Find all real constants c for which there exist strictly increasing sequence $a$ of positive integers such that $(a_{2n-1}+a_{2n})/{a_n}=c$ for all positive intŠµgers n.
Problem
Source: Bosnia Herzegovina TST 2017, problem 3
Tags: algebra
RagvaloD
22.07.2017 12:49
Obvious, that $c$ is rational and $>2$. Let $c=\frac{m}{k}$, $m>2k$ and $(m,k)=1$ $a_1+a_2=ca_1 \to a_2= (c-1)a_1$ $a_3+a_4=ca_2=c(c-1)a_1$ $a_5+a_6+a_7+a_8=c(a_3+a_4)=c^2(c-1)a_1$ ... $a_{2^n+1}+...a_{2^{n+1}}=c(a_{2^{n-1}+1}+...+a_{2^n})=...=c^n (c-1)a_1 = \frac{m^n (m-k)a_1}{k^n}$ So $k^n|a_1$ for every $n$ so $k=1$ and $c$ is integer $\geq 3$
Vietnamisalwaysinmyheart
25.07.2017 17:54
I got all the answer is $c\geq 4$, and is a natural number. Here it goes:
The first thing to show is $c\geq 4$. This is not too hard. Using the old trick:
Consider $(\Delta_n)$ is a positive integer sequence satisfy: $\Delta_n=a_{n+1}-a_n$, for convenient, denote:
$\Delta_0=0$.
Since: $a_{2n-1}+a_{2n}=ca_n$ and: $a_{2n+1}+a_{2n+2}=ca_{n+1}$. Subtract and replace by $\Delta$, we have:
$c\Delta n=\Delta_{2n-1}+2\Delta_{2n}+\Delta_{2n+1}$
Hence, the minium of $\Delta_{2n-1},\Delta_{2n},\Delta_{2n+1}$, must not exceed $\frac{c}{4}\Delta n$, i.e, we can find $i_1$, such that: $\Delta_{i_1}\leq\frac{c}{4}\Delta_n$. Continue this progress, we find that, there exist a sequence: $(i_k)$ such that: $\Delta_{i_k}\leq(\frac{c}{4})^k.\Delta_n$.
Now, if: $c<4$, then the above obviously converges to $0$, which is impossible!
For $c=4$, we can use the sequence: $1,3,5,7,...$ (i.e odd natural numbers sequence)
Consider $c>4$. I will prove that: $c$ must be an integer.
Assume $c$ is not integer, then $c=\frac{p}{q}$ with $q>2$ and: $p,q$ are two relative primes number.
The idea is to use infinite descendant: $\frac{p}{q}\Delta_n=\Delta_{2n-1}+2\Delta_{2n}+\Delta_{2n+1}$, since the right-hand side is natural, we must have: $q|\Delta_n$ forall $n$. Continue, we must have: $q^k|\Delta_n$ forall $k$, which can't be true with $q\geq 2$
Thus, $c$ is an integer greater that $4$. The last part, I will construct a sequence satisfy:
$a_1=1,a_2=c,a_3=\frac{c(c-1)}{2}-1,a_4=\frac{c(c-1)}{2}+1$, easily check that: $a_1<a_2<a_3<a_4$ under the condition.
Now, if $a_n$ has been constructed, there are two cases: $ca_n$ is odd, takes: $a_{2n-1}=\frac{ca_n-1}{2},a_{2n}=\frac{ca_n+1}{2}$, and if $ca_n$ is even, takes: $a_{2n-1}=\frac{ca_n}{2}-1,a_{2n}=\frac{ca_n}{2}+1$
We only need to check the condition that: $(a_n)$ is strictly increasing. I will prove by induction.
Of course: $a_{2n-1}< a_{2n}$ forall $n$, by our construction.
Suppose at some $n$, we have: $a_{2n-1}\leq a_{2n-2}$
Then, since: $a_{2n-1}\geq \frac{ca_n}{2}-1$, $a_{2n-2}\leq\frac{ca_{n-1}}{2}+1$, we have:
$c(a_n-a_{n-1})\geq 4$, but this is of course wrong since: $c\geq 5$, $a_n-a_{n-1}\geq 1$
Thus, we have the answer: All positive integer greater or equal to $4$
jerker
05.10.2018 04:14
RagvaloD wrote: Obvious, that $c$ is rational and $>2$. Let $c=\frac{m}{k}$, $m>2k$ and $(m,k)=1$ $a_1+a_2=ca_1 \to a_2= (c-1)a_1$ $a_3+a_4=ca_2=c(c-1)a_1$ $a_5+a_6+a_7+a_8=c(a_3+a_4)=c^2(c-1)a_1$ ... $a_{2^n+1}+...a_{2^{n+1}}=c(a_{2^{n-1}+1}+...+a_{2^n})=...=c^n (c-1)a_1 = \frac{m^n (m-k)a_1}{k^n}$ So $k^n|a_1$ for every $n$ so $k=1$ and $c$ is integer $\geq 3$ it is just not enough...