On number of ordered pairs with given lcm and an interesting sequence of fractions

For a positive integer $$n,$$ let $$f(n)$$ denote the number of ordered pairs with lcm $$n$$. Since the total number of ordered pairs from the set $$\{1,2,\ldots,n\}$$ is $$n^2$$ it seems interesting to to see how the sequence $$\left\langle \frac{f(n)}{n^2} \right\rangle$$ behaves. I did some simulations and it seems the sequence should converge to zero but I am not analytically sure. Can somebody kindly help me out?

• Since this (long-winded) response is (in my opinion) inappropriate as an answer, I am commenting. Initially, I thought that if the prime factorization of $n = \prod_{i=1}^r \,p_i^{(\alpha_i)},$ then the question would navigate towards considering $n = a\times b,$ with $a$ = the product of some subset of the $r$ primes, with each prime $p_i$ raised to the exponent $\alpha_i.$ ...See next comment. Aug 6, 2020 at 2:58
• If $n = \prod_{i=1}^r \,p_i^{(\alpha_i)}$, the number of ordered pairs with lcm.$n$ can be shown to be $\prod_{i=1}^r \{(2.\alpha_i+1)}$ Aug 6, 2020 at 3:01
• However, I now think that this navigation is oversimplified. With $n = a\times b,$ a given prime $p_i$ can be shared by both $a$ and $b.$ That is, $p_i^{(\beta_i)} | a$ and $p_i^{(\gamma_i)} | b$, with $\alpha_i = \max(\beta_i, \gamma_i).$ This sharing of prime factors throws a monkey wrench into my intuition and prevents me from trying to use my intuition to move forward on the problem. Aug 6, 2020 at 3:04
• Just to be clear, re my reaction to your comment: the problem is too complicated for me, so I am forced to (currently) have no opinion re your comment. By the way, I upvoted; interesting problem. Aug 6, 2020 at 3:07
• $f(n)$ is the number of divisors of $n^2$. It's tabulated at oeis.org/A048691 with links to the literature. The number of divisors of $n^2$ is trivially bounded above by $2n-1$, so the quotient you ask about certainly goes to zero as $n$ goes to infinity. Aug 6, 2020 at 4:01

First, for any prime $$p$$ the ordered pairs are $$(1,p), (p,1), (p,p)$$, so we have $$f(p)=3, \frac{f(p)}{p^2}=\frac {3}{p^2}$$ which goes to $$0$$ for large $$p$$, so $$0$$ is an accumulation point. To show it is the limit, we need to show the highest values also go to $$0$$.
Consider $$f(n)$$ for $$n$$ the power of a prime, $$n=p^k$$. We have $$f(n)=2k+1$$ because one of the elements of the ordered pair must be $$p^k$$ while the other can be any power of $$p$$ from $$0$$ to $$k$$. Now note that your function $$f(n)$$ is multiplicative. If $$n=ab$$ with $$a,b$$ coprime, $$f(n)=f(a)\cdot f(b)$$. If the prime factorization of $$n$$ is $$n=p^aq^br^c$$ we have $$f(n)=(2a+1)(2b+1)(2c+1)$$ because you can distribute the powers of each prime independently. You need $$a,b,c$$ powers of each prime in one of the two factors, then $$0$$ to the maximum in the other.
The $$n$$ that are maxima so far of $$f(n)$$ will be similar to the highly composite numbers but will have more large primes and less factors of $$2$$ because adding a new prime triples the number of pairs instead of only doubling the number of factors. For example $$30$$ is a new maximum of $$f(n)$$ with $$27$$ pairs, while it only has $$8$$ factors which is no more than $$24$$ has.
A rough way to see that $$\frac {f(n)}{n^2}$$ converges to $$0$$ is to consider the primorials, the products of the first $$k$$ primes. These have $$3^k$$ ordered pairs, which is a lot, but they are not the smallest numbers with that many ordered pairs-you want more small factors than that. The primorials are roughly $$k^k$$, so we are asking about $$\frac {3^k}{k^{2k}}$$, which goes to zero. I am sure that the same thing will happen with the maxima of $$f(n)$$ but don't know how to demonstrate it.
• betraying the limit of my intuition: If you deem it worthwhile, please addendum/edit your answer to demonstrate that $f(n)$ is multiplicative. Aug 6, 2020 at 3:15
• Please explain the case of $n = p^{k}$. It's divisors will be $1, p,p^2,...,p^k$ which are total k+1 in number. In that case how can f(n) be 2k + 1? Shouldn't it be (k+1)? Other multiples of $p$ can only have a non trivial gcd with$p^{k}$ but cannot yield the LCM as $p^k$ Aug 6, 2020 at 14:16
• @Laxmi: You can have $(p^k,p^m)$ with $m$ ranging from $0$ to $k$, which is $k+1$ of them. You can also have $(p^m,p^k)$ with $m$ in the same range, making a total of $2k+2$. We have counted $(p^k,p^k)$ twice, so the final answer is $2k+1$ Aug 6, 2020 at 14:25