# The maximum of the complementary of a semigroup with restricted conditions

A. Linero Bas and D. Nieves Roldán

###### Abstract.

We consider the set

$\mathcal{A}=\left\\{10\cdot a+11\cdot b\ |\gcd(a,b)=1,a\geq 1,b\geq
2a+1\right\\}.$

We will prove that $\mathcal{A}$ is unbounded and that there exists a natural
number $M\notin\mathcal{A}$ for which

$\left\\{M+m:m\geq 1,m\in\mathbb{N}\right\\}\subset\mathcal{A}.$

Indeed, such number is $M=1674$.

Keywords: Semigroup; boundedness; restricted conditions; coin problem; max-
type difference equations.
Mathematics Subject Classification: 39A23; 11A99; 20M99.

##  1\. Introduction

In this paper, we focus on the set

$\mathcal{A}=\left\\{10\cdot a+11\cdot b\ |\gcd(a,b)=1,a\geq 1,b\geq
2a+1\right\\}.$

In concrete, we will prove that it is unbounded and has bounded gaps. In fact,
it is shown that 1674 is the biggest natural number not included in it.

This set appears when analysing the periodic character of the solutions of the
max-type difference equation

(1.1) $x_{n+4}=\max\\{x_{n+3},x_{n+2},x_{n+1},0\\}-x_{n}.$

In fact, Linero and Nieves, [3], show that the set of periods,
$\mathrm{Per}(F_{4})$, of Equation (1.1) is
$\mathrm{Per}(F_{4})=\mathcal{A}\cup\\{1,8,11\\}$. Therefore, a natural
question that arises is whether there exists or not a number $M$ such that
every integer number greater than $M$ belongs to $\mathcal{A}$.

Beforehand, it is interesting to mention that if we had not had the
restriction $b\geq 2a+1$, the answer would follow directly by the Diophantic
Frobenius Problem. In concrete, the problem -also referred as Coin Problem in
the literature- consists in finding the greatest number that can not be
expressed as a linear combination with positive coefficients of a set
$(a_{1},\ldots,a_{n})$ of natural numbers with $\gcd(a_{1},\ldots,a_{n})=1$.
In the particular case of two natural numbers, that greatest number is given
by $a_{1}a_{2}-a_{1}-a_{2}$ (for more information, the reader is referred to
[1]). In our case, $a_{1}=10$ and $a_{2}=11$, the greatest number that cannot
be expressed as a linear combination of $10$ and $11$ is $89$. But as we have
highlighted, the conditions $b\geq 2a+1$ and $\gcd(a,b)=1$ complicates the
problem.

The main section of this work, namely, Section 2, starts analysing the prime
numbers in $\mathcal{A}$ and the multiples of 11 that belong to the set. Then,
the technique developed to prove the unboundedness of $\mathcal{A}$ and to
describe in detail which numbers are in the set is based on the division of
the set of natural numbers, not multiple of 11, in ten different classes,

$C_{m}:=\\{10m+11k,k\geq 0\\},$

where we fix the value $m\in\\{1,2,\ldots,10\\}$. Finally, we collect in Table
2.11 all the numbers in $\mathcal{A}$.

##  2\. Unboundedness of $\mathcal{A}$

Let us denote the set of prime numbers by $\mathcal{P}$. In the following
list, we present the first elements $p$ of $\mathcal{P}$, until $401$, and we
encircle those admitting a decomposition $p=10\cdot a+11\cdot b$ with
$\gcd(a,b)=1$ and $b\geq 2a+1$:

$2,3,5,7$, $11$, $13,17,19,23,29,31,37,41,$ $43$,
$47,53,59,61,67,71,73,79,83,$ $89,$ $97$, $101,103,$ $107$, $109$, $113,127,$
$131$, $137,$ $139$, $149,$ $151$, $157,$ $163$, $167$, $173$, $179,181,191$,
$193$, $197$, $199,211,223,$ $227$, $229$, $233,$ $239$, $241$, $251$, $257$,
$263$, $269$, $271$, $277,$ $281$, $283$, $293$, $307$, $311$, $313$, $317$,
$331$, $337$, $347$, $349$, $353$, $359$, $367$, $373$, $379$, $383$, $389$,
$397$, $401$.

For instance, $397\in\mathcal{A}$ since $397=297+100=11\cdot 27+10\cdot 10$,
whereas $277\notin\mathcal{A}$ as the decompositions $277=187+90=11\cdot
17+10\cdot 9$ and $277=77+200=11\cdot 7+10\cdot 20$ are not allowed, since
$b<2a+1$ in both cases. We observe that we have encircled all the primes
greater than $277$. In fact, this observation can be confirmed by the
following result.

######  Proposition 1.

If $p\in\mathcal{P}$, with $p\geq 281$, then $p\in\mathcal{A}.$

###### Proof.

According to the above table, we can assume that $p\geq 401$. Notice that $p$
can be written in the form $p=(p-10m)+10m$, where $m$ is the smallest positive
integer (in fact, unique, with $m\in\\{1,2,\ldots,11\\}$) holding that $p-10m$
is a multiple of $11$. Then $p=11\left(\frac{p-10m}{11}\right)+10m$. Let
$d=\gcd\left(\frac{p-10m}{11},m\right).$ Then, on one hand, $d|p$ due to the
fact that $d$ divides $\frac{p-10m}{11}$ and $m$; on the other hand, since
$d\leq m$, and $p$ is prime, $p>12$, we deduce that $d=1$. Next, in order to
definitively prove that $p\in\mathcal{A}$, we need to show that
$\frac{p-10m}{11}\geq 2m+1.$ This is equivalent to see that $p\geq 32m+11$,
which obviously occurs because $m\leq 11$ and $p\geq 363$. ∎

We collect some information about the formation of elements of $\mathcal{A}$
once we know that a prime number $p$ belongs to the set.

######  Proposition 2.

Let $p\in\mathcal{P}\cap\mathcal{A}$, with $p\geq 43.$ Assume that
$p=10a+11b$, with $\gcd(a,b)=1$, $b\geq 2a+1$, $a\geq 1$. Then:

  * (a)

$pq\in\mathcal{A}$ for all $q\geq 1$ with $\gcd(p,q)=1$ and $aq\geq
12,(b-2a)q\geq 33$. In particular, $pq\in\mathcal{A}$ for all $q\geq 33$ with
$\gcd(p,q)=1$.

  * (b)

$p^{k}q\in\mathcal{A}$ for all $k\geq 2$ and for all $q\geq 1$ with
$\gcd(p,q)=1$.

###### Proof.

(a) We have $p=10a+11b$, with $\gcd(a,b)=1$, $b\geq 2a+1$, $a\geq 1$. For all
integer $r$, with $aq\geq 1+11r,$ we can write

(2.1) $pq=(10a+11b)q=10(aq)+11(bq)=10(aq-11r)+11(bq+10r).$

Take $r=1$, that is, $pq=10(aq-11)+11(bq+10)$. Obviously, $bq+10\geq
2\left(aq-11\right)+1.$ Let $d_{1}=\gcd\left(aq-11,bq+10\right)$. Notice that
$d_{1}|pq$. If $d_{1}$ contains some divisor $q^{\prime}$ of $q$, then
$q^{\prime}|(aq-11)$ and $q^{\prime}|(bq+10)$; then, $q^{\prime}|11$ and
$q^{\prime}|10$, so $q^{\prime}=1$. Consequently, $d_{1}$ only contains
divisors of $p$, and being $p$ prime, we deduce that $d_{1}\in\\{1,p\\}$.

– If $d_{1}=1$, we finish by imposing the additional property of being $aq\geq
12.$

– If $d_{1}=p$, since $\gcd(p,q)=1$ and $d|p$, we deduce that $d|(aq)$ and
$d|(bq)$, so $d|a$ and $d|b$ according to the above observation on the
divisors of $q$. Finally, taking into account that $\gcd(a,b)=1$, we conclude
that $d=1$, and $pq\in\mathcal{A}$ whenever $aq\geq 1+11p$. We refine our
bound for the values $q$ in the following way. we take $r=-1$ in (2.1) to set
$pq=10(aq+11)+11(bq-10).$ Notice that we must have $bq-10\geq 2aq+23$, that
is, $bq\geq 2aq+33$. Put $d_{2}=\gcd(aq+11,bq-10)$. Similarly to the above
study for $d_{1}$, it is easy to deduce that $d_{2}\in\\{1,p\\}$. If $d_{2}=1$
we finish the proof by requiring $aq\geq 12$ and $bq\geq 2aq+33$ (in
particular, if $q\geq 33$, we simultaneously get the two inequalities).
Otherwise, for $d_{1}=d_{2}=p$, we would obtain $aq-11=pu_{1}$ and
$aq+11=pu_{2}$ for some positive integers $u_{1},u_{2}$, and then
$22=p(u_{2}-u_{1})$, which would imply that $p$ divides $2$ or $11$, contrary
to the fact that $p\geq 43$.

(b) For $k\geq 2$, we write
$p^{k}q=pp^{k-1}q=(10a+11b)p^{k-1}q=10(ap^{k-1}q)+11(bp^{k-1}q)$, or
$p^{k}q=10(ap^{k-1}q-11)+11(bp^{k-1}q+10).$ Denote
$a^{\prime}=ap^{k-1}q-11,b^{\prime}=bp^{k-1}q+10.$ Of course, $b^{\prime}\geq
2a^{\prime}+1,$ as it can be easily verified. Put
$d=\gcd(a^{\prime},b^{\prime})$. We are going to prove that $d=1$, which will
end the proof.

If $d$ contains some divisor $q^{\prime}$ of $q$ in its factorization, then
$q^{\prime}|a^{\prime}$ and $q^{\prime}|b^{\prime}$, thus $q^{\prime}|11$ and
$q^{\prime}|10$. Therefore, $q^{\prime}=1$ and $d|p^{k}$. If $p|d$, then $p$
would divide both $a^{\prime}$ and $b^{\prime}$, and consequently $p|10$ and
$p|11$ (notice that $k-1\geq 1$), so $p=1$, a contradiction. We conclude that
$d=1$. ∎

In the above results we have excluded $p=11\in\mathcal{A}$. Next, we obtain
the elements of $\mathcal{A}$ of the form $11^{k}q$, $k\geq 1$, and
$\gcd(11,q)=1$.

######  Proposition 3.

Consider $11$, $a=0,b=1$. It holds:

  * (a)

$11^{k}q\in\mathcal{A}$ for all $k\geq 3$ and for all $q\geq 1$ with
$\gcd(11,q)=1$.

  * (b)

$11^{2}q\in\mathcal{A}$ for all $q\geq 3$ with $\gcd(11,q)=1$. Moreover,
$11^{2}$ and $11^{2}\cdot 2$ do not belong to $\mathcal{A}.$

  * (c)

$11q\in\mathcal{A}$ whenever $\gcd(11,q)=1$ and

$q\in\\{1\\}\cup\left(\\{q:q\geq 33\\}\setminus\\{43,54,76,120\\}\right).$

###### Proof.

Write $11^{k}q=11\cdot\left(11^{k-1}q-10\right)+10\cdot(11).$ Whereas $k\geq
3$ or $k=2,q\geq 3$, the inequalities $11^{k-1}q-10>0$ and $11^{k-1}q-10\geq
23$ are correct, and the decomposition is permitted. Then, obviously
$\gcd\left(11,11^{k-1}q-10\right)=1$ and $11^{k}q\in\mathcal{A}$. It remains
two numbers to be analyzed, namely, $11^{2},11^{2}\cdot 2.$ Since
$11^{2}=110+11$ and $11^{2}\cdot 2=110+132=220+20$, we deduce that both
numbers do not belong to $\mathcal{A}$. This completes Cases (a) and (b).

(c) Concerning the numbers of the form $11q$, we distinguish several cases. If
$1\leq q\leq 10$, then the only possible decomposition is given by
$11q=11q+10\cdot 0$, thus we exclude these values from $\mathcal{A}$, except
$11$.

If $11\leq q\leq 32$, the numbers $11q$ admit the decompositions:
$0+11q,10\cdot 11+11(q-10),10\cdot 22+11(q-20)$, each of which is not
admissible as an element of $\mathcal{A}$, either because $\gcd(0,q)=q>1$ or
because the inequality $b\geq 2a+1$ is violated. Hence, there are not elements
of $\mathcal{A}$ in the set $\\{11q:11\leq q\leq 32\\}$.

If $q\geq 33$, we claim that all the elements $11q$ belong to $\mathcal{A}$,
except for $q\in\\{43,54,76,120\\}$. As a first step, we show that
$11q\in\mathcal{A}$ whenever $q\geq 353$. This is clear, from Cases (a)-(b),
when $q$ is a multiple of $11$. If $q$ and $11$ are coprime, write
$11q=10\cdot(121)+11\cdot(q-110)$. Notice that $q-110\geq 2\cdot 121+1$, since
$q\geq 353$. Also, $\gcd(121,q-110)=1$, due to the fact that $q$ and $11$ are
coprime numbers.

Next, we focus on the range $33\leq q\leq 352$. Now, it makes sense to write
$11q=10\cdot 11+11\cdot(q-10)$ since $q-10\geq 23.$ If $q-10\neq\dot{11}$,
then $\gcd(11,q-10)=1$ and we finish. Then, it remains to analyze the case in
which $q-10=11m$, $33\leq q\leq 352$. This gives us the following set of
candidates to be elements of $\mathcal{A}$ of type $q=10+11m$,

$\displaystyle C_{q}^{0}:$ $\displaystyle=$
$\displaystyle\big{\\{}43,54,65,76,87,98,109,120,131,142,153,164,175,186,197,$
$\displaystyle
208,219,230,241,252,263,274,285,296,307,318,329,340,351\big{\\}}.$

For these elements $q\in C_{q}^{0}$, except $43,54,$ we write
$11q=10\cdot(22)+11\cdot(q-20)$ (necessarily, $q\geq 65$ if we try to obtain
an element of $\mathcal{A}$), and we consider $d_{0}=\gcd(22,q-20)$. If $q$ is
odd, then $q-20$ so is, and $d_{0}\in\\{1,11\\}$ in this situation. If
$11|d_{0}$, then $11|(q-20)$ or $d_{0}|(11m-10)$; as a consequence,
$d_{0}|10$, which is impossible. Hence, $11q$ is an admissible number if $q$
is odd, $65\leq q\leq 353$. So, we discard odd numbers $q$ in $C_{q}^{0}$,
except $43$, and the set is reduced to
$C_{q}^{1}:=\big{\\{}43,54,76,98,120,142,164,186,208,230,252,274,296,318,340\big{\\}}.$

Next, consider the even elements of $C_{q}^{1}$, of type $54+22j,0\leq j\leq
13,$ and write
$11q=10\cdot(33)+11\cdot(q-30)=10\cdot(33)+11\cdot(24+22j),\,j=0,1,\ldots,13$
(as we are forced to obey the inequality $q-30\geq 2\cdot 33+1$, we consider
$q\geq 97$). If $d_{1}=\gcd(33,24+22j)$, since $24+22j=11(2+2j)+2$ and
$33=3\cdot 11$, we deduce that $d_{1}\in\\{1,3\\}.$ Therefore, if $24+22j$ is
not a multiple of $3$, we find that $d_{1}=1$ and the corresponding value
$11q$ belongs to $\mathcal{A}.$ This happens whenever
$j\notin\\{0,3,6,9,12\\}$. Subjected to the constraint $q\geq 97$, the above
study allows us to say that $11q$ is an element of $\mathcal{A}$, except,
maybe, if $q$ is confined in the new subset
$C_{q}^{2}:=\big{\\{}43,54,76,120,186,252,318\big{\\}}.$

Now, we consider the decomposition $11q=10\cdot(55)+11\cdot(q-50)$, where it
is necessary to assume that $q\geq 161$, so it applies for the numbers
$186,252,318$. Since $q-50$ is then $136,202,268$, respectively, we find that
$\gcd(55,q-50)=1$ for these numbers, and they belong to $\mathcal{A}$. By this
analysis, $C_{q}^{2}$ is reduced to
$C_{q}^{3}:=\big{\\{}43,54,76,120\big{\\}}.$ For these numbers, it is easy to
check that neither of the possible combinations $10a+11b$ satisfy
simultaneously the two conditions $\gcd(a,b)=1$ and $b\geq 2a+1$:

$\displaystyle 11\cdot 43=473$ $\displaystyle=$ $\displaystyle 10\cdot
11+11\cdot 33=10\cdot 22+11\cdot 23$ $\displaystyle=$ $\displaystyle 10\cdot
33+11\cdot 13=10\cdot 44+11\cdot 3,$ $\displaystyle 11\cdot 54=594$
$\displaystyle=$ $\displaystyle 10\cdot 11+11\cdot 44=10\cdot 22+11\cdot 34$
$\displaystyle=$ $\displaystyle 10\cdot 33+11\cdot 24=10\cdot 44+11\cdot
14=10\cdot 55+11\cdot 4,$ $\displaystyle 11\cdot 76=836$ $\displaystyle=$
$\displaystyle 10\cdot 11+11\cdot 66=10\cdot 22+11\cdot 56=10\cdot 33+11\cdot
46$ $\displaystyle=$ $\displaystyle 10\cdot 44+11\cdot 36=10\cdot 55+11\cdot
26=10\cdot 66+11\cdot 16$ $\displaystyle=$ $\displaystyle 10\cdot 77+11\cdot
6,$ $\displaystyle 11\cdot 120=1320$ $\displaystyle=$ $\displaystyle 10\cdot
11+11\cdot 110=10\cdot 22+11\cdot 100=10\cdot 33+11\cdot 90$ $\displaystyle=$
$\displaystyle 10\cdot 44+11\cdot 80=10\cdot 55+11\cdot 70=10\cdot 66+11\cdot
60$ $\displaystyle=$ $\displaystyle 10\cdot 77+11\cdot 50=10\cdot 88+11\cdot
40=10\cdot 99+11\cdot 30$ $\displaystyle=$ $\displaystyle 10\cdot 110+11\cdot
20=10\cdot 121+11\cdot 10.$

∎

######  Corollary 1.

It holds that $1320=\max\\{11\cdot n:n\geq 1,11\cdot n\notin\mathcal{A}\\}$.

Once we have described the main properties concerning prime numbers as well as
a detailed study on which multiples of $11$ belong to $\mathcal{A},$ we are
now interested in proving that the set
$\mathcal{NA}:=\mathbb{N}\setminus\mathcal{A}$ is bounded. In fact our
objective is to calculate the maximum of $\mathcal{NA}$.

The strategy consists in dividing the set of natural numbers, not multiple of
$11$, in ten different classes, $\mathcal{C}_{m}:=\left\\{10m+11k,\,k\geq
0\right\\}$, where we fix the value $m\in\\{1,2,\ldots,10\\}$. For each class
$\mathcal{C}_{m}$, we will show that $\mathcal{NA}\cap\mathcal{C}_{m}$ is
bounded, and from the inspection of each subset
$\mathcal{NA}\cap\mathcal{C}_{m}$ we will deduce the maximum of
$\mathcal{NA}.$ To develop the study of those classes, we use the following
basic fact:

> Given a natural number $N$, $N\neq\dot{11}$, there exists a unique
> $m\in\\{1,2,\ldots,10\\}$ such that $N-10m=\dot{11}.$

###  2.1. The class $\mathcal{C}_{1}$

Notice that a number $n\in\mathcal{C}_{1}$ can be written as $n=10+11b$, where
$a=1$ and $\gcd(b,1)=1$. To assure that $n$ is in $\mathcal{A}$, it is
necessary to require $b\geq 3$. Therefore, $n\in\mathcal{A}$ for all $n\geq
43$. Additionally, $n=10,21,32$ belong to $\mathcal{NA}$ and
$N_{1}:=\max\left\\{\mathcal{NA}\cap\mathcal{C}_{1}\right\\}=32.$

###  2.2. The class $\mathcal{C}_{2}$

Let $n\in\mathcal{C}_{2}$, $n=20+11b,\,b\geq 0.$ Here, we take $a=2$.

$\bullet$ If $b$ is odd, with $b\geq 5$, then $n\in\mathcal{A}$ since
$\gcd(b,a)=1$. On the other hand, for $b=1$ and $b=3$, we find that $31$ and
$53$ belong to $\mathcal{NA}.$

$\bullet$ Assume that $b$ is even, $b=2j$, with $j\geq 1$ (if $j=0$ we have
$n=20\in\mathcal{NA}$). Since $\gcd(a,b)=2$, we need to manage another
decomposition, $n=130+11(b-10)$, that is, $n=10\cdot 13+11\cdot(2j-10).$ This
is possible if $2j-10\geq 27$, so $j\geq 19$. When $j\leq 18$, the numbers
$n=130+11(2j-10)$ belong to $\mathcal{NA}.$

We stress that $438,460,482,504,526\in\mathcal{A}$ as we have the
decompositions

$\displaystyle 438$ $\displaystyle=$ $\displaystyle 10\cdot 13+11\cdot 28,$
$\displaystyle 460$ $\displaystyle=$ $\displaystyle 10\cdot 13+11\cdot 30,$
$\displaystyle 482$ $\displaystyle=$ $\displaystyle 10\cdot 13+11\cdot 32,$
$\displaystyle 504$ $\displaystyle=$ $\displaystyle 10\cdot 13+11\cdot 34,$
$\displaystyle 526$ $\displaystyle=$ $\displaystyle 10\cdot 13+11\cdot 36,$

but, on the contrary, $416\in\mathcal{NA}$ since the number admits the
following decompositions

$416=10\cdot 2+11\cdot 36=10\cdot 13+11\cdot 26=10\cdot 24+11\cdot 16=10\cdot
35+11\cdot 6,$

and neither verifies simultaneously the two conditions
$\gcd(a^{\prime},b^{\prime})=1,b^{\prime}\geq 2a^{\prime}+1$ on the
coefficients of $10a^{\prime}+11b^{\prime}$.

We continue the analysis of this subcase, $n=10\cdot 13+11\cdot(2j-10),\,$
$j\geq 19$. If $2j-10$ is not a multiple of $13$, or $j-5\neq\dot{13}$, we
ensure that $n$ belongs to $\mathcal{A}$. Instead, if $j=5+13k$, with $k\geq
2$, then $n=10\cdot 13+11\cdot(26k)$. Notice that $26|n$, with
$n=26\left(5+11k\right).$ By descending in the decomposition, we can write
$n=10\cdot 35+11\cdot(26k-20)$.

Then, if $26k-20$ is not divisible neither by $5$ nor $7$, we finish once we
guarantee that $26k-20\geq 71$, which can be accomplished when $k\geq 4$. In
this case, the remaining cases $k=2,3$ provide the numbers $n=702,988$ both of
them belonging to $\mathcal{NA}$, as can be easily checked by hand.

If, on the contrary, $26k-20$ contains $5$ or $7$ in its factor decomposition,
with $k\geq 5$, as a first observation we mention that $n$ will be divisible
by $26\cdot 5$ or $26\cdot 7$. Secondly, realize that in the sequence
$\left\\{x_{r}:r\geq 0\right\\}=\left\\{13+11r:r\geq
0\right\\}=\left\\{13,24,35,46,57,68,79,\ldots\right\\}$ we find $x_{36}=409$,
a prime number in $\mathcal{A}$. If in the decompositions

(2.2) $n=10\cdot(13+11r)+11(26k-10r)$

we are not able to achieve $\gcd(13+11r,26k-10r)=1$ for $r=2,\ldots,36$, at
least we know that $n$ can be divided by $26$, $5$ or $7$, $79$ and $409$, so
$n=409\cdot t$, with $t\geq 26\cdot 79$. By Proposition 2, this implies that
$n\in\mathcal{A}$. In order to carry out this reasoning, for $r=36$ it is
necessary that in (2.2) either $26k-360\geq 2\cdot(409)+1$ if
$\gcd(409,26k-360)=1$ or $26k-360\geq 1$, otherwise; in both cases, it
suffices to take $k\geq 46$. For the remaining values, $5\leq k\leq 45$, let
consider $r=6$, $n=10\cdot 79+11\cdot(26k-60)$. Notice that
$\gcd(79,26k-60)\neq 1$ if and only if $k=57+79s$ with $s\in\mathbb{N}$. Also,
$26k-60\geq 2\cdot 79+1$ is equivalent to $k\geq 9$. Thus, for $9\leq k\leq
45$, the decomposition will be acceptable and $n\in\mathcal{A}$. Finally, for
$5\leq k\leq 8$, as we assumed that $26k-20$ was divisible by 5 or 7, the only
possible value is $k=5$. So, $n=1560$ and neither of its possible
decompositions verify simultaneously the two required conditions
$\gcd(a^{\prime},b^{\prime})=1$ and $b^{\prime}\geq 2a^{\prime}+1$. Recall
that we had assumed that $26k-20$ was divisible by $5$ or $7$. This refines
the analysis to the values $k=5,10,15,20,25,30,35,40,45$ (divisibility of
$13k-10$ by $5$) and $k=11,18,25,32,39$ (divisibility of $13k-10$ by $7$),
when we replace them in $n=130+286k$. With the help of a personal computer and
a rather rudimentary algorithm, we obtain the values

$1560,2990,4420,5850,7280,8710,10140,11570,13000$

and

$3276,5278,7280,9282,11284,$

respectively, and we check that, among them, $1560$ is the unique element in
$\mathcal{NA}$. It is easy to see that neither of the its possible
decompositions verify simultaneously the two required conditions
$\gcd(a^{\prime},b^{\prime})=1$ and $b^{\prime}\geq 2a^{\prime}+1$. For the
rest of numbers we find the following admissible pairs
$[a^{\prime},b^{\prime}]$:

$[57,220],[57,350],[79,460],[57,610],[57,740],[79,850],[57,1000],[57,1130]$

(for $2990,4420,5850,7280,8710,10140,11570,13000$), and

$[79,226],[57,428],[57,610],[79,772],[57,974]$

(for $3276,5278,7280,9282,11284$).

This concludes the case $b$ even in $\mathcal{C}_{2}$. From our analysis, we
highlight that

$N_{2}:=\max\left\\{\mathcal{NA}\cap\mathcal{C}_{2}\right\\}=1560.$

###  2.3. The class $\mathcal{C}_{3}$

We find the maximum of $\mathcal{NA}$ in
$\mathcal{C}_{3}=\left\\{30+11k,\,k\geq 0\right\\}.$ Let
$n\in\mathcal{C}_{3}$, $n=30+11b,\,b\geq 0.$ First, notice that
$n=30,41,52,63,74,85,96,$ are the values corresponding to $b=0,\ldots,6,$
respectively, and it is simple to see that they belong to $\mathcal{NA}$.
Hence, we start with $b\geq 7$ and $a=3$.

$\bullet$ If $b$ is not a multiple of $3$ with $b\geq 7$, we find that
$\gcd(b,3)=1$ and all the elements $30+11b$, with $b\neq\dot{3}$ are in
$\mathcal{A}.$

$\bullet$ Assume that $b=3j$, with $j\geq 3$. Since $\gcd(3,3j)=3$, we employ
the decomposition $n=10\cdot 14+11\cdot(3j-10)$. It has sense if we suppose
that $j\geq 4$ and, in order to be $n$ in $\mathcal{A}$, we must impose
$3j-10\geq 2\cdot 14+1,$ that is, $j\geq 13$. For the remaining values, $3\leq
j\leq 12$, we find the numbers
$n\in\left\\{129,162,195,228,261,294,327,360,393,426\right\\}$ which, as can
be checked, belong to $\mathcal{NA}$. (for instance, neither of the
decompositions $426=30+396=10\cdot 3+11\cdot 36=10\cdot 14+11\cdot 26=10\cdot
25+11\cdot 16=10\cdot 36+11\cdot 6$) is admissible in order to $426$ belong to
$\mathcal{A}$).

If $\gcd(14,3j-10)=1$, for $j\geq 13$ we guarantee that $n\in\mathcal{A}$.
This holds whenever $j=13$ and for the odd numbers $j\geq 15$ such that $j\neq
15+14u,u\geq 0$, namely

$j\in\\{17,19,21,23,25,27,31,33,35,37,39,41,\ldots\\}.$

But if $j\geq 14$ is even or $j=15+14k,$ with $k\geq 0$, we obtain that
$\gcd(14,3j-10)>1$ ($n$ can be divided either by $2$ or by $7$) and we need to
employ another decomposition of $n$. If we vary $a^{\prime}$ in
$10a^{\prime}+11b^{\prime}$ in an ascendant way, we obtain the successive
decompositions

(2.3) $n=10\cdot(14+11r)+11\cdot(3j-10-10r),\,r\geq 0.$

For $r=1$, we have $n=10\cdot 25+11(3j-20)$. In order to obtain elements of
$\mathcal{A}$, it must be $3j-20\geq 51$, to wit $j\geq 24$. Since we know
that $j=2k$ with $k\geq 7$ or $j=15+14k,$ with $k\geq 0$, the restriction
$j\geq 24$ enforces us to check the values $j=14,15,16,18,20,22$. For them,
their respective numbers $n$ are $492,525,558,624,690,756$, and it can be
verified that all of them are elements of $\mathcal{NA}$. Since we are
interested in the location of the maximum of $\mathcal{NA}$, for the
convenience of the reader we present the decompositions of $756$ which assure
us that this number is not in $\mathcal{A}$:

$\displaystyle 756$ $\displaystyle=$ $\displaystyle 10\cdot 3+11\cdot
66=10\cdot 14+11\cdot 56=10\cdot 25+11\cdot 46=10\cdot 36+11\cdot 36$
$\displaystyle=$ $\displaystyle 10\cdot 47+11\cdot 26=10\cdot 58+11\cdot
16=10\cdot 69+11\cdot 6.$

Now, once we restrict ourselves to $j\geq 24$, we note that if $3j-20$ is not
a multiple of $5$ then $n=30+33j$ is an element of $\mathcal{A}$. Since either
$j=2k$ ($k\geq 12$) or $j=15+14k$ (with $k\geq 1$), $3j-20$ is a multiple of
$5$ whenever either $k=5w$ for $w\geq 3$ or $j=15+70w$ for $w\geq 1$, and in
these cases is necessary to use a new decomposition.

We then pass to use $r=3$, i.e, $n=10\cdot 47+11(3j-40)$. This implicitly
means that $3j-40\geq 2\cdot 47+1,$ so it must be $j\geq 45$. In that case, we
need to analyze independently the remaining lest values of $j$, namely
$j=30,40$ (recall that we have in this point that $j=10w,w\geq 3$ or
$j=15+70w,w\geq 1$, and $n=30+33j$), which provide the numbers $n=1020,1350$,
respectively. Both of them are elements of $\mathcal{NA}$.

(for instance

$\displaystyle 1350$ $\displaystyle=$ $\displaystyle 10\cdot 3+11\cdot
120=10\cdot 14+11\cdot 110=10\cdot 25+11\cdot 100$ $\displaystyle=$
$\displaystyle 10\cdot 36+11\cdot 90=10\cdot 47+11\cdot 80=10\cdot 58+11\cdot
70=\cdots,$

and neither of them satisfy the conditions to be included in $\mathcal{A}$).
In the sequel, we consider $j=10w$ with $w\geq 5$, or $j=15+70w,w\geq 1$.

Since the case $r=3$ has associate the value $g:=\gcd(47,3j-40)\in\\{1,47\\}$,
if $3j-40$ is not a multiple of $47$, we will include $n=30+33j$ in
$\mathcal{A}$. Otherwise, $3j-40$ is a multiple of $47$ whenever either
$30w-40,w\geq 5$ is multiple of $47$ or $3(15+70w)-40,w\geq 1$ so is; that is,
whenever either $3w-4=\dot{47},w\geq 5$ or $1+42w=\dot{47},w\geq 1$. In the
first case, $w=17+47u$, $u\geq 0$, and in the second one, $w=19+47u,u\geq 0$.
Hence, either $j=10(17+47u)=170+470u,u\geq 0$, or
$j=15+70(19+47u)=1345+3290u,u\geq 0$. For these values of $j$, we must use a
new $n=10\cdot(14+11r)+11(3j-10-10r)$.

In the sequence $\left\\{x^{(3)}_{r}:r\geq 0\right\\}=\left\\{14+11r:r\geq
0\right\\}=$ $\left\\{14,25,36,47,\ldots\right\\}$ we find $x^{(3)}_{27}=311$,
a prime number in $\mathcal{A}$. From this fact, if in the decompositions
$n=10\cdot(14+11r)+11(3j-10-10r)$ we are not able to attain a value
$r=1,2,3,\ldots,27$ such that $\gcd(14+11r,3j-10-10r)=1$, at least we know
that $n$ can be divided by $3$, $2$ or $7$, $47$ and $311$, so $n=311\cdot t$,
with $t\geq 6\cdot 47$. Then, Proposition 2 gives $n\in\mathcal{A}$. In order
to be able to use this argument, for $r=27$ it is necessary that in (2.3)
either $3j-10-10\cdot 27\geq 2\cdot(311)+1$ if $\gcd(311,3j-10-10\cdot 27)=1$
or $3j-10-10\cdot 27\geq 1$, otherwise; in both cases, it suffices to take
$j\geq 301$. Taking into account that $j=170+470u,u\geq 0$, or
$j=1345+3290u,u\geq 0$, this choice is always suitable, except for the value
of $j=170$. Here, we find the number $n=30+33\cdot 170=5640$ which, in fact,
also belongs to $\mathcal{A}$ since $5640=10\cdot 91+11\cdot 430$, with
$\gcd(91,430)=1$ and $430\geq 184$. This ends the case $b=3j$ in
$\mathcal{C}_{2}$. From our analysis, it is worth mentioning that

$N_{3}:=\max\left\\{\mathcal{NA}\cap\mathcal{C}_{3}\right\\}=1350.$

###  2.4. The class $\mathcal{C}_{4}$

We want to discuss if there are elements of $\mathcal{NA}$ in
$\mathcal{C}_{4}=\left\\{40+11k,\,k\geq 0\right\\},$ and to find the maximum
of $\mathcal{NA}\cap\mathcal{C}_{4}.$ Let $n\in\mathcal{C}_{4}$,
$n=40+11b,\,b\geq 0.$

The first values of $\mathcal{C}_{4}$ are $n=40,51,62,73,84,95,106,117,128,$
corresponding to $b=0,\ldots,8$ respectively, and it is easily seen that all
of them belong to $\mathcal{NA}$. Hence, we start our analysis with $b\geq 9$.
Realize that $a=4$.

$\bullet$ If $b$ is odd, $b\geq 9$, then automatically
$n=40+11b\in\mathcal{A}$ as the conditions $\gcd(a,b)=1$ and $b\geq 2a+1$ are
guaranteed.

$\bullet$ If $b$ is even, $b=2j$, $j\geq 5$, we need to change the
decomposition. Firstly, we consider $n=10\cdot 15+11\cdot(2j-10)$. Since it
must be $2j-10\geq 31$, we need $j\geq 21$. Thus, if $5\leq j\leq 20$, all the
numbers $40+22j$ i.e.,
$\\{150,172,194,216,238,260,282,304,326,348,370,392,414,436,458,480\\}$ belong
to $\mathcal{NA}$, (for instance, $480=10\cdot 4+11\cdot 40=10\cdot 15+11\cdot
30=10\cdot 26+11\cdot 20=\cdots$).

From here, we will assume that $j\geq 21$. If $2j-10$ is neither a multiple of
$3$ nor a multiple of $5$, the number $n$ will be in $\mathcal{A}$. Otherwise,
either $j=5k,k\geq 5$ if $5|(2j-10)$, or $j=8+3k,\ k\geq 5$, if $3|(2j-10)$.
Similarly to the studies developed for $\mathcal{C}_{2}$ and
$\mathcal{C}_{3}$, we can write $n$ as follows:

(2.4) $n=10\cdot(15+11r)+11(2j-10-10r),\,r\geq 0.$

For $r=2$, $n=10\cdot 37+11\cdot(2j-30)$ will be in $\mathcal{A}$ if
$2j-30\geq 75$, that is, if $j\geq 53$. As we consider that $j=5k,k\geq 5$ or
$j=8+3k,k\geq 5$, the decomposition cannot be performed if
$j=25,30,35,40,45,50$ or for $j=23,26,29,32,35,38,41,44,47,50$. For those
values, their corresponding $n$ are elements of $\mathcal{NA}$.

we obtain the numbers $\\{590,$ $700,$ $810,$ $920,$ $1030,$ $1140\\}$ and
$\\{612,678,$ $744,$ $810,$ $876,$ $942,$ $1008,$ $1074,$ $1140\\},$
respectively. All those numbers are elements of $\mathcal{NA}$.

Until now, he have detected the values of $\mathcal{NA}$ having the form
$40+22j$, when $j<53$. Now, if $j\geq 53$ and $2j-30$ is coprime with $37$, we
finish: every number will belong to $\mathcal{A}$. Otherwise, if $37|(2j-30)$,
then either $37|(10k-30)$, with $k\geq 11$ (when $j=5k$) or $37|(6k-14)$, with
$k\geq 15$ (when $j=8+3k$); in the first case, $37|(k-3)$, so
$k=40+37u,\,u\geq 0$, and in the second one, $37|(3k-7)$, so $k=27+37u,\,u\geq
0$. Then our analysis will continue by considering $j\geq 53$ and
$j=5k=200+185u,\geq 0,$ or $j=8+3k=89+111u,\geq 0.$ The sequence
$\left\\{x^{(4)}_{r}:r\geq 0\right\\}=\left\\{15+11r:r\geq
0\right\\}=\left\\{15,26,37,48,59,\ldots\right\\}$ contains at least a prime
number included in $\mathcal{A}$, namely $x^{(4)}_{32}=367$. Consequently, if
in the decompositions (2.4) we cannot find a value $r=1,\ldots,32$ such that
$\gcd(15+11r,2j-10-10r)=1$, at least we know that $n$ can be divided by $37$
and $59$, and $367$, so $n=367\cdot t$, with $t\geq 37\cdot 59$. Then,
Proposition 2 implies that $n\in\mathcal{A}$. In order to develop rightly this
reasoning, for $r=32$ it is necessary that in (2.4) either $2j-10-10\cdot
32\geq 2\cdot(367)+1$ if $\gcd(367,2j-10-10\cdot 32)=1$ or $2j-10-10\cdot
32\geq 1$, otherwise; in both cases, it is sufficient to take $j\geq 533$.
Since $j=200+185u,u\geq 0$, or $j=89+111u,u\geq 0$, the choice $j\geq 533$ is
appropriate, except for the values $j=89,200,311,385,422$. For them, the
corresponding numbers $n=40+22j$ are $1998,4440,6882,8510,9324$, belonging to
$\mathcal{A}$. This ends the case $b=2j$ in $\mathcal{C}_{4}$ and we stress
that

$N_{4}:=\max\left\\{\mathcal{NA}\cap\mathcal{C}_{4}\right\\}=1140.$

###  2.5. The class $\mathcal{C}_{5}$

We study the set $\mathcal{NA}$ in $\mathcal{C}_{5}=\left\\{50+11k,\,k\geq
0\right\\}$. Let $n\in\mathcal{C}_{5}$, $n=50+11b,\,b\geq 0.$ The first values
of $\mathcal{C}_{5}$ are $n=50,$ $61,$ $72,$ $83,$ $94,$ $105,$ $116,$ $127,$
$138,$ $149,$ $160,$ corresponding to $b=0,1,\ldots,10$ respectively, and it
is direct to check that all of them belong to $\mathcal{NA}$. Hence, we start
our analysis with $b\geq 11$. Realize that, at the beginning, $a=5$.

$\bullet$ If $b$ is not a multiple of $5$, then $\gcd(5,b)=1$. Additionally,
as we have supposed that $b\geq 11$, the condition $b\geq 2a+1$ is also
fulfilled, and therefore every number $n=50+11b$, with $b\geq 11$ and
$b\neq\dot{5}$, are included in $\mathcal{A}$.

$\bullet$ From now on, we assume that $b=5j,$ with $j\geq 3$. Being
unsuccessful the decomposition $n=10\cdot 5+11\cdot(5j)$ in the search of
elements of $\mathcal{A}$, we use the new decomposition $n=10\cdot
16+11\cdot(5j-10).$

Notice that if $j$ is odd, then $\gcd(16,5j-10)=1$. Also, we need $5j-10\geq
33$ or $j\geq 9$. Then, $n=50+55j\in\mathcal{A}$ for all $j\geq 9$, $j$ odd.
We exclude the cases $j=3,5,7$. For them, the corresponding $n$ are
$215,325,435$, all of them in $\mathcal{NA}$, as can be easily verified.

Now we assume that $j$ is even, $j\geq 4$, thus $j=2k$ and $b=10k$, $k\geq 2$.
We apply the decomposition $n=10\cdot 27+11\cdot(5j-20).$ If $3$ is not a
divisor of $5j-20$, or equivalently, if $j-4\neq\dot{3}$, then $n$ will be in
$\mathcal{A}$ whenever $j\geq 15$. With respect to the remaining even values
$j=6,8,12,14$, with $j-4\neq\dot{3}$, we find
$n=50+55j\in\\{380,490,710,820\\}$, all of them in $\mathcal{NA}$, (for
instance, $820=10\cdot 5+11\cdot 70=10\cdot 16+11\cdot 60=10\cdot 27+11\cdot
50=10\cdot 38+11\cdot 40=10\cdot 49+11\cdot 30=\cdots$).

Otherwise, assume $j=2k,k\geq 2$, $j-4=\dot{3}$, hence $k=2+3u,\,$ and
$j=4+6u,\,u\geq 0$. Then $n=10\cdot 27+11\cdot(30u),$ with $u\geq 0$. Next,
consider the decomposition $n=10\cdot 49+11\cdot(30u-20).$ If $3u-2$ is not a
multiple of $7$, we could ensure that $n\in\mathcal{A}$ if $30u-20\geq 99$,
that is, if $u\geq 4$. Hence, if $30u-20=5j-40$ is not a multiple of $7$, with
$u\geq 4$, then $n$ is in $\mathcal{A}$. In this analysis we have excluded the
values $u=0,1,2$ (for which $3u-2$ is not a multiple of $7$), or the
corresponding values $n=50+55(4+6u)\in\\{270,600,930\\}$, which are in
$\mathcal{NA}$, (for instance, $930=10\cdot 5+11\cdot 80=10\cdot 16+11\cdot
70=10\cdot 27+11\cdot 60=10\cdot 38+11\cdot 50=10\cdot 49+11\cdot 40=\cdots$).

If $3u-2$ is a multiple of $7$, with $u\geq 3$, we have $u=3+7\ell,$ with
$\ell\geq 0$, and consequently $j=4+6u=22+42\ell$, $\ell\geq 0$. In
particular, $j\geq 22$. Notice that we can write $n=50+55j$ in the following
forms:

(2.5) $n=10\cdot(27+11r)+11(30u-10r),\,r\geq 0.$

The sequence $\left\\{x^{(5)}_{r}:r\geq 0\right\\}=\left\\{27+11r:r\geq
0\right\\}=\left\\{27,38,49,60,\ldots\right\\}$ contains at least a prime
number included in $\mathcal{A}$, namely $x^{(5)}_{22}=269$. Then, if in the
decompositions (2.5) we cannot find a value $r=1,\ldots,22$ such that
$\gcd(27+11r,30u-10r)=1$, at least we know that $n$ can be divided by $3,7$
and $71$, as well as $269$, so $n=269\cdot t$, with $t\geq 21\cdot 71$. Thus,
Proposition 2 implies that $n\in\mathcal{A}$. In order to develop an
appropriate reasoning, for $r=22$ it is necessary that in (2.5) either
$30u-10\cdot 22\geq 2\cdot(269)+1$ if $\gcd(269,30u-10\cdot 22)=1$ or
$30u-10\cdot 22\geq 1$, otherwise; in both cases, it is sufficient to take
$u\geq 26$. Since $u=3+7\ell,$ this is achieved if we take $\ell\geq 4$. For
the remaining cases, $\ell=0,1,2,3$, we have $n=1260,3570,5880,8190.$ Among
them, $3570,5880,8190\in\mathcal{A}$. On the contrary, $1260\in\mathcal{NA}$,
since $1260=10\cdot 5+11\cdot 110=10\cdot 16+11\cdot 100=10\cdot 27+11\cdot
90=10\cdot 38+11\cdot 80=10\cdot 49+11\cdot 70=10\cdot 60+11\cdot 60=\cdots$.
This ends the case $b=5j$ in $\mathcal{C}_{5}$. According to our study, we
emphasize that

$N_{5}:=\max\left\\{\mathcal{NA}\cap\mathcal{C}_{5}\right\\}=1260.$

###  2.6. The class $\mathcal{C}_{6}$

We discuss the existence of values of $\mathcal{NA}$ in
$\mathcal{C}_{6}=\left\\{60+11k,\,k\geq 0\right\\},$ and try to find the
maximum value in $\mathcal{NA}\cap\mathcal{C}_{6}.$ Let $n\in\mathcal{C}_{6}$,
$n=60+11b,\,b\geq 0.$ It is immediate to check that the first values of
$\mathcal{C}_{6}$, for $b=0,1,\ldots,12$ belong to $\mathcal{NA}$. Hence, we
start our analysis with $b\geq 13$. Realize that, at the beginning, $a=6$.

$\bullet$ If, simultaneously, $b\neq\dot{2}$ and $b\neq\dot{3}$, then
$60+11b\in\mathcal{A}$ whenever $b\geq 2a+1$, and this is guaranteed by our
initial hypothesis on $b\geq 13$.

$\bullet$ Assume that either $b=2j$ for $j\geq 7$, or $b=3j$ for $j\geq 5$.
Now, it is necessary to consider another decomposition, say $n=10\cdot
17+11\cdot(b-10).$

– If $b-10\neq\dot{17},$ we have $n\in\mathcal{A}$ once we check that the
condition $b-10\geq 2\cdot 17+1$ is satisfied. This holds for $b\geq 45$.

(i) In this case, if in turn $b=2j$ for $j\geq 7$, then we have
$n\in\mathcal{A}$ for $j\geq 23$; and, being $2j-10\neq\dot{17}$, the excluded
cases in our analysis are $7\leq j\leq 21$. For them, the corresponding values
of $n=10\cdot 17+11\cdot(2j-10)$ are
$214,236,258,280,302,324,346,368,390,412,434,456,478,500,522.$ None of them is
realized as an element of $\mathcal{A}$, all of them belong to $\mathcal{NA}$,
(for instance,$522=10\cdot 6+11\cdot 42=10\cdot 17+11\cdot 32=\cdots$, here
$32<2\cdot 17+1$). Notice that, even, for $j=22$ we obtain
$n=544\in\mathcal{NA}$.

(ii) If, additionally, $b=3j$, with $j\geq 5$, then $n\in\mathcal{A}$ for
$j\geq 15$ (notice $b\geq 45$); and imposing the restriction
$3j-10\neq\dot{17}$, the excluded cases now are $j=5,6,7,8,10,11,12,13,14$,
and its associate values $n=10\cdot 17+11\cdot(3j-10)$ are
$225,258,291,324,390,423,456,489,522$ which are included in $\mathcal{NA}$.

– From now on, we consider that $b-10=\dot{17}$. We have two subcases: either
$b=2j$, $j\geq 7$, or $b=3j$ with $j\geq 5$. In the first one, we have
$2j-10=\dot{17}$ if and only if $j-5=\dot{17}$, therefore $j=22+17u,u\geq 0$;
in the second case, we find $3j-10=\dot{17}$ if and only if $j=9+17u,u\geq 0$.
Notice that we can write $n=60+11b$ in the following forms:

(2.6) $n=10\cdot(6+11r)+11(b-10r),\,r\geq 0.$

The sequence $\left\\{x^{(6)}_{r}:r\geq 0\right\\}=\left\\{6+11r:r\geq
0\right\\}=\left\\{6,17,28,39,50,\ldots\right\\}$ contains at least a prime
number included in $\mathcal{A}$, namely $x^{(6)}_{17}=193$. So, if in (2.6)
we cannot find a value $r=1,\ldots,17$ such that $\gcd(6+11r,b-10r)=1$, at
least we know that $n$ can be divided by $2$ or $3$, by $17$ and $61$, as well
as $193$, so $n=193\cdot t$, with $t\geq 34\cdot 61$. Then, Proposition 2
implies that $n\in\mathcal{A}$. For $r=17$ it is necessary that in (2.6)
either $b-10\cdot 17\geq 2\cdot(193)+1$ if $\gcd(193,b-10\cdot 17)=1$ or
$b-10\cdot 17\geq 1$, otherwise; in both cases, it is sufficient to take
$b\geq 557$.

According to the subcases above mentioned: (a) If $j=22+17u,u\geq 0$, then
$b=2j=44+34u$, and $b\geq 557$ for $u\geq 16$; consequently,
$n=544+374u\in\mathcal{A}$. For the remaining cases $0\leq u\leq 15$ we obtain
the following values of $n$, $544,$ $918,$ $1292,$ $1666,$ $2040,$ $2414,$
$2788,$ $3162,$ $3536,$ $3910,$ $4284,$ $4658,$ $5032,$ $5406,$ $5780,$
$6154.$ It is easy to check that $544,918\in\mathcal{NA}$, and with the help
of an algorithmic routine implemented in our personal computer, it can be
checked that the rest of values can be decomposed in the form
$10a^{\prime}+11b^{\prime}$, where the pairs $[a^{\prime},b^{\prime}]$ are
given respectively by

$\displaystyle[39,82],[39,116],[61,130],[39,184],[39,218],[61,232],[61,266],$
$\displaystyle[39,320],[61,334],[39,388],[39,422],[61,436],[39,490],[39,524].$

(b) If $j=9+17u,u\geq 0$, then $b=3j=27+51u$, and $b\geq 557$ for $u\geq 11$;
in this case, $n=357+561u\in\mathcal{A}$. For $0\leq u\leq 10$ we obtain the
following values of $n$,
$357,918,1479,2040,2601,3162,3723,4284,4845,5406,5967;$ we find that
$357,918\in\mathcal{NA}$, whereas the rest of values belong to $\mathcal{A}$
and its respective pairs $[a^{\prime},b^{\prime}]$ in the decomposition
$10a^{\prime}+11b^{\prime}$ are given by

$[28,109],[61,130],[28,211],[61,232],[28,313],[61,334],[28,415],[61,436],[28,517].$

This completes the study of $\mathcal{NA}\cap\mathcal{C}_{6}$. We have found
that

$N_{6}:=\max\left\\{\mathcal{NA}\cap\mathcal{C}_{6}\right\\}=918.$

###  2.7. The class $\mathcal{C}_{7}$

We discuss the existence of values of $\mathcal{NA}$ in
$\mathcal{C}_{7}=\left\\{70+11k,\,k\geq 0\right\\},$ and try to find the
maximum in $\mathcal{NA}\cap\mathcal{C}_{7}.$ Let $n=70+11b,\,b\geq 0.$
Firstly, we take $a=7$. It is necessary to consider $b\geq 15$ so $b\geq 2a+1$
is fulfilled. Thus, the first values of $\mathcal{C}_{7}$, for $b\leq 14$ are
$n=70,$ $81,$ $92,$ $103,$ $114,$ $125,$ $136,$ $147,$ $158,$ $169,$ $180,$
$191,$ $202,$ $213,$ $224,$ which are in $\mathcal{NA}$. Hence, we start our
analysis with $b\geq 15$.

$\bullet$ If $b$ is not a multiple of $7$, then $\gcd(7,b)=1$ and
$n=70+11b\in\mathcal{A}$.

$\bullet$ Assume that $b=7j$, with $j\geq 3$. We have to use the decomposition
$n=10\cdot 18+11\cdot(7j-10)$, whenever $7j-10\geq 37,$ or $j\geq 7$. For the
cases $j=3,\ldots,6$ we find the values $n=301,378,455,532$, all of them being
elements of $\mathcal{NA}$. From this point, we take $j\geq 7$.

– If $7j-10$, $j\geq 7$, is not multiple of $2$ nor $3$, we have that
$n=180+11(7j-18)\in\mathcal{A}$ since $\gcd(18,7j-10)=1$ and $7j-10\geq 37$.

– Instead, if $7j-10$ is either a multiple of $2$ or $3$, the condition on the
greatest common divisor is not satisfied. Take into account that: (a) if
$7j-10=\dot{2},j\geq 7$, then $j=8+2u,u\geq 0;$ (b) if $7j-10=\dot{3},j\geq
7$, we find $j=7+3u,u\geq 0$. In the next step, we consider $n=10\cdot
29+11\cdot(7j-20)$, which provides numbers in $\mathcal{A}$ if $7j-20\geq 59,$
or $j\geq 12$. For $j=7,\ldots,11$ we only need to check the values $j=7,8,10$
due to (a) and (b), for which the corresponding values are $609,686,840$,
which belong to $\mathcal{NA}$.

If, apart from the above requisites on $j$, we suppose that $7j-20$ is not a
multiple of $29$, we obtain that $n=10\cdot 29+11\cdot(7j-20)\in\mathcal{A}$
for all $j\geq 12$. If, on the contrary, we assume that $7j-20=\dot{29}$, the
above restrictions (a), (b) imply that either: (i) $7\cdot(8+2u)-20=\dot{29}$,
that is, $18+7u=\dot{29}$, so $u=14+29\ell,\,\ell\geq 0$; or (ii)
$7(7+3u)-20=\dot{29}$, that is, $u=29\ell,\,\ell\geq 0.$ As usual, we write
$n$ as follows:

(2.7) $n=10\cdot(18+11r)+11(7j-10-10r),\,r\geq 0.$

The sequence $\left\\{x^{(7)}_{r}:r\geq 0\right\\}=\left\\{18+11r:r\geq
0\right\\}=\left\\{18,29,40,51,\ldots\right\\}$ exhibits prime numbers which
are in $\mathcal{A}$, for instance $x^{(7)}_{11}=139$. In this way, if in
(2.7) we cannot find a value $r=1,\ldots,11$ such that
$\gcd(18+11r,7j-10-10r)=1$, at least we know that $n$ can be divided by $7$,
$2$ or $3$, by $29$, by $73$, as well as $139$, so $n=139\cdot t$, with $t\geq
14\cdot 29$. Then, Proposition 2 implies that $n\in\mathcal{A}$. In order to
our reasoning makes sense, for $r=11$ we need that in (2.7) either
$7j-10-10\cdot 11\geq 2\cdot(139)+1$ if $\gcd(139,7j-10-10\cdot 11)=1$ or
$7j-10-10\cdot 11\geq 1$, otherwise; in both cases, it is sufficient to take
$j\geq 57$. Finally: (i) if $j=8+2u=36+58\ell$, $\ell\geq 0$, it suffices to
take $\ell\geq 1$ to ensure that the value $n$ is in $\mathcal{A}$; the unique
exception appears for $\ell=0$ or $j=36$, and for this value we obtain
$n=2842$, a new number in $\mathcal{A}$; (ii) if $j=7+3u=7+87\ell$, $\ell\geq
0$, we take again $\ell\geq 1$ to guarantee $j\geq 57$; in this case, the
exception is given by $\ell=0$, or $j=7$, and for this value we obtain
$n=609$, which is not in $\mathcal{A}$.

This finishes the study of the set $\mathcal{NA}\cap\mathcal{C}_{7}$. We have
obtained

$N_{7}:=\max\left\\{\mathcal{NA}\cap\mathcal{C}_{7}\right\\}=840.$

###  2.8. The class $\mathcal{C}_{8}$

We pass to discuss the existence of values of $\mathcal{NA}$ in
$\mathcal{C}_{8}=\left\\{80+11k,\,k\geq 0\right\\},$ as well as to try to find
the maximum value in $\mathcal{NA}\cap\mathcal{C}_{8}.$ Let
$n\in\mathcal{C}_{8}$, $n=80+11b,\,b\geq 0.$ To start, we take $a=8$. To get
the necessary condition $b\geq 2a+1$, we must consider $b\geq 17$. Therefore,
the first values of $\mathcal{C}_{8}$, for $b=0,1,\ldots,16$ are elements of
$\mathcal{NA}$, namely $n=80,$ $91,$ $102,$ $113,$ $124,$ $135,$ $146,$ $157,$
$168,$ $179,$ $190,$ $201,$ $212,$ $223,$ $234,$ $245,$ $256.$ Hence, we start
our analysis with $b\geq 17$.

$\bullet$ If $b$ is odd, then automatically $n=80+11b\in\mathcal{A}$ for all
$b\geq 17$.

$\bullet$ Assume that $b$ is even, $b=2j,\,j\geq 9$. Consider the
decomposition

(2.8) $n=10\cdot 19+11\cdot(2j-10).$

If we apply $b^{\prime}\geq 2a^{\prime}+1$, with $a^{\prime}=19$ and
$b^{\prime}=2j-10$, it must be $j\geq 25.$ Thus, for the remaining values of,
$9\leq j\leq 24$, we obtain that the respective values of $n$ are $278,$
$300,$ $322,$ $344,$ $366,$ $388,$ $410,$ $432,$ $454,$ $476,$ $498,$ $520,$
$542,$ $564,$ $586,$ $608,$ and all of them are in $\mathcal{NA}$.
Consequently, from now on we assume that $b=2j$ with $j\geq 25$. From the
inspection of (2.8), we deduce that if $2j-10$ is not a multiple of $19$, then
we assure that $n\in\mathcal{A}$ for all $j\geq 25$. However, if
$2j-10=\dot{19}$, that is, $j=24+19u,\,u\geq 1$, we have to continue with the
process of looking for a suitable decomposition $10a^{\prime}+11b^{\prime}$.
For instance,

(2.9) $n=10\cdot(19+11r)+11\cdot(2j-10-10r),\,r\geq 0.$

The sequence $\left\\{x^{(8)}_{r}:r\geq 0\right\\}=\left\\{19+11r:r\geq
0\right\\}=\left\\{19,30,41,52,\ldots\right\\}$ includes prime numbers which
are in $\mathcal{A}$, for instance $x^{(8)}_{8}=107$. In this way, if in the
decompositions (2.9) we are not able to find a value $r=1,\ldots,8$ such that
$\gcd(19+11r,2j-10-10r)=1$, at least we know that $n$ can be divided by $2$,
$19$, as well as $107$, so $n=107\cdot t$, with $t\geq 38$. Then, by
Proposition 2 we know that $n\in\mathcal{A}$. In order to sustain our
argument, for $r=8$ it is necessary that in (2.9) either $2j-10-10\cdot 8\geq
2\cdot(107)+1$ if $\gcd(107,2j-10-10\cdot 8)=1$ or $2j-10-10\cdot 8\geq 1$,
otherwise; in both cases, it suffices to take $j\geq 153$. Taking into account
that $j=24+19u,$ $u\geq 1$, the inequality $j\geq 153$ is reached when $u\geq
7$, and then we can establish that $n=10\cdot 19+11(2j-10)$ is in
$\mathcal{A}$; the unique exceptions appear for $1\leq u\leq 6$, and for them
we have $n=1026,1444,1862,2280,2698,3116,3534;$ in this list,
$1026\in\mathcal{NA}$, since $1026=10\cdot 8+11\cdot 86=10\cdot 19+11\cdot
76=10\cdot 30+11\cdot 66=10\cdot 41+11\cdot 56=\cdots$, whereas the remaining
numbers are elements of $\mathcal{A}$ as it can be easily verified.

$\displaystyle[41,56],[41,94],[41,132],[41,170],[41,208],$
$\displaystyle[63,226],[41,284],[41,322],[41,360].$

Thus, we have finished the study of case $b=2j$ corresponding to the set
$\mathcal{NA}\cap\mathcal{C}_{8}$. Then

$N_{8}:=\max\left\\{\mathcal{NA}\cap\mathcal{C}_{8}\right\\}=1026.$

###  2.9. The class $\mathcal{C}_{9}$

We consider the set $\mathcal{C}_{9}=\left\\{90+11k,\,k\geq 0\right\\},$ and
try to guess the maximum value in $\mathcal{NA}\cap\mathcal{C}_{9}.$ Let
$n\in\mathcal{C}_{9}$, $n=90+11b,\,b\geq 0.$ To start, we take $a=9$. To get
the necessary condition $b\geq 2a+1$, we must consider $b\geq 19$. Therefore,
the first values of $\mathcal{C}_{9}$, for $b=0,1,\ldots,18$ are elements of
$\mathcal{NA}$, given by $n=90,$ $101,$ $112,$ $123,$ $134,$ $145,$ $156,$
$167,$ $178,$ $189,$ $200,$ $211,$ $222,$ $233,$ $244,$ $255,$ $266,$ $277,$
$288.$ Hence, we start our analysis with $b\geq 19$.

$\bullet$ If $b\neq\dot{3}$, then directly $n=10\cdot 9+11\cdot
b\in\mathcal{A}$ for all $b\geq 19$.

$\bullet$ Assume that $b=3j$, with $j\geq 7$ and consider $n=10\cdot
31+11\cdot(3j-20).$

– If, additionally, $3j-20$ is not a multiple of $31$, then $n$ is in
$\mathcal{A}$ if $3j-20\geq 63$ or $j\geq 28$. So, we must analyze the
exceptions $7\leq j\leq 28$. For these values, we find that $321,$ $354,$
$387,$ $420,$ $453,$ $486,$ $519,$ $552,$ $585,$ $618,$ $684,$ $750,$ $816,$
$882,$ $915,$ $948$ belong to $\mathcal{NA}$, and on the other hand $\\{651,$
$717,$ $783,$ $849,$ $981,$ $1014\\}\subset\mathcal{A}$, with associate pairs
$[a^{\prime},b^{\prime}]$ given respectively by
$[20,47],[20,53],[20,59],[20,71],$ $[31,64].$

– If $3j-20=\dot{31}$, $j\geq 7$, then $j=17+31u,\,u\geq 0.$ In this point, we
consider the general formulation of the decompositions in $\mathcal{C}_{9}$,

(2.10) $n=10\cdot(9+11r)+11\cdot(3j-10r),\,r\geq 0.$

The sequence $\left\\{x^{(9)}_{r}:r\geq 0\right\\}=\left\\{9+11r:r\geq
0\right\\}=\left\\{9,20,31,42,53,\ldots\right\\}$ includes prime numbers which
are in $\mathcal{A}$, for instance $x^{(9)}_{8}=97$. In this way, if in the
decompositions (2.10) we are not able to find a value $r=1,\ldots,8$ such that
$\gcd(9+11r,3j-10r)=1$, at least we know that $n$ can be divided by $3$, $31$,
as well as $97$, so $n=97\cdot t$, with $t\geq 93$. Then, Proposition 2 gives
$n\in\mathcal{A}$. For $r=8$ it is necessary that in (2.10) either $3j-10\cdot
8\geq 2\cdot(97)+1$ if $\gcd(97,3j-10\cdot 8)=1$ or $3j-10\cdot 8\geq 1$,
otherwise; in both cases, it suffices to take $j\geq 92$. Taking into account
that $j=17+31u,$ $u\geq 0$, the inequality $j\geq 92$ is true when $u\geq 3$.
In this case, the numbers $n$ will be in $\mathcal{A}$. For $u=0,1,2$, we have
$n=651,1674,2697$, being $651$ and $2697$ in $\mathcal{A}$. Finally, $1674$
provides us the maximum of $\mathcal{NA}\cap\mathcal{C}_{9}$, (notice that
$1674=10\cdot 9+11\cdot 144=10\cdot 20+11\cdot 134=10\cdot 31+11\cdot
124=10\cdot 42+11\cdot 114=10\cdot 53+11\cdot 104=\cdots,$ and realize that
$104<2\cdot 53+1$. This concludes the part for a multiple of $3$, $b=3j$, and
ends the case. We have found that

$N_{9}:=\max\left\\{\mathcal{NA}\cap\mathcal{C}_{9}\right\\}=1674.$

###  2.10. The class $\mathcal{C}_{10}$

We consider the set $\mathcal{C}_{10}=\left\\{100+11k,\,k\geq 0\right\\},$ and
try to find the maximum of $\mathcal{NA}\cap\mathcal{C}_{10}.$ Let
$n\in\mathcal{C}_{10}$, $n=100+11b,\,b\geq 0.$ To start, we take $a=10$. To
get the necessary condition $b\geq 2a+1$, we must consider $b\geq 21$. Thus,
the first values of $\mathcal{C}_{10}$, for $b=0,\ldots,20$ are elements of
$\mathcal{NA}$, given by $n=100,$ $111,$ $122,$ $133,$ $144,$ $155,$ $166,$
$177,$ $188,$ $199,$ $210,$ $221,$ $232,$ $243,$ $254,$ $265,$ $276,$ $287,$
$298,$ $309,$ $320.$ We start our analysis with $b\geq 21$. Write $n=10\cdot
32+11\cdot(b-20).$

$\bullet$ If $b$ is odd, then $\gcd(32,b-20)=1$ and we can establish that
$n\in\mathcal{A}$ if the condition $b^{\prime}\geq 2a^{\prime}+1$ is
satisfied, where $a^{\prime}=32,b^{\prime}=b-20$. This supposes to take $b\geq
85$, $b$ odd. With the help of a computer, we find out whether the excluded
values $b=21,\ldots,83$ originate elements of $\mathcal{A}$ or elements of
$\mathcal{NA}$. For these values we obtain
$\\{375,485,595,705\\}\subset\mathcal{NA}$ and the rest of the elements are in
$\mathcal{A}$, whose respective pairs $[a^{\prime},b^{\prime}]$ are given by

$\displaystyle[10,21],[10,23],[10,27],[10,29],[10,31],[10,33],[10,37],[10,39],[10,41],[10,43],$
$\displaystyle[10,47],[10,49],[10,51],[10,53],[10,57],[10,59],[10,61],[10,63],[21,55],[10,67],$
$\displaystyle[10,69],[10,71],[10,73],[21,65],[10,77],[10,79],[10,81],[10,83].$

$\bullet$ If $b$ is even, $b\geq 22$, or $b=2j$ with $j\geq 11$. Similarly to
the odd case, we suppose that $b\geq 86$ and consider alone the exceptions
$b=22+2\cdot s,0\leq s\leq 31.$ From the corresponding values of $n=10\cdot
32+11\cdot(b-20)$, only
$\\{694,716,760,782,848,892,914,958,1024\\}\subset\mathcal{A}$, with
respective pairs $[a^{\prime},b^{\prime}]$ given by

$[21,44],[21,46],[21,50],[21,52],[21,58],[21,62],[21,64],[21,68],[21,74],$

and with the rest of the elements included in $\mathcal{NA}$ (by way of
example, $1002=10\cdot 10+11\cdot 82=10\cdot 21+11\cdot 72=10\cdot 32+11\cdot
62=10\cdot 43+11\cdot 52=\cdots$).

From this point, we are restricted to the case $b=2j\geq 86$, or $j\geq 43$.
Next, we consider the combination $n=10\cdot 43+11\cdot(b-30).$ If
$b-30=2j-30\neq\dot{43}$, then $\gcd(43,b-30)=1$ and we finish if we guarantee
that $2j-30\geq 87$, or $j\geq 58$. This means that we have to study
separately the values $43\leq j\leq 57$. For them, we find
$\\{1068,1134,1200,1266,1288,1332\\}\subset\mathcal{NA}$ (in this direction,
let us show the decompositions of $1332$: $10\cdot 10+11\cdot 112=10\cdot
21+11\cdot 102=10\cdot 32+11\cdot 92=10\cdot 43+11\cdot 82=10\cdot 54+11\cdot
72=\cdots$), and the rest are elements of $\mathcal{A}$ with the following
respective pairs $[a^{\prime},b^{\prime}]$:

$\displaystyle[21,76],[21,80],[21,82],[21,86],[21,88],[21,92],[21,94],[21,100],[21,104].$

If $b-30=2j-30=\dot{43}$, then $j-15=\dot{43}$, so $j=58+43u,\,u\geq 0$. In
order to assign a combination $10a^{\prime}+11b^{\prime}$ to $n$, we need to
continue searching for new values $a^{\prime},b^{\prime}$. Then, we consider

(2.11) $n=10\cdot(10+11r)+11\cdot(b-10r),\,r\geq 0.$

The sequence $\left\\{x^{(10)}_{r}:r\geq 0\right\\}=\left\\{10+11r:r\geq
0\right\\}$ contains several prime numbers which are in $\mathcal{A}$. We fix
our attention in $x^{(10)}_{9}=109$. In this way, if in the decompositions
(2.11) we are not able to find a value $r=1,\ldots,9$ such that
$\gcd(10+11r,2j-10r)=1$ for the number $n=10\cdot 10+11\cdot(2j),$ with $j\geq
58$, $j=58+43u,u\geq 0$, at least we know that $n$ can be divided by $2$ and
$43$, as well as $109$, so $n=109\cdot t$, with $t\geq 86$. Then, Proposition
2 gives $n\in\mathcal{A}$. In order to well guiding our reasoning, for $r=9$
it is necessary that in (2.11) either $2j-10\cdot 9\geq 2\cdot(109)+1$ if
$\gcd(109,2j-10\cdot 9)=1$ or $2j-10\cdot 9\geq 1$, otherwise; in both cases,
it suffices to take $j\geq 155$. Bearing in mind that $j=58+43u,$ $u\geq 0$,
the inequality $j\geq 155$ is valid when $u\geq 3$. In this case, the numbers
$n$ will be in $\mathcal{A}$. For the remaining values of $u$,
$u\in\\{0,1,2\\}$, we obtain $n\in\\{1376,2322,3268\\}$, which present the
following respective decompositions $10\cdot 21+11\cdot 106,$ $10\cdot
65+11\cdot 152$, and $10\cdot 21+11\cdot 278$, thus they are in $\mathcal{A}$.
This ends the discussion on the even numbers $b=2j$, and closes the case. We
have found that

$N_{10}:=\max\left\\{\mathcal{NA}\cap\mathcal{C}_{10}\right\\}=1332.$

###  2.11. The final bound. A table of the elements of $\mathcal{A}$

By collecting all the study developed for the sets $\mathcal{C}_{m},1\leq
m\leq 10$, if we denote by $N_{m}$ the maximum value in
$\mathcal{NA}\cap\mathcal{C}_{m}$, $1\leq m\leq 10$, we find:

$N_{1}=32,$ | $N_{2}=1560,$ | $N_{3}=1350,$ | $N_{4}=1140,$ | $N_{5}=1260,$
---|---|---|---|---
$N_{6}=918,$ | $N_{7}=840,$ | $N_{8}=1026,$ | $N_{9}=1674,$ | $N_{10}=1332.$

On the other hand, the maximum value of $\mathcal{NA}$ being a multiple of
$11$ is $N_{11}=1320$ as Proposition 3 and Corollary 1 show. Therefore,

$M=\max\\{\mathcal{NA}\\}=\max\\{N_{m}:1\leq m\leq 11\\}=\boxed{1674}.$

Once we know the maximum of $\mathcal{NA}$, with the help of a mathematical
software and a few of patient we obtain all the elements in $\mathcal{A}$,
which we gather in the following table:

Intervals | Numbers in $\mathcal{A}$
---|---
$n\in[1,100]$ | $1,8,11,43,54,65,75,76,87,97,98$
$n\in[101,200]$ | $107,109,118,119,120,131,139,140,141,142,151,153,$
| $161,163,164,171,173,175,182,183,184,185,186,193,197$
$n\in[201,300]$ | $203,204,205,206,207,208,217,219,226,227,229,230,235,237,239,$
| $241,246,247,248,249,250,251,252,257,259,263,267,268,269,271,$
| $272,273,274,279,281,283,285,289,290,292,293,295,296,299$
$n\in[301,400]$ | $303,305,307,311,312,313,314,315,316,317,318,323,329,331,332,$
| $333,334,335,336,337,338,339,340,343,345,347,349,351,353,355,$
| $356,358,359,361,362,363,365,367,369,371,373,374,376,377,379,$
| $381,382,383,384,385,389,391,395,396,397,398,399,400$
$n\in[401,500]$ | $[401,500]\setminus\\{408,410,412,414,416,420,423,426,430,$
| $432,434,435,436,452,453,454,455,456,458,473,$
| $474,476,478,480,485,486,490,492,496,498,500\\}$
$n\in[501,600]$ | $[501,600]\setminus\\{518,519,520,522,525,532,540,542,544,546,$
| $552,558,562,564,584,585,586,590,594,595,600\\}$
$n\in[601,700]$ | $[601,700]\setminus\\{606,608,609,612,618,624,$
| $628,650,672,678,684,686,690,700\\}$
$n\in[701,800]$ | $[701,800]\setminus\\{702,705,710,738,744,750,756\\}$
$n\in[801,900]$ | $[801,900]\setminus\\{804,810,820,826,836,840,870,876,882\\}$
$n\in[901,1000]$ | $[901,1000]\setminus\\{915,918,920,930,936,942,948,980,988\\}$
$n\in[1001,1100]$ | $[1001,1100]\setminus\\{1002,1008,1020,1026,1030,1068,1074\\}$
$n\in[1101,1300]$ | $[1101,1300]\setminus\\{1134,1140,1200,1260,1266,1274,1288\\}$
$n\in[1301,1700]$ | $[1301,1700]\setminus\\{1320,1332,1350,1560,1674\\}$
$n>1674$ | All the values

######  Remark 1.

After Table 2.11 was done by performing a routine program in a standard
mathematical software, Professor Pedro A. García Sánchez commented us the
existence of an algebraic platform of free access, GAP (see [4], a system for
computational discrete algebra), and the corresponding package for calculating
the numbers of $\mathcal{A}$ (see [2]); for instance, the reader interested in
algorithm procedures can compute them by implementing the following
instructions:

s:=NumericalSemigroup(10,11);
belong:=x->ForAny(Factorizations(x,s),p->p[2]>=2*p[1]+1 and Gcd(p)=1);
Filtered(Intersection(s,[0..2000]),belong); $\hfill\Box$

##  3\. Conclusion

To sum up we have proved that the set

$\mathcal{A}=\left\\{10\cdot a+11\cdot b\ |\gcd(a,b)=1,a\geq 1,b\geq
2a+1\right\\}$

is unbounded. Moreover, we have found the greatest number that does not belong
to $\mathcal{A}$, that is, $M=1674$ and we have determined in detail all the
element of $\mathcal{A}$, which are collected in Table 2.11.

As far as we know, this problem has not been previously studied in the
literature. Observe that our result is a particular case from the analysis of
a certain difference equation of order $k=4$, Equation (1.1). Obviously, such
equation can be extended to an arbitrary order $k\geq 5$. In [3], the authors
propose a conjecture establishing a relationship between the general set of
periods and the semigroup generated by numbers $3k-1$ and $3k-2$; the exact
knowledge of such set of periods perhaps would need some additional conditions
to the elements of the semigroup. In this sense, it is an open problem to
determine the extra conditions that should be considered and, consequently, to
determine the gaps which those conditions generate in the semigroup and to
analyse the boundedness character of these gaps. To this regard, the authors
are aware of the existence of certain additional conditions, to wit, $1\geq
a\geq 10$, that yield to the unboundedness character of the complementary,
since it can be proved that there exists a strictly increasing sequence of
positive numbers $(n_{j})_{j}$ such that $2^{n_{j}}\notin\mathcal{A}$.

Therefore, some general interesting questions for algebrists would be: the
study of the boundedness character of the complementary of semigroups
generated by two coprimes numbers, $p$ and $q$, whose elements are originated
by combinations $a\cdot p+b\cdot q$ being $a,b$ natural integers satisfying
$\gcd(a,b)=1$, as well as to find in the bounded case a formula for the
biggest positive integer not representable with the assumptions of being
coprimes $a,b$; to extend the previous problem to the case where $p,q$ are not
necessarily coprime; to analyse the problem of adding extra conditions to the
numbers $a,b$ apart from the fact of being coprime, as for example, linear
inequalities.

## Acknowledgements

We sincerely thank Prof. Pedro A. García Sánchez from University of Granada,
Spain, and Prof. Christopher O’Neill from San Diego Statal University, CA,
USA, for providing us some useful comments and to inform us about the
existence of GAP system.

This paper has been partially supported by Grant MTM2017-84079-P
(AEI/FEDER,UE), Ministerio de Ciencias, Innovación y Universidades, Spain.

## References

  * [1] J. L. Ramírez Alfonsín, _The diophantine Frobenius problem_ , Oxford Lecture Series in Mathematics and its Applications 30, Oxford University Press, Oxford, 2005.
  * [2] M. Delgado, P.A. García-Sánchez, J. Morais, _NumericalSgps - a GAP package_ , v. 1.2.2, 03/03/2020, https://www.gap-system.org/Packages/numericalsgps.html.
  * [3] A. Linero Bas, D. Nieves Roldán, _Periods of a max-type equation_ , preprint.
  * [4] The GAP Group, _GAP – Groups, Algorithms, and Programming, Version 4.11.1_ ; 2021, https://www.gap-system.org.