Ryan asked about a variation of the coin problem, which was whether for any coprime natural numbers $x,y$ every sufficiently large natural number is $ax+by$ for some coprime natural numbers $a,b$. Batominovski showed that this conjecture is indeed true and the maximum unrepresentable natural number is $O((xy)^{1+ε})$ for any $ε > 0$. But an obvious lower bound is $Ω(xy)$, leaving a gap. Then I conjectured that it is in fact asymptotically $Θ( xy \ln(xy) )$ as $x,y \to \infty$, and his plot for $xy \in [1..2500]$ seems to support this wild guess.

Can anyone prove or disprove my conjecture?

  • $\begingroup$ With $m(x,y)$ as defined in my post in that link, I would rather say the conjecture should be $$\limsup_{xy\to\infty}\,\frac{m(x,y)}{xy\,\ln(xy)}=1\,,$$ since $$\liminf_{xy\to\infty}\,\frac{m(x,y)}{xy}=1\,.$$ $\endgroup$ – Batominovski Jul 24 '15 at 7:56
  • $\begingroup$ @Batominovski: Wait my question already requires $x,y$ to be coprime. Is the $\liminf$ still $1$? $\endgroup$ – user21820 Jul 24 '15 at 8:06
  • $\begingroup$ That is because $m(x,y)=xy-1$ if $x$ or $y$ is $1$ but not both. $\endgroup$ – Batominovski Jul 24 '15 at 8:07
  • $\begingroup$ @Batominovski: How is the updated question? =D $\endgroup$ – user21820 Jul 24 '15 at 8:09
  • $\begingroup$ I guess that is what we should be after. :) $\endgroup$ – Batominovski Jul 24 '15 at 8:11

With a heuristic argument, I can show that, if $z$ is an integer with $z \gtrsim \zeta(2)\,xy\,\ln(xy)$, then $z$ can be written as $ax+by$ with $a,b\in\mathbb{Z}_{\geq 0}$ such that $\gcd(a,b)=1$. This would mean that $$xy-1\leq m(x,y)\lesssim \zeta(2)\,xy\,\ln(xy)\,,$$ if $m(x,y)$ is the largest integer not in $\big\{ax+by\,\big|\,a,b\in\mathbb{Z}_{\geq 0}\,\text{ and }\gcd(a,b)=1\big\}$. (Here, $\zeta$ is the Riemann zeta-function.)

Note that integer solutions $(a,b)$ to the equation $ax+by=z$ are of the form $$(a,b)=\left(a_0+ky,b_0-kx\right)$$ for some $a_0,b_0\in\mathbb{Z}$, where $k\in\mathbb{Z}$ is arbitrary. Of these, there are approximately $\frac{z}{xy}$ solutions $(a,b)$ with $a,b\geq 0$.

Consider a prime $p\nmid \gcd(xy,z)$. The probability that $p$ divides $a_0+ky$ is $\frac{1}{p}$ and the probability that $p$ divides $b_0-kx$ is $\frac{1}{p}$. Hence, approximately, the probability that $p$ divides both $a_0+ky$ and $b_0-kx$ is $\frac{1}{p^2}$. (Note that the event that $a_0+ky$ is divisible by $p$ and the event that $b_0-kx$ is divisible by $p$ are not quite independent; hence, this argument is not valid, but as I said, this is a heuristic argument.)

Consider a prime $p\mid \gcd(xy,z)$. Then, either $p\mid x$ or $p\mid y$, but not both. Without loss of generality, assume that $p\mid x$. Then, as $p\mid z$, we have $p\mid b_0$ and $p$ always divides $b_0-kx$. The probability that $p$ also divides $a_0+ky$ is $\frac{1}{p}$.

Consequently, among nonnegative integer solutions $(a,b)$ to $ax+by=z$, approximately $$\frac{z}{xy}\,\prod_{\substack{{p\text{ prime}}\\{p\nmid \gcd(xy,z)}}}\,\left(1-\frac{1}{p^2}\right)\,\prod_{\substack{{p\text{ prime}}\\{p\mid \gcd(xy,z)}}}\,\left(1-\frac{1}{p}\right)=\frac{z}{xy}\,\frac{\displaystyle\prod_{p\text{ prime}}\,\left(1-\frac{1}{p^2}\right)}{\displaystyle \prod_{\substack{{p\text{ prime}}\\{p\mid \gcd(xy,z)}}}\,\left(1+\frac{1}{p}\right)}$$ pairs have the property that $\gcd(a,b)=1$. Hence, we need $$\frac{z}{xy}\,\frac{\displaystyle\prod_{p\text{ prime}}\,\left(1-\frac{1}{p^2}\right)}{\displaystyle \prod_{\substack{{p\text{ prime}}\\{p\mid \gcd(xy,z)}}}\,\left(1+\frac{1}{p}\right)}\gtrsim 1\,.$$ However, note that $$\prod_{\substack{{p\text{ prime}}\\{p\mid \gcd(xy,z)}}}\,\left(1+\frac{1}{p}\right)\leq\sum_{j=1}^{\gcd(xy,z)}\,\frac{1}{j}\leq\sum_{j=1}^{xy}\,\frac{1}{j}\approx \ln(xy)$$ and $$\prod_{p\text{ prime}}\,\left(1-\frac{1}{p^2}\right)=\frac{1}{\zeta(2)}\,.$$ Since $z\gtrsim\zeta(2)\,xy\,\ln(xy)$, we obtain $$\frac{z}{xy}\,\frac{\displaystyle\prod_{p\text{ prime}}\,\left(1-\frac{1}{p^2}\right)}{\displaystyle \prod_{\substack{{p\text{ prime}}\\{p\mid \gcd(xy,z)}}}\,\left(1+\frac{1}{p}\right)}\gtrsim \frac{z}{\zeta(2)\,xy\,\ln(xy)}\gtrsim 1\,.$$

Here is a plot to illustrate that this upper bound of $m(x,y)$ works well for most values of $x$ and $y$. In this plot, $x,y\in\{1,2,\ldots,100\}$. The magenta curve is given by $m(x,y)=\zeta(2)\,xy\,\ln(xy)$, the red curve is given by $m(x,y)=xy\,\ln(xy)$, and the cyan line is the lower bound $m(x,y)=xy-1$. The scattered blue plot is the actual $xy$-versus-$m(x,y)$ plot. Hence, I now believe that $\displaystyle\limsup_{xy\to\infty}\,\frac{m(x,y)}{xy\,\ln(xy)}$ is very close to $\zeta(2)$.

enter image description here

For $x,y\in\{1,2,\ldots,100\}$ with $x\leq y$, there are only $18$ pairs with $m(x,y)>\zeta(2)\,xy\,\ln(xy)$. For the sake of convenience, write $f(x,y)$ for $\zeta(2)\,xy\,\ln(xy)$. These pairs $(x,y)$ along with the associated $m(x,y)$ are listed below:
(1) $m(3,7)=110$ with $f(3,7)\approx 105.17$,
(2) $m(4,15)=462$ with $f(4,15)\approx 404.10$,
(3) $m(4,37)=1386$ with $f(4,37)\approx 1216.57$,
(4) $m(4,41)=1386$ with $f(4,41)\approx 1375.79$,
(5) $m(4,63)=2310$ with $f(4,63)\approx 2292.08$,
(6) $m(8,37)=2772$ with $f(8,37)\approx 2770.64$,
(7) $m(8,73)=6270$ with $f(8,73)\approx 6119.19$,
(8) $m(11,23)=2310$ with $f(11,23)\approx 2302.82$,
(9) $m(11,58)=6930$ with $f(11,58)\approx 6777.82$,
(10) $m(14,45)=7410$ with $f(14,45)\approx 6679.75$,
(11) $m(19,20)=3990$ with $f(19,20)\approx 3713.05$,
(12) $m(21,61)=15510$ with $f(21,61)\approx 15077.6$,
(13) $m(22,25)=6090$ with $f(22,25)\approx 5708.67$,
(14) $m(28,45)=14820$ with $f(28,45)\approx 14796.1$,
(15) $m(35,86)=40590$ with $f(35,86)\approx 39658.0$,
(16) $m(41,55)=30030$ with $f(41,55)\approx 28639.4$,
(17) $m(61,89)=87780$ with $f(61,89)\approx 76796.6$, and
(18) $m(67,89)=90090$ with $f(67,89)\approx 85270.6$.
The pair $(4,15)$ is the one most deviated from the $xy$-vs-$f(x,y)$ curve, followed by $(61,89)$ and $(14,45)$, respectively. I originally thought that these extraordinary pairs $(x,y)$ would have many prime divisors, especially small primes, but apparently, the data indicate that the prime divisors of $xy$ are irrelevant.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.