Upsilon^{(1)}(\mathfrak{u}),\dots,\Upsilon^{(6)}(\mathfrak{u})$ of the
intermediate triangles are shown. In both images, the initial
$\mathfrak{u}=\Upsilon^{(0)}(\mathfrak{u})$ is covered by
$\Upsilon^{(6)}(\mathfrak{u})$, but they can be seen for comparison on the
accompanying maps.

In order to prove the third part of Theorem 4, let $\rho\geq 0$ be integer and
let $\mathscr{C}(\rho)$ be the hexagonal circle of differences of radius
$\rho$ and center $a_{0}$ on a layer generated by
$\mathfrak{u}=(a_{0},a_{1},a_{2},\dots)$. Explicitly, the numbers on the first
edge of the circle are the elements on the eastern edge of the (P-G) triangle:

$E_{0}=\big{(}a_{\rho}=d_{\rho}^{(0)},d_{\rho-1}^{(1)},d_{\rho-2}^{(2)},\dots,d_{0}^{(\rho)}\big{)},$

where the differences $d_{k}^{(j)}$ are defined by (1). Likewise are obtained
all the six edges $E_{m}$, $0\leq m\leq 5$, of $\mathscr{C}(\rho)$, where
$E_{m}$ is the eastern edge of the (P-G) triangle generated by
$\Upsilon^{(m)}(\mathfrak{u})$ instead of $\mathfrak{u}$, that is,

$E_{m}=\Big{(}\Upsilon^{(m)}(a_{\rho})=d_{\rho}^{(0)}(m),d_{\rho-1}^{(1)}(m),d_{\rho-2}^{(2)}(m),\dots,d_{0}^{(\rho)(m)}\Big{)},\
\ \text{ for $m=0,1,\dots,5$,}$

where

$d_{k}^{(j+1)}(m):=\big{|}d_{k+1}^{(j)}(m)-d_{k}^{(j)}(m)\big{|}\quad\text{
and }\quad d_{k}^{(0)}(m):=\Upsilon^{(m)}(a_{k})\quad\text{ for $j,k\geq
0$.}\\\ $

Then

$\mathscr{C}(\rho):=E_{0}\cup E_{1}\cup E_{2}\cup E_{3}\cup E_{4}\cup
E_{5}\,.$

Now, suppose that $\mathfrak{u}$ is the generator of a helicoid that has just
one distinct layer, that is, we assume that
$\Upsilon^{(6)}(\mathfrak{u})=\mathfrak{u}$. If $a\geq 0$ is the $\rho$th
element of $\mathfrak{u}$, then the assumption says, in particular, that the
$\rho$th element of $\Upsilon^{(6)}(\mathfrak{u})$ is also equal to $a$. Then,
let us analyze the process of generating the layer just on the circle
$\mathscr{C}(\rho)$.

######  Remark 5.1.

Suppose that $\rho>0$ and $a$ is a champion of the initial sequence
$\mathfrak{u}$, that is, $a>0$ and $a$ is strictly larger than all the
elements if $\mathfrak{u}$ of lower indices.

  1. (1)

As the construction of the (P-G) triangle and its five subsequent
continuations involves only taking absolute values of differences, the numbers
on $\mathscr{C}(\rho)$ cannot be larger than $a$. Moreover, the sequence of
numbers on $\mathscr{C}(\rho)$ cannot increase again if it has dropped at any
point to a lower value, because, by assumption, $a$ is a champion.

  2. (2)

Since the number that arrives to cover on the upper level the original $a$
after the $6$th rotation is still $a$, it then follows that all numbers on
circle $\mathscr{C}(\rho)$ are equal.

  3. (3)

Again, since $a$ is a champion, the only possibility for this to happen is
when all the numbers on the smaller adjacent hexagonal circle
$\mathscr{C}(\rho-1)$ are $0$’s.

  4. (4)

Further, it follows that all the numbers on $\mathscr{C}(\rho-2)$ are also
only $0$’s.

By iterating, it follows from the above remark that if $a$ is a champion, then
all the numbers on the circle $\mathscr{C}(\rho)$ are equal to $a$ and all the
numbers in the interior of $\mathscr{C}(\rho)$ are $0$’s. Therefore, there is
no other champion in $\mathfrak{u}$ besides $a$, and this concludes the proof
of Theorem 4.

Figure 8. The hexagons on the base level of two helicoids generated by the
sequence $\mathfrak{u}$ of non-negative integers whose base-$3$ representation
contains no $2$ [19, A005836]. The image on the left uses $20$ and the image
on the right $30$ elements of the sequence in decreasing order, both followed
by the same ten bits: $0,1,0,0,0,0,0,1,0,0$. Distinct integers are shown in
different colors. The helicoids have four and two distinct levels,
respectively. Under the hexagons, the corresponding generating sequences
$\Upsilon^{(0)}(\mathfrak{u})$,
$\Upsilon^{(1)}(\mathfrak{u}),\dots,\Upsilon^{(6)}(\mathfrak{u})$ of the
intermediate triangles are shown. In both images, the initial generators
$\mathfrak{u}=\Upsilon^{(0)}(\mathfrak{u})$ are covered by
$\Upsilon^{(6)}(\mathfrak{u})$, but they can be seen for comparison in the
accompanying maps.

###  5.2. Trial of sequences with a single champion

The necessary condition for sequences to have at most one champion in order
for the helicoids they generate to have just one distinct layer proves to be
insufficient. In fact, there may exist one-champion sequences that generate
helicoids with a record number of distinct layers. In Figures 5–8, the
helicoids generated by different finite integer sequences are shown, each
having a single champion, their first element. To use comparable units in
reasonably sized images that can be displayed in print, we have chosen
decreasing sequences $\mathfrak{u}$ of $20$ or $30$ integers, all followed by
the same sequence of $10$ random bits: $0,1,0,0,0,0,0,1,0,0$. Distinct numbers
are represented by different colors. Under each helicoid, the generating
sequences of the partial equilateral triangles, namely
$\Upsilon^{(0)}(\mathfrak{u})$,
$\Upsilon^{(1)}(\mathfrak{u}),\dots,\Upsilon^{(6)}(\mathfrak{u})$,
$\Upsilon^{(0)}(\mathfrak{u})$, can be seen stacked on top of each other,
making it easier to compare and determine if the helicoid has multiple
distinct sheets on distinct levels.

One finds that the results are mixed. There are sequences that generate
helicoids with one distinct level, as the one in Figure 5, or with exactly two
distinct levels, as those in Figure 6 (left), Figure 7 and Figure 8 (right).
There are also helicoids with more distinct levels, like the one that is
generated by $5$th powers in Figure 6 (right), where is shown level four out
of the nine distinct ones it has, or the one in Figure 8 (left), which has
four distinct levels. The lead sequence, also used in reversed order in Figure
8, is $0,1,3,4,9,10,12,13,27,28,30,31,\dots$, the sequence that starts with
$0$ and is generated by the greedy algorithm so that it contains no arithmetic
progressions of length $3$. Its elements are also characterized as being sums
of distinct powers of $3$ or as having only $0$ and $1$ in their base-$3$
representation [19, A005836]. The longer instances with the first
$N=314,315,\dots,320$ elements in reversed order followed by the above
sequence of ten bits generate helicoids with $84$ distinct layers, the cycle
having just one element.

A complete characterization of the integer sequences based on the number of
distinct levels that their associated helicoids have requires further
investigation. For instance, it would be interesting to know if there are
helicoids with an arbitrarily large number of distinct levels. A promising
candidate to try seems to be a series of length-balanced decreasing sequences
of powers. For instance, to add to the points already mentioned, we note that
the sequence of just ten $9$th powers in decreasing order followed by
$0,1,0,0,0,0,0,1,0,0$ produces a helicoid with $262$ distinct layers, of which
$P=198$ are in a precycle and $C=64$ are in an endless repeated cycle. And the
analogues with $10$th and $11$th powers have $P+C=140+128=268$ and
$P+C=512+32=544$ distinct layers, respectively.

######  Question.

Is there a sequence of finite sequences of positive integers that generates
helicoids with an unlimited number of distinct levels?

##  6\. Proof of Theorem 5

Let us start by noting that it is sufficient to prove the belonging
relationship in (5) for proportions $R_{\mathfrak{w}}(k)$, as it will also
imply the one for $R_{\mathfrak{e}}(k)$. This follows from the rotation-
reflection symmetry, as we have seen in the discussion Section 5, and also by
following the same reasoning below with the sequences in
$\mathfrak{u}=\mathfrak{u}(N)=(a_{0},\dots,a_{N-1})\in\mathcal{L}_{0}(N)$
indexed from right to left instead of left to right, which obviously would
lead to the same conclusion, this time for $R_{\mathfrak{e}}(k)$.

Let $\varepsilon\in(0,1/2)$ be fixed. We first prove that there exists an
integer $N_{\varepsilon}$ such that if $N\geq N_{\varepsilon}$ then

$\begin{split}\mbox{\small$\displaystyle\frac{1}{2^{N}}$}\\#\Big{\\{}\mathfrak{u}\in\mathcal{L}_{0}(N):\mbox{\small$\displaystyle\frac{1}{N}$}\\#\big{\\{}0\leq
j\leq
N-1:a_{j}=1\big{\\}}\in\big{[}1/2-\varepsilon,\,1/2+\varepsilon\big{]}\Big{\\}}\geq
1-\varepsilon.\end{split}$ (19)

For each $\mathfrak{u}\in\mathcal{L}_{0}(N)$, denote the set of indices with
components equal to $1$ by

$S(\mathfrak{u}):=\big{\\{}0\leq j\leq
N-1:a_{j}=1\big{\\}}\subset\\{0,1,\dots,N-1\\}.$

Then the left-hand side of inequality (19) can be rewritten as

$\begin{split}&\mbox{\small$\displaystyle\frac{1}{2^{N}}$}\\#\Big{\\{}\mathfrak{u}\in\mathcal{L}_{0}(N):\mbox{\small$\displaystyle\frac{1}{N}$}\\#S(\mathfrak{u})\in\big{[}1/2-\varepsilon,\,1/2+\varepsilon\big{]}\Big{\\}}\\\
=&\mbox{\small$\displaystyle\frac{1}{2^{N}}$}\\#\Big{\\{}\mathfrak{u}\in\mathcal{L}_{0}(N):\mbox{\small$\displaystyle\frac{N}{2}$}-N\varepsilon\leq\\#S(\mathfrak{u})\leq\mbox{\small$\displaystyle\frac{N}{2}$}+N\varepsilon\Big{\\}}\\\
=&\mbox{\small$\displaystyle\frac{1}{2^{N}}$}\sum_{\frac{N}{2}-N\varepsilon\leq
l\leq\frac{N}{2}+N\varepsilon}\\#\Big{\\{}\mathfrak{u}\in\mathcal{L}_{0}(N):\\#S(\mathfrak{u})=l\Big{\\}}\,.\end{split}$

Noticing in passing that the unrestricted sum of all the cardinalities in the
last sum is $2^{N}$, the inequality (19) that we want to prove is

$\begin{split}\mbox{\small$\displaystyle\frac{1}{2^{N}}$}\sum_{\frac{N}{2}-\varepsilon
N\leq l\leq\frac{N}{2}+\varepsilon N}\binom{N}{l}\geq
1-\varepsilon\,\end{split}$

for sufficiently large $N$. For this, it is sufficient to prove that in the
sum above, the missed tails at the beginning and at the end, which are equal,
due to the symmetry of the binomial coefficients, are small, that is,

$2\sum_{0\leq l\leq M}\binom{N}{l}\leq\varepsilon 2^{N}\,,$

where we denoted $M:=\big{\lfloor}\frac{N}{2}-\varepsilon N\big{\rfloor}$.
Since the binomial coefficients are increasing in the range of summation
above, we replace the sum with a trivial upper bound and reformulate our
object as the following convenient statement that is sufficient to be proven:

$(M+1)\binom{N}{M}\leq\varepsilon 2^{N-1}\,,$ (20)

for sufficiently large $N$. To estimate the binomial coefficients we use
Stirling’s approximation formula for factorials given by Robbins [22] in the
form of two tight upper and lower bounds:

$\sqrt{2\pi n}\left({\frac{n}{e}}\right)^{n}e^{\frac{1}{12n+1}}<n!<{\sqrt{2\pi
n}\left({\frac{n}{e}}\right)^{n}e^{\frac{1}{12n}}.}$

Then the left-hand side of (20) is

$(M+1)\binom{N}{M}=(M+1)\frac{N!}{M!(N-M)!}<a\frac{N^{N+1/2}}{M^{M-1/2}(N-M)^{N-M+1/2}},$
(21)

for a positive constant $a<1/200$ if $N>100$. Here, we change the variable $M$
into $R$, where $R:=N/2-M$. Note that since $M=N/2-\varepsilon N+\theta$,
where $|\theta|\leq 1$, it follows that

$R=\varepsilon N+O(1).$ (22)

Then the new form of the inequality (20) that we want to prove for $N$
sufficiently large is

$a\cdot\frac{N^{N+1/2}}{\Big{(}\frac{N}{2}-R\Big{)}^{N/2-R-1/2}\Big{(}\frac{N}{2}+R\Big{)}^{N/2+R+1/2}}\leq\varepsilon
2^{N-1},$

which can still be rearranged further into the more convenient form

$2a\cdot\frac{N^{1/2}}{\Big{(}1-\frac{2R}{N}\Big{)}^{N/2-R-1/2}\Big{(}1+\frac{2R}{N}\Big{)}^{N/2+R+1/2}}\leq\varepsilon.$
(23)

Let us note that here $r:=2R/N$ is small because (22) implies

$r=\frac{2R}{N}=\frac{2\varepsilon
N+O(1)}{N}=2\varepsilon+O\Big{(}\mbox{\small$\displaystyle\frac{1}{N}$}\Big{)}.$
(24)

Introducing the new variable $r$ in (23), we find that it is equivalent with

$2a\cdot\frac{N^{1/2}}{\big{(}1-r^{2}\big{)}^{N/2}(1-r)^{-R-1/2}(1+r)^{R+1/2}}\leq\varepsilon.$
(25)

It remains to show that here the denominator has an order of magnitude higher
than that of the numerator. To do this, we evaluate the logarithm of the
denominator, which is

$\begin{split}=&\log\Big{(}\big{(}1-r^{2}\big{)}^{N/2}(1-r)^{-R-1/2}(1+r)^{R+1/2}\Big{)}\\\
=&\mbox{\small$\displaystyle\frac{N}{2}$}\log\big{(}1-r^{2}\big{)}+\Big{(}R+\mbox{\small$\displaystyle\frac{1}{2}$}\Big{)}\big{(}\log(1+r)-\log(1-r)\big{)}.\end{split}$

Then, by taking into account the size of $R$ and $r$ from (22) and (24), and
particularly the fact that $r$ is small, we replace the logarithms with their
power series approximations and see that the logarithm of the denominator in
(25) is further equal to

$\begin{split}&\mbox{\small$\displaystyle\frac{N}{2}$}\big{(}-r^{2}+O(r^{4})\big{)}+\Big{(}R+\mbox{\small$\displaystyle\frac{1}{2}$}\Big{)}\Big{(}2r+\mbox{\small$\displaystyle\frac{2r^{3}}{3}$}+O(r^{5})\Big{)}\\\
=&N\sum_{k=1}^{\infty}2^{2k}\left(\frac{1}{2k-1}-\frac{1}{2k}\right)\varepsilon^{2k}+O(\varepsilon).\end{split}$
(26)

It then follows that there exists an absolute constant $c>0$ such that (25) is
satisfied for sufficiently large $N$ if

$\begin{split}\frac{\log
N}{c\varepsilon^{2}N}<\frac{N^{1/2}}{c\varepsilon^{2}N}=\frac{1}{c\varepsilon^{2}N^{1/2}}<\varepsilon.\end{split}$

And, in order for this last requirement to be fulfilled, it is enough to take
$N>N_{\varepsilon}$, with
$N_{\varepsilon}:=\max\big{\\{}100,\,\big{\lfloor}\frac{1}{c\varepsilon^{6}}\big{\rfloor}\big{\\}}$.

To conclude the proof of Theorem 5 for just the left edge of the (P-G)
triangle, we apply the operator $\Upsilon$ on the generating sequences
$\mathfrak{u}$. Since we know by Theorems 2 and 3 that $\Upsilon$ is an
involution, therefore also a bijection, it follows from the above argument
that, aside from an exceptional set of the same size as that of
$\mathfrak{u}$’s, on the western edge, the sequences $\mathfrak{w}_{0}$
contain approximately the same number of $0$’s and $1$’s as well.

Then, in the same way, the same conclusion can be drawn for the next segments
$\mathfrak{w}_{1},\mathfrak{w}_{2},\dots$, which are parallel to
$\mathfrak{w}_{0}$, by reasoning with the partial subsequences of
$\mathfrak{u}$ that ignore a few starting elements. In the following, we need
to quantify precisely for how many segments we can be sure that the almost-
equal-proportions-result actually holds for.

Let $K$ be the number of the rays
$\mathfrak{w}_{0},\mathfrak{w}_{1},\dots,\mathfrak{w}_{K-1}$ in question. The
size of $K$ that we can afford will be determined later. These rays are
generated by the partial subsequences of $\mathfrak{u}$ that are given by
$\mathfrak{u}_{k}=\mathfrak{u}_{k}(N):=(a_{k},a_{k+1},\dots,a_{N-1})$ for
$0\leq k\leq K-1$. Note that the number of the sequences $\mathfrak{u}_{k}(N)$
is $2^{N-k}$ for any fixed $k$ and $N$.

Let $\varepsilon\in(0,1/2)$ be fixed and let $k$ and $K$ be such that $0\leq
k\leq K-1<N$ for some large $N$. In order to fulfill the additional
requirements on the multiple rays, we choose a larger value for the term on
the right side of the inequality (19), which means, a narrower target. Thus,
since the size of $\mathfrak{u}_{k}$ is $N-k$, its counterpart form becomes:

$\begin{split}\mbox{\small$\displaystyle\frac{1}{2^{N-k}}$}\\#\Big{\\{}\mathfrak{u}\in\mathcal{L}_{0}(N-k):\mbox{\small$\displaystyle\frac{\\#\big{\\{}k\leq
j\leq
N-1:a_{j}=1\big{\\}}}{N-k}$}\in\big{[}1/2-\varepsilon,\,1/2+\varepsilon\big{]}\Big{\\}}\geq
1-\mbox{\small$\displaystyle\frac{\varepsilon}{K}$}.\end{split}$ (27)

Let us note that this would suffice to completely prove Theorem 5. Indeed, let
$\mathcal{E}_{N}(k)$ be the part of the exceptional set $\mathcal{E}_{N}$ in
the statement of Theorem 5 that corresponds to the sequences $\mathfrak{u}$
that do not meet the requirement (7) of approximately-equal-proportion of
$1$’s and $0$’s for ray $\mathfrak{w}_{k}$. Then, the exceptional sets are
smaller and smaller in size as $k$ increases, and despite the fact that the
sets $\mathcal{E}_{N}(k)$ are not disjoint, we have

$\mathcal{E}(N)=\bigcup_{k=0}^{K-1}\mathcal{E}_{N}(k)\quad\text{ and
}\quad\\#\mathcal{E}(N)\leq\sum_{k=0}^{K-1}\\#\mathcal{E}_{N}(k).$ (28)

We know, via the one-to-one correspondence $T$ from Theorem 3, that the
generating sequence $\mathfrak{u}$ and the left edge $\mathfrak{w}$ of any
typical (P-G) triangle of zeros and ones are permutations of one another, so
that, the cardinality of the analogous exceptional sets of sequences that do
not fulfill the approximately equal number of $1$’s and $0$’s condition is the
same.

Therefore, once proven (27), using the inequality in (28), for the exceptional
set $\mathcal{E}(N)$ in the statement of Theorem 5, we find that

$\begin{split}\frac{1}{2^{N}}\\#\mathcal{E}(N)\leq\frac{1}{2^{N}}\sum_{k=0}^{K-1}\\#\mathcal{E}_{N}(k)\leq\sum_{k=0}^{K-1}\frac{1}{2^{N-k}}\\#\mathcal{E}_{N}(k)\leq
K\cdot\frac{\varepsilon}{K}=\varepsilon,\end{split}$

that is, $\\#\mathcal{E}(N)\leq\varepsilon 2^{N}$, as needed. Then we only
have to prove (27), indicating the range of $N$ in which it holds, and how
large we are allowed to take $K$.

Following the same steps as before in arguing relation (19), by rewriting the
cardinality of the sets inside (27), and then continuing by simplifying the
resulting expression, the inequality on the tails becomes

$\frac{1}{2^{N-k-1}}\sum_{0\leq l\leq
M}\binom{N-k}{l}\leq\frac{\varepsilon}{K}\,,$

where $M=\big{\lfloor}\frac{N-K}{2}-\varepsilon(N-K)\big{\rfloor}$. It then
follows that it suffices to see that for large $N$ we have

$(M+1)\binom{N}{M}\leq\frac{\varepsilon}{K}2^{N-1-K}\,,$ (29)

the strongest condition that, once met, all the corresponding $k$ inequalities
will also be satisfied for $0\leq k\leq K-1$.

Further, the estimates of the binomial coefficients is done as before, and we
arrive to ponder the inequality

$\frac{2^{K}N^{1/2}}{\big{(}1-r^{2}\big{)}^{N/2}(1-r)^{-R-1/2}(1+r)^{R+1/2}}\leq\frac{\varepsilon}{AK},$
(30)

for some constant $A>0$ that is independent of $\epsilon,N$ and $K$. Here $r$
is small and $R$ captures the new $M$ from (29). Precisely, as in (22) and
(24), we have:

$r=2\varepsilon+O\Big{(}\mbox{\small$\displaystyle\frac{1}{N}$}\Big{)}\quad\text{
and }\quad R=\varepsilon N+O(1).$ (31)

The logarithm of the numerator on the left-hand side of (30) is

$K\log 2+\frac{1}{2}\log N$ (32)

and, using (31), the logarithm of the denominator is the same as in (26):

$N\sum_{k=1}^{\infty}\frac{2^{2k-1}}{k(2k-1)}\varepsilon^{2k}+O(1)>c_{0}\varepsilon^{2}N.$
(33)

for some absolute constant $c_{0}>0$.

As a consequence, using (32) and (33), we find that inequality (30) holds true
for sufficiently large $N$ as long as the following condition is also
verified.

$c_{0}\varepsilon^{2}N-K\log 2-\frac{1}{2}\log N>\log\frac{K}{\varepsilon}$

But, for this to happen, it is necessary to exist absolute constants
$c_{1},c_{2}>0$ such that

$c_{1}\varepsilon^{2}N>K\quad\text{ and }\quad
c_{2}\varepsilon^{2}N>\log{K}-\log{\varepsilon}.$

Both these conditions are fulfilled if we take $K=\delta_{\varepsilon}N$,
where $\delta=\delta_{\varepsilon}>0$ is a suitable small constant that
depends only on $\varepsilon$ and the absolute constants above, and $N$ is
larger than a threshold $N_{\varepsilon,\delta}$ that relies on the same
dependencies and additionally on the choice of $\delta$. This concludes the
proof of Theorem 5.

### Acknowledgement

The authors acknowledge the contributions of Sundarraman Madhusudanan, a co-
author of [4], in the proof of Lemma 3.1.

## References

  * [1] Bhat RN.  Distribution of square-prime numbers.  Missouri J. Math. Sci., 2022, 34(1):121–126.  https://doi.org/10.35834/2022/3401121.
  * [2] Bhat RN.  Sequences, series and uniform distribution of sp numbers.  arXiv preprint arXiv:2210.04622v3, 2022.  https://doi.org/10.48550/arXiv.2210.04622.
  * [3] Bhat RN, Cobeli C, Zaharescu A.  On quasi-periodicity in proth-gilbreath triangles.  arXiv preprint arXiv:2307.11776, 2023.  https://doi.org/10.48550/arXiv.2307.11776.
  * [4] Bhat RN, Madhusudanan S.  Algebraic results on sp-numbers along with a generalization.  arXiv preprint arXiv:2211.09009v1, 2022.  https://doi.org/10.48550/arXiv.2211.09009.
  * [5] Caragiu M, Zaharescu A, Zaki M.  On Ducci sequences with algebraic numbers.  Fibonacci Q., 2011, 49(1):34–40.  https://www.fq.math.ca/Papers1/49-1/CaragiuZaharescuZaki.pdf.
  * [6] Caragiu M, Zaharescu A, Zaki M.  An analogue of the Proth-Gilbreath conjecture.  Far East J. Math. Sci. (FJMS), 2013, 81(1):1–12.  http://www.pphmj.com/abstract/7973.htm.
  * [7] Caragiu M, Zaharescu A, Zaki M.  On Ducci sequences with primes.  Fibonacci Q., 2014, 52(1):32–38.  https://www.fq.math.ca/Papers1/52-1/CaragiuZaharescuZaki.pdf.
  * [8] Cobeli C, Prunescu M, Zaharescu A.  A growth model based on the arithmetic $Z$-game.  Chaos Solitons Fractals, 2016, 91:136–147.  https://doi.org/10.1016/j.chaos.2016.05.016.
  * [9] Cobeli C, Zaharescu A.  Promenade around Pascal triangle – number motives.  Bull. Math. Soc. Sci. Math. Roum., Nouv. Sér., 2013, 56(1):73–98.  https://www.jstor.org/stable/43679285.
  * [10] Cobeli C, Zaharescu A.  A game with divisors and absolute differences of exponents.  J. Difference Equ. Appl., 2014, 20(11):1489–1501.  https://doi.org/10.1080/10236198.2014.940337.
  * [11] Cobeli C, Zaharescu A.  Flurries of Ducci waves.  Bull. Math. Soc. Sci. Math. Roum., Nouv. Sér., 2023, 66(2):177–188.  https://ssmr.ro/bulletin/pdf/66-2/articol_4.pdf.
  * [12] Cobeli CI, Crâşmaru M, Zaharescu A.  A cellular automaton on a torus.  Port. Math., 2000, 57(3):311–323.  https://www.emis.de/journals/PM/57f3/pm57f305.pdf.
  * [13] Gilbreath N.  Processing process: the Gilbreath conjecture.  J. Number Theory, 2011, 131(12):2436–2441.  https://doi.org/10.1016/j.jnt.2011.06.008.
  * [14] Guy RK.  The strong law of small numbers.  Am. Math. Mon., 1988, 95(8):697–712.  https://doi.org/10.2307/2322249.
  * [15] Guy RK.  Unsolved problems in number theory.  Probl. Books Math. New York, NY: Springer-Verlag, 3rd ed. edition, 2004\.
  * [16] Killgrove RB, Ralston KE.  On a conjecture concerning the primes.  Math. Tables Aids Comput., 1959, 13:121–122.  https://doi.org/10.2307/2001963.
  * [17] Montgomery HL.  Ten lectures on the interface between analytic number theory and harmonic analysis, volume 84 of Reg. Conf. Ser. Math.  Providence, RI: American Mathematical Society, 1994.
  * [18] Odlyzko AM.  Iterated absolute values of differences of consecutive primes.  Math. Comput., 1993, 61(203):373–380.  https://doi.org/10.2307/2152962.
  * [19] OEIS Foundation Inc .  The On-Line Encyclopedia of Integer Sequences, 2023.  Published electronically at http://oeis.org.
  * [20] Proth F.  Sur la série des nombres premiers.  Nouvelle Correspondance Mathématique, 1878, 4:236–240.  https://gdz.sub.uni-goettingen.de/download/pdf/PPN598948236_0004/LOG_0088.pdf.
  * [21] Prunescu M.  Symmetries in the Pascal triangle: $p$-adic valuation, sign-reduction modulo $p$ and the last non-zero digit.  Bull. Math. Soc. Sci. Math. Roum., Nouv. Sér., 2022, 65(4):431–447.  https://ssmr.ro/bulletin/pdf/65-4/articol_6.pdf.
  * [22] Robbins H.  A remark on Stirling’s formula.  Am. Math. Mon., 1955, 62:26–29.  https://doi.org/10.2307/2308012.