a) Does there exist a sequence $a_1<a_2<\dots$ of positive integers, such that there is a positive integer $N$ that $\forall m>N$, $a_m$ has exactly $d(m)-1$ divisors among $a_i$s? b) Does there exist a sequence $a_1<a_2<\dots$ of positive integers, such that there is a positive integer $N$ that $\forall m>N$, $a_m$ has exactly $d(m)+1$ divisors among $a_i$s?
Problem
Source: Iran TST 3, Day 1, Problem 3
Tags: number theory proposed, number theory
15.05.2013 19:14
I do for a,b similar, maybe i do not true, check for me Let $a_1=p_{1}$ Let $a_{m}=lcm(a_{1},a_{2},..,a_{d(m)+e}).p_{m+2}^{k_{m}}$ Here: e=-1 for a and =1 for b $k_{m}$ is enough large number such that $a_{m}>a_{r}$ with k>r If $a_{k}\mid a_{m}$ for $m> k>d(m)+e$ then $p_{k+2} \mid a_{m}$ that is not true. Then that sequence $a_{m}$ exist.
01.08.2016 11:45
Exactly that
29.04.2017 17:07
a) Since $\gcd(2^m-1,2^n-1)=2^{\gcd(m,n)}-1$, the sequence $a_1=2$ and $a_n=2^n-1$ for $n\geq2$ has this property. b) Yes. Let $a_1,\dots,a_N$ be a sequence of primes, and for $k>N$, let $a_k=p_k\cdot\operatorname{lcm}(a_1,\dots,a_{d(k)})$, where $p_k>a_{k-1}$ is prime. Clearly now $a_1,\dots,a_{d(m)},a_m$ are the only terms dividing $a_m$.
29.09.2017 17:14
what is d(m) ?