Proposed by Ilija Jovčevski.

steppewolf wrote: Proposed by Ilija Jovčevski. Can you tell me the country so I can put it.

Macedonia

The sequence $\{ a_n \}$ is the sequence of Highly Composite Numbers

$3^{10} \nmid a_n$ for only finitely many $n$. $\tau(a_n) > \tau(m)$ for any $m < a_n$, so $\nu_2(a_n) \ge \nu_3(a_n)$. Then, from some point $a_{n+1} \le \frac{4a_n}{3}$. Otherwise $\tau(\frac{9a_n}{8}) \le \tau(a_n)$ and $\tau(\frac{4a_n}{3}) \le \tau(a_n)$ would imply $\nu_3(a_n) \le 9$.

@above, nice solution! Side note: the first claim is proven in 2012 China TST, for completeness. Edit: Using generalized Bertrand Postulate (at least one prime in $(n, n+\mathcal{O}(n^{6/11+\epsilon})$ ), or Yitang Zhang's result on twin primes, one can prove that $\lim_{n \rightarrow \infty} \frac{a_{n+1}}{a_n}=1$.

Solution spoon-fed to me by p_square The problem asks us to prove that for all large highly composite $n$, there is a highly composite number between $n$ and $\tfrac 32n$. We will use a few well known facts about Highly Composite numbers. If $n=2^{e_1}\cdot 3^{e_2} \cdot \dots$ then $e_1\geq e_2\geq \dots$ For all primes $p$ and a positive integer $k$ there are only finitely many HC numbers $n$ with $\nu_p(n)<k$. Claim : For all primes $p,q$ and a sufficiently large HC number $n$ we have the identity : $$ \lim_{n \rightarrow \infty} \frac {\nu_p(n)}{\nu_q(n)} = \frac { \log q }{ \log p }$$ Proof : Fix primes $p$ and $q$ and $a,b \in \mathbb N$ such that $p^a>q^b$. Now there exists a $N$ such that all HC numbers $>N$, say $n$ is divisible by $p^aq^b$. (by the second fact) Let $x= \nu_p (n)+1$ and $y= \nu_q(n)+1$. We get $$ \tau(n) > \tau ( n\cdot \tfrac {q^b}{p^a}) \implies bx-ay < ab$$This implies $bx-ay$ is bounded above. In other words we have : $$ \lim_{n \rightarrow \infty} \frac {\nu_p(n)}{\nu_q(n)} \leq \frac ab$$ Take $\tfrac ab$ to be arbitrarily close to $\tfrac { \log q }{ \log p }$ and this proves the upper bound. Similarly we can prove the lower bound. So the claim holds. Fix primes $5$ and $7$. We will use the deep number-theoretic fact that $\tfrac 75 > \tfrac 32$. The claim implies that for all large enough HC $n$ we have $\nu_5(n)> \nu_7(n)+1$. A short computation shows that $\tau (\tfrac 75n)> \tau (n)$, so we are done. $\blacksquare$.

Aryan-23 your solution is very beautiful.But one thing bothers me.You really should have put $\varlimsup\limits_{n \to \infty}\frac{\nu_p(n)}{\nu_q(n)} \leq \frac{a}{b}$ because you don't know that the limit exists in advance.Analogously if we take $p^{a'}<q^{b'}$ we obtain $\varliminf\limits_{n \to \infty}\frac{\nu_p(n)}{\nu_q(n)} \geq \frac{a'}{b'}$.From here it follows the existence of your limit which is $\frac{\log p}{\log q}$.

dangerousliri wrote: Let $a_1=2$ and, for every positive integer $n$, let $a_{n+1}$ be the smallest integer strictly greater than $a_n$ that has more positive divisors than $a_n$. Prove that $2a_{n+1}=3a_n$ only for finitely many indicies $n$. Proposed by Ilija Jovčevski, North Macedonia Your Problems are always nice.

satoshinakamoto wrote: $3^{10} \nmid a_n$ for only finitely many $n$. $\tau(a_n) > \tau(m)$ for any $m < a_n$, so $\nu_2(a_n) \ge \nu_3(a_n)$. Then, from some point $a_{n+1} \le \frac{4a_n}{3}$. Otherwise $\tau(\frac{9a_n}{8}) \le \tau(a_n)$ and $\tau(\frac{4a_n}{3}) \le \tau(a_n)$ would imply $\nu_3(a_n) \le 9$. I don't understand your solution. Please, can you write that more clearly?

Notice that the sequence $\{a_n\}$ are just the set of positive integers which has the most number of divisors out of the positive integers less than or equal to it. Therefore, in the prime factorization of $\{a_n\}$, the index of a smaller prime is always larger than the index of a larger prime. Claim. Let $C$ be a constant, for all sufficiently large $i$ we have $v_3(a_i)> C$. Proof. Indeed, suppose for $b_1,...,b_n,...$ we have $v_3(a_{b_i})\leq C$. Notice that $$(a-1)(b+2)-(a+1)(b+1)=a-2-2b$$Therefore, if $v_2(a_{b_i})\geq 2C+2$ then $\frac{3}{4}a_{b_i}$ will have more prime factors than $a_{b_i}$, contradiction. Therefore, the number of distinct prime divisors of $a_{b_i}$ must be unbounded, i.e. for every prime $p$ there exists $i$ such that $p|a_{b_i}$. However, this is clearly impossible. Indeed, notice that $$(a+C+2)b-(a+1)(b+1)=(C+2)b-a-1$$Therefore, if $p>3^{C+1}$ then replacing $a_{b_i}$ by $\frac{3^{C+1}}{p}a_{b_i}$ we got something which has more prime factors than $a_{b_i}$, contradiction. $\blacksquare$ Now suppose there exists infinitely many $n$ such that $2a_{n+1}=3a_n$. Then \begin{align*} d(\frac{4a_n}{3})&\leq d(a_n)\\ d(\frac{9a_n}{8})&\leq d(a_n) \end{align*}Therefore, letting $v_2(a_n)=a$ and $v_2(b_n)=b$, then \begin{align*} (a+3)b&\leq (a+1)(b+1)\\ (a-2)(b+3)&\leq (a+1)(b+1)\\ \end{align*}Therefore, $2b\leq a+1$ and $2a\leq 3b+7$. Hence $$3b+7\geq 2a\geq 2(2b-1)=4b-2$$Hence $b\leq 9$, which contradicts Claim 1.

Let $n$ be big enough and $2a_{n+1}=3a_n$. $v_2(a_n)=x, v_3(a_n)=y$ $\tau (a_n)<\tau (a_{n+1}) \le \tau (\frac{9}{8} a_n) \implies x(y+2) \le (x-2)(y+3) \implies 2y+6\le x$ $\tau (a_n)<\tau (a_{n+1}) \le \tau (\frac{32}{27} a_n) \implies x(y+2) \le (x+6)(y-2) \implies 2x\le 3y-6$ Thus we have $4y+12 \le 3y-6 \implies y+18\le 0$. Contradiction. $\textbf{Remark:}$ We used $n$ being big enough to make $\frac{9}{8} a_n ,\frac{32}{27} a_n$ integers.

dangerousliri wrote: Let $a_1=2$ and, for every positive integer $n$, let $a_{n+1}$ be the smallest integer strictly greater than $a_n$ that has more positive divisors than $a_n$. Prove that $2a_{n+1}=3a_n$ only for finitely many indicies $n$. Proposed by Ilija Jovčevski, North Macedonia FTSOC,assume not. First we prove a couple of Claims about $a_n$(denoted by 'dead' numbers') Claim 1: If $n=\prod_i p_i^{e_i}$ is an dead integer then:- (a) We have $e_1\ge e_2\ge e_3\ge\cdots$. (b) $n$ must be divisible by the first $k$ primes. Proof: Note that permuting the exponents in the prime factorization does not change the value of $d(n)$, so if the $e_i$'s were non increasing, we could permute them to be non increasing, decreasing the value of $n$ while maintaining the value of $d(n)$,which is alive a contradiction to the graveyard. For part(b),Let $n=\prod^m_{j=1} p_j^{\alpha_j}$ be dead. Then $d(a)=d(n)$ where $a=\prod^m_{j=1} q_j^{\alpha_j}$ and $a \le n$. So $a=n$ and the claim is true. $\blacksquare$ Let $p_i$ be the ith prime.Define a process Reviving:Given a dead number $\prod p_i^{a_i}$,we call another dead number totally revived with respect to $p_{M}$ if we shift exponents ${a_i} \rightarrow {a_i}+X$ and $p_i^{X}>p$ Claim 2: For all $ m, s \in \mathbb{Z}^{+} $ there exists an $ N \in \mathbb{Z} $ such that for all good $ n > N $ we have that $ p_m^s \vert n. $ Proof: We'll use induction where the base case is clear by the last claim.Suppose that for $s=n$ it works and we will show that $s=n+1$ also works. Suppose that the nth number is $M=\prod p_i^{a_i}$ and we totally revive it to get another dead number $T_r=\prod p_i^{A_i} $ Now observe that $\max(d(\frac{p_s}{p_1}M),...........,d(\frac{p_s}{p_1 p_2........p_i}M)) \ge d(T_r)$,contradiction so we are done.$\blacksquare$ Claim 3:-There are only finitely many pairs of consecutive highly divisible integers of the form $(a, b)$ with $a\mid b$. Proof:- Choose a large number which is dead. Then $\operatorname{max}\left( d(\tfrac{4}{3}n), d(\tfrac{3}{2}n) \right)>d(n)$ so there is another dead integer $X$ which is $n<X<2n$,so this is proved. Then,from some point $a_{n+1} \le \frac{4a_n}{3}$. Otherwise $\tau(\frac{9a_n}{8}) \le \tau(a_n)$ and $\tau(\frac{4a_n}{3}) \le \tau(a_n)$ would imply $\nu_3(a_n) \le 9$, contradiction.$\blacksquare$