0

Suppose we are given a vector $u=(a_1, a_2, ..., a_n)$ with positive integers $a_j$ and a positive integer $q$. How many vectors $v=(b_1, b_2, ..., b_n)$ with positive integers $b_j$ exist such that $(u , v)=q$? My attempt: consider a function $$l(z)=\prod_{k=1}^{n} \frac{1}{1-z^{a_k}} = (1+z^{1a_1 }+z^{2a_1} +z^{3a_1}+...)(1+z^{1a_2}+z^{2a_2}+z^{3a_2}+...)...(1+z^{1a_n}+z^{2a_n}+z^{3a_n}+...)$$ Then the coefficient of $z^q$ is equal to what we are trying to evaluate and it equals to $[z^q]l(z)=\sum_{a_1b_1+a_2b_2+...+a_nb_n=q} 1$, but I would like to find a closed form of this sum... Is it possible in general? Help me please

pioo
  • 661
  • There isn't a really good closed formula. The general approach is to let $a=\operatorname{lcm}(a_1,\dots,a_n).$ Then you get a closed formula for the generating function: $$l(z)=\frac{p(z)}{(1-z^A)^n}$$ But $p(x)$ is a big polynomial, and there isn't a good general form for it. – Thomas Andrews Apr 23 '25 at 20:16
  • You can also get a linear recurrence, but again, with a lot of terms. Specifically, if your count is $c_q,$ will depend on the it will be linear in terms of the previous $a_1+\cdots+a_n$ terms. – Thomas Andrews Apr 23 '25 at 20:24
  • @ThomasAndrews why are you sure there is no good closed formula? The question is actually how many ways to fit some quantity of $a_1$-s, $a_2$-s, ..., $a_n$-s in the number $q$... I think there will be some closed formula, but maybe using some special divisor functions or maybe Euler's function etc. I think some number theory must be applied to solve this, but I am not sure – pioo Apr 23 '25 at 20:36
  • Just experience. I've seen such problems many times. For specific $a_i$ you can find a closed form, but it can get very monstrous. I know what the resulting closed forms look like for specific small $n$ and specific $a_i,$ and how it expands as $n$ increases. – Thomas Andrews Apr 23 '25 at 20:50
  • You certainly get an asymptote, by writing $l(z)$ as partial fractions and figuring out $\frac{\alpha}{(1-z)^n}.$ You get $\alpha=\frac1{a_1a_2\cdots a_n}$ and get that your count $c_q$ has asymptote $$c_q\sim \frac1{a_1\cdots a_n}\binom{q+n-1}{n-1},$$ at least if the gcd of all the $a_i$ is $1.$ If an asymptote is all you want, that should do. – Thomas Andrews Apr 23 '25 at 20:56
  • 1
    That $\alpha$ can be seen as an estimation the probability that a solution $c_1+\cdots+c_n=q$ has the property that $a_i\mid c_i$ for all $i.$ – Thomas Andrews Apr 23 '25 at 20:59
  • I think this is a duplicate of https://math.stackexchange.com/questions/910809/counting-bounded-integer-solutions-to-sum-ia-ix-i-leqq-n/. – Mike Earnest Apr 23 '25 at 21:36
  • @MikeEarnest no it is not the same unfortunately – pioo Apr 23 '25 at 22:11
  • @ThomasAndrews yes, looks like you are right – pioo Apr 23 '25 at 22:12

0 Answers0