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I don't know Swedish and I'm not finding the article "G. Blom and C. E. Froberg, On money changing" translated into English... so I tried to read the original (Swedish) with the help of Google Translate but, from the results, I fear that some semantics could get lost...
The article prove a denumerant's upper and lower bound:
$\frac{n^{k-1}}{(k-1)!\prod_{i=1}^{k}a_i} \leq d(n;a_1,\dots,a_k) \leq \frac{(n+s_k)^{k-1}}{(k-1)!\prod_{i=1}^{k}a_i}$
where
$s_1=1, s_2=a_2$ and $s_k=a_2+\frac{1}{2}(a_3+\dots+a_k), \forall k\geq 3$.
I would like to know: is it required (hypothesis) that all $a_i$ must be coprime i.e. $(a_1,\dots,a_k)=1$?
I first read about this theorem in the book "J. L. R. Alfonsin, The Diophantine Frobenius Problem" (pag. 74) but what I'm asking is not specified.

UPDATE
I would have written in the comments but it was too long.
Despite @Carl-FredrikNybergBrodda's competent reassurance about the translation (i.e. I'm assuming that there are no explicit statements about co-primality) something in the back of my mind was not in peace... unfortunately very often there are implicit hypotheses in math articles (especially dated ones).
The reasons why this happen... well... are different but it is usually because are considered trivial by the context (or by the author).
So yesterday I started reading the mathematical component of the article trying to reconstruct the proof and from what I seemed to understand... is that the implicit hypothesis is stronger than $(a_1,...,a_k)=1$ i.e. it's $a_1=1$.
This idea is supported by those observations:

1. This article is about coins and there is always the currency of unitary value (at least $"a.s."$).
2. All the explicit example start with $a_1=1$
3. pag. 60 the proof is by induction so there are two "base case" because the general expression for $s_i$ does not cover $s_1$ (by the way in the article the sequence $s$ is called $\Delta$), therefore it is necessary to check $m=1$ and $m=2$ before the inductive step $m\geq 3$.
   The case $m=2$ check the truthfulness of:
   $\frac{1}{(2-1)!}\frac{n^{2-1}}{a_2}\leq D(2,n)\leq \frac{1}{(2-1)!}\frac{(n+s_2)^{2-1}}{a_2}$
   but if it was $a_1\neq 1$ should be:
   $\frac{1}{(2-1)!}\frac{n^{2-1}}{a_1 a_2}\leq D(2,n)\leq \frac{1}{(2-1)!}\frac{(n+s_2)^{2-1}}{a_1 a_2}$

Maybe I didn't understand the proof or maybe the case $a_1 = 1$ is not restrictive (but I don't see why).

I don't know Swedish and I'm not finding the article "G. Blom and C. E. Froberg, On money changing" translated into English... so I tried to read the original (Swedish) with the help of Google Translate but, from the results, I fear that some semantics could get lost...
The article prove a denumerant's upper and lower bound:
$\frac{n^{k-1}}{(k-1)!\prod_{i=1}^{k}a_i} \leq d(n;a_1,\dots,a_k) \leq \frac{(n+s_k)^{k-1}}{(k-1)!\prod_{i=1}^{k}a_i}$
where
$s_1=1, s_2=a_2$ and $s_k=a_2+\frac{1}{2}(a_3+\dots+a_k), \forall k\geq 3$.
I would like to know: is it required (hypothesis) that all $a_i$ must be coprime i.e. $(a_1,\dots,a_k)=1$?
I first read about this theorem in the book "J. L. R. Alfonsin, The Diophantine Frobenius Problem" (pag. 74) but what I'm asking is not specified.

UPDATE
I would have written in the comments but it was too long.
Despite @Carl-FredrikNybergBrodda's competent reassurance about the translation (i.e. I'm assuming that there are no explicit statements about co-primality) something in the back of my mind was not in peace... unfortunately very often there are implicit hypotheses in math articles (especially dated ones).
The reasons why this happen... well... are different but it is usually because are considered trivial by the context (or by the author).
So yesterday I started reading the mathematical component of the article trying to reconstruct the proof and from what I seemed to understand... is that the implicit hypothesis is stronger than $(a_1,...,a_k)=1$ i.e. it's $a_1=1$.
This idea is supported by these observations:

1. This article is about coins and there is always the currency of unitary value (at least $"a.s."$).
2. All the explicit example start with $a_1=1$
3. pag. 60 the proof is by induction so there are two "base case" because the general expression for $s_i$ does not cover $s_1$ (by the way in the article the sequence $s$ is called $\Delta$), therefore it is necessary to check $m=1$ and $m=2$ before the inductive step $m\geq 3$.
   The case $m=2$ check the truthfulness of:
   $\frac{1}{(2-1)!}\frac{n^{2-1}}{a_2}\leq D(2,n)\leq \frac{1}{(2-1)!}\frac{(n+s_2)^{2-1}}{a_2}$
   but if it was $a_1\neq 1$ should be:
   $\frac{1}{(2-1)!}\frac{n^{2-1}}{a_1 a_2}\leq D(2,n)\leq \frac{1}{(2-1)!}\frac{(n+s_2)^{2-1}}{a_1 a_2}$

Maybe I didn't understand the proof or maybe the case $a_1 = 1$ is not restrictive (but I don't see why).

I don't know Swedish and I'm not finding the article "G. Blom and C. E. Froberg, On money changing" translated into English... so I tried to read the original (Swedish) with the help of Google Translate but, from the results, I fear that some semantics could get lost...
The article prove a denumerant's upper and lower bound:
$\frac{n^{k-1}}{(k-1)!\prod_{i=1}^{k}a_i} \leq d(n;a_1,\dots,a_k) \leq \frac{(n+s_k)^{k-1}}{(k-1)!\prod_{i=1}^{k}a_i}$
where
$s_1=1, s_2=a_2$ and $s_k=a_2+\frac{1}{2}(a_3+\dots+a_k), \forall k\geq 3$.
I would like to know: is it required (hypothesis) that all $a_i$ must be coprime i.e. $(a_1,\dots,a_k)=1$?
I first read about this theorem in the book "J. L. R. Alfonsin, The Diophantine Frobenius Problem" (pag. 74) but what I'm asking is not specified.

UPDATE
I would have written in the comments but it was too long.
Despite @Carl-FredrikNybergBrodda's competent reassurance about the translation (i.e. I'm assuming that there are no explicit statements about co-primality) something in the back of my mind was not in peace... unfortunately very often there are implicit hypotheses in math articles (especially dated ones).
The reasons why this happen... well... are different but it is usually because are considered trivial by the context (or by the author).
So yesterday I started reading the mathematical component of the article trying to reconstruct the proof and from what I seemed to understand... is that the implicit hypothesis is stronger than $(a_1,...,a_k)=1$ i.e. it's $a_1=1$.
This idea is supported by those observations:

1. This article is about coins and there is always the currency of unitary value (at least $"a.s."$).
2. All the explicit example start with $a_1=1$
3. pag. 60 the proof is by induction so there are two "base case" because the general expression for $s_i$ does not cover $s_1$ (by the way in the article the sequence $s$ is called $\Delta$), therefore it is necessary to check $m=1$ and $m=2$ before the inductive step $m\geq 3$.
   The case $m=2$ check the truthfulness of:
   $\frac{n}{a_2}\leq D(2,n)\leq \frac{n}{a_2}+1$
   but if it was $a_1\neq 1$ should be:
   $\frac{n}{a_1 a_2}\leq D(2,n)\leq \frac{n}{a_1 a_2}+1$

Maybe I didn't understand the proof or maybe the case $a_1 = 1$ is not restrictive (but I don't see why).

I don't know Swedish and I'm not finding the article "G. Blom and C. E. Froberg, On money changing" translated into English... so I tried to read the original (Swedish) with the help of Google Translate but, from the results, I fear that some semantics could get lost...
The article prove a denumerant's upper and lower bound:
$\frac{n^{k-1}}{(k-1)!\prod_{i=1}^{k}a_i} \leq d(n;a_1,\dots,a_k) \leq \frac{(n+s_k)^{k-1}}{(k-1)!\prod_{i=1}^{k}a_i}$
where
$s_1=1, s_2=a_2$ and $s_k=a_2+\frac{1}{2}(a_3+\dots+a_k), \forall k\geq 3$.
I would like to know: is it required (hypothesis) that all $a_i$ must be coprime i.e. $(a_1,\dots,a_k)=1$?
I first read about this theorem in the book "J. L. R. Alfonsin, The Diophantine Frobenius Problem" (pag. 74) but what I'm asking is not specified.

UPDATE
I would have written in the comments but it was too long.
Despite @Carl-FredrikNybergBrodda's competent reassurance about the translation (i.e. I'm assuming that there are no explicit statements about co-primality) something in the back of my mind was not in peace... unfortunately very often there are implicit hypotheses in math articles (especially dated ones).
The reasons why this happen... well... are different but it is usually because are considered trivial by the context (or by the author).
So yesterday I started reading the mathematical component of the article trying to reconstruct the proof and from what I seemed to understand... is that the implicit hypothesis is stronger than $(a_1,...,a_k)=1$ i.e. it's $a_1=1$.
This idea is supported by those observations:

1. This article is about coins and there is always the currency of unitary value (at least $"a.s."$).
2. All the explicit example start with $a_1=1$
3. pag. 60 the proof is by induction so there are two "base case" because the general expression for $s_i$ does not cover $s_1$ (by the way in the article the sequence $s$ is called $\Delta$), therefore it is necessary to check $m=1$ and $m=2$ before the inductive step $m\geq 3$.
   The case $m=2$ check the truthfulness of:
   $\frac{1}{(2-1)!}\frac{n^{2-1}}{a_2}\leq D(2,n)\leq \frac{1}{(2-1)!}\frac{(n+s_2)^{2-1}}{a_2}$
   but if it was $a_1\neq 1$ should be:
   $\frac{1}{(2-1)!}\frac{n^{2-1}}{a_1 a_2}\leq D(2,n)\leq \frac{1}{(2-1)!}\frac{(n+s_2)^{2-1}}{a_1 a_2}$

Maybe I didn't understand the proof or maybe the case $a_1 = 1$ is not restrictive (but I don't see why).

I don't know Swedish and I'm not finding the article "G. Blom and C. E. Froberg, On money changing" translated into English... so I tried to read the original (Swedish) with the help of Google Translate but, from the results, I fear that some semantics could get lost...
The article prove a denumerant's upper and lower bound:
$\frac{n^{k-1}}{(k-1)!\prod_{i=1}^{k}a_i} \leq d(n;a_1,\dots,a_k) \leq \frac{(n+s_k)^{k-1}}{(k-1)!\prod_{i=1}^{k}a_i}$
where
$s_1=1, s_2=a_2$ and $s_k=a_2+\frac{1}{2}(a_3+\dots+a_k), \forall k\geq 3$.
I would like to know: is it required (hypothesis) that all $a_i$ must be coprime i.e. $(a_1,\dots,a_k)=1$?
I first read about this theorem in the book "J. L. R. Alfonsin, The Diophantine Frobenius Problem" (pag. 74) but what I'm asking is not specified.

I don't know Swedish and I'm not finding the article "G. Blom and C. E. Froberg, On money changing" translated into English... so I tried to read the original (Swedish) with the help of Google Translate but, from the results, I fear that some semantics could get lost...
The article prove a denumerant's upper and lower bound:
$\frac{n^{k-1}}{(k-1)!\prod_{i=1}^{k}a_i} \leq d(n;a_1,\dots,a_k) \leq \frac{(n+s_k)^{k-1}}{(k-1)!\prod_{i=1}^{k}a_i}$
where
$s_1=1, s_2=a_2$ and $s_k=a_2+\frac{1}{2}(a_3+\dots+a_k), \forall k\geq 3$.
I would like to know: is it required (hypothesis) that all $a_i$ must be coprime i.e. $(a_1,\dots,a_k)=1$?
I first read about this theorem in the book "J. L. R. Alfonsin, The Diophantine Frobenius Problem" (pag. 74) but what I'm asking is not specified.

UPDATE
I would have written in the comments but it was too long.
Despite @Carl-FredrikNybergBrodda's competent reassurance about the translation (i.e. I'm assuming that there are no explicit statements about co-primality) something in the back of my mind was not in peace... unfortunately very often there are implicit hypotheses in math articles (especially dated ones).
The reasons why this happen... well... are different but it is usually because are considered trivial by the context (or by the author).
So yesterday I started reading the mathematical component of the article trying to reconstruct the proof and from what I seemed to understand... is that the implicit hypothesis is stronger than $(a_1,...,a_k)=1$ i.e. it's $a_1=1$.
This idea is supported by those observations:

1. This article is about coins and there is always the currency of unitary value (at least $"a.s."$).
2. All the explicit example start with $a_1=1$
3. pag. 60 the proof is by induction so there are two "base case" because the general expression for $s_i$ does not cover $s_1$ (by the way in the article the sequence $s$ is called $\Delta$), therefore it is necessary to check $m=1$ and $m=2$ before the inductive step $m\geq 3$.
   The case $m=2$ check the truthfulness of:
   $\frac{n}{a_2}\leq D(2,n)\leq \frac{n}{a_2}+1$
   but if it was $a_1\neq 1$ should be:
   $\frac{n}{a_1 a_2}\leq D(2,n)\leq \frac{n}{a_1 a_2}+1$

Maybe I didn't understand the proof or maybe the case $a_1 = 1$ is not restrictive (but I don't see why).