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Let's say that the linear form $ax+by$ represents $n$ if $ax+by=n$ for some positive integer $x$ and $y$.

Call a pair $(a,b)\in\Bbb N\times\Bbb N$ with $\mathsf{gcd}(a,b)=1$ excellent if linear form $ax+by$ has following property: For each composite $n<(a-1)(b-1)$ (the Frobenius number of $(a,b)$) represented by the linear form there is exactly one collection of divisors starting from some $t_j\geq t_1>a,b$ to $t_s\geq t_j$ at every $i\geq1$ those $t_{i}$ with $t_j|t_{i}$ in $a,b<t_1<\dots<t_s\leq n$ is represented by the linear form and no other divisors are represented.

Do excellent pairs exist at all?

If they do, then is it true that for every sufficiently large integer $l$, there is a excellent pair $(a,b)$ with $a,b\in[l,2l]$?

Note that every excellent pair is a good pair in Problem related to Frobenius coin problem and so excellent pair is a stronger condition.

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    $\begingroup$ Could you try to rephrase this a bit. I find it quite hard to parse with the nested conditionals. $\endgroup$ – user9072 Dec 1 '15 at 20:51
  • $\begingroup$ @quid let me find the best way to put this $\endgroup$ – Brout Dec 1 '15 at 20:52
  • $\begingroup$ @quid there is only one if now $\endgroup$ – Brout Dec 1 '15 at 20:56
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    $\begingroup$ Is the following the same what you mean: "Call a pair $a,b\in\Bbb N$ with $\mathsf{gcd}(a,b)=1$ a reasonable coprime pair if linear form $ax+by$ has the following property: For each composite $n< g(a,b)$ (the frobenius number) represented by the linear form there is exactly one maximal divisor $s|n$ with $s>a,b$ that is represented by the linear form and for this $s$ every $t|s$ with $t>a,b$ is also represented. " $\endgroup$ – user9072 Dec 1 '15 at 21:11
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    $\begingroup$ @quid I missed your comment and have now corrected to your language you have proposed $\endgroup$ – Brout Dec 2 '15 at 9:58
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As I read the question, you want find certain pairs $(a,b)$ to use the linear form $ax+by$ to represent positively only certain numbers, and only in a certain way. In particular, for any $n$ with $a,b<n<g(a,b)=ab-a-b$, it is either not representable as $ax+by$ for integers $x$ and $y$ where both $x$ and $y$ is at least $1$, or $n$ is representable and at most one of its maximal divisors is also representable.

Well, you aren't going to find many such $a,b$. Let's pick $s=a+b$. Then $2s, 3s, 6s$ are also representable, so by your condition $6s \gt ab -a -b$, or $7(a+b)>ab$. So the smaller of $a$ and $b$ is less than $14$, and as $b$ gets large $a$ is bounded above by $7$. And we haven't explored what happens when you pick $\lceil b/a \rceil a$. I predict there will be only finitely many such pairs.

Edit 2015.12.02: I don't see why there is so much difficulty. I shall attempt a very clear explanation.

Suppose $7(a+b)\leq ab$. Then $6(a+b) \leq ab -a -b$. Also $6(a+b), 3(a+b)$, and $2(a+b)$ are representable. Thus $(a,b)$ is not an excellent pair. Thus a pair is excellent implies $7(a+b) \gt ab$, which gives a bound on the smaller of $a$ and $b$, namely if $a$ is smaller then $14 \gt 7(1 + a/b) \gt a$. When $a$ is fixed and not too small $7a/(a-7) \gt b$. The remaining excellent pairs must have the smaller number at most 7. When the smaller number is 2 or 3, one can choose a larger coprime number to get an excellent pair. This happens because the interval of representability is too small for the condition to fail. I leave the cases 4 through 7 to the interested reader. End Edit 2015.12.02.

Gerhard "Leaving The Rest To You" Paseman, 2015.12.01

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  • $\begingroup$ Indeed, for $a\lt b$, what are your plans for $6b$? Gerhard "Maybe This Is Even Simpler" Paseman, 2015.12.01 $\endgroup$ – Gerhard Paseman Dec 1 '15 at 22:16
  • $\begingroup$ Indeed, choices are slim if you have $b \lt 2ka \lt 3ka \lt 6ka \lt ab -a-b$ for some positive integer $k$. Gerhard "Finite Is Looking Really Good" Paseman, 2015.12.01. $\endgroup$ – Gerhard Paseman Dec 1 '15 at 22:24
  • $\begingroup$ Oops. (2,2k+1) works, although somewhat trivially, as does (3,b) for somewhat similar reasons. I think there is little hope for (6,b) though. Gerhard "Done With This For Now" Paseman, 2015.12.01 $\endgroup$ – Gerhard Paseman Dec 1 '15 at 22:32
  • $\begingroup$ I think there are sufficiently many in a partcular $[n,2n]$ if one exists in that particular $[n,2n]$ $\endgroup$ – Brout Dec 1 '15 at 22:40
  • $\begingroup$ $b$ in $(3,b)$ is a number greater than and coprime to $3$, in which case one is looking at representability of numbers in $(b, 2b-3)$ which gives no numbers in that interval with proper divisors in that interval. Looking at slightly larger intervals for larger values of $a$ starts to show the possibilities. So there are infinitely many pairs, but many of them aren't interesting from a divisibility lattice standpoint. Gerhard "Been Staring At Lattices Lately" Paseman, 2015.12.01 $\endgroup$ – Gerhard Paseman Dec 1 '15 at 22:43

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