n=2,$a_2=a_1+1$,then we can't write such number

nmd27082001 wrote: n=2,$a_2=a_1+1$,then we can't write such number Please give example with values.

I was looking for $a_1=3$, $a_2=5$, $a_3=7$ but I couldn't

$a_1=3,a_2=5,a_3=7$ $b_1=3.5,b_2=7,b_3=7$

muuratjann wrote: I was looking for $a_1=3$, $a_2=5$, $a_3=7$ but I couldn't 3,6,9 works. but in the case of a,a+1 we can't find a solution.

aviateurpilot wrote: but in the case of a,a+1 we can't find a solution. $a_1=2,a_2=3$ then $b_1=2,b_2=4$ works $a_1=0.7,a_2=1.7$ then $b_1=0.85,b_2=1.7$ works

This problem is really difficult, any suggestions anyone?

RagvaloD wrote: aviateurpilot wrote: but in the case of a,a+1 we can't find a solution. $a_1=2,a_2=3$ then $b_1=2,b_2=4$ works $a_1=0.7,a_2=1.7$ then $b_1=0.85,b_2=1.7$ works so you mean bi are real numbers?

Yes, $a_i,b_i$ are positive reals. But $\frac{b_i}{b_j}$ is integer if $b_i \geq b_j$

RagvaloD wrote: Yes, $a_i,b_i$ are positive reals. But $\frac{b_i}{b_j}$ is integer if $b_i \geq b_j$ Thanks. you should add this information in the problem.

Let such a sequence of $b_i$ be a good sequence $B$, and consider good sequences $B_1,B_2,\ldots,B_n$ such that in $B_i$ we have $b_i=a_i$ and $b_j=2^k b_i$ where $k$ is the smallest integer such that $b_j\ge a_j$. These sequences are good because the ratio of any two elements is a power of 2. By definition we have in $B_i$ that $b_j=2^{\left\{\log_2\left(\frac{a_i}{a_j}\right)\right\}}a_j$, so the product of the elements of $B_i$ is $2^{\sum_{j=1}^n\left\{\log_2\left(\frac{a_i}{a_j}\right)\right\}}a_1a_2\ldots a_n$. Thus, the product of all the elements in $B_1,B_2,\ldots,B_n$ is \[2^{\sum_{i=1}^n\sum_{j=1}^n\left\{\log_2\left(\frac{a_i}{a_j}\right)\right\}}(a_1a_2\ldots a_n)^n\le 2^{\binom n2}(a_1a_2\ldots a_n)^n\]as we have $\left\{\log_2\left(\frac{a_i}{a_j}\right)\right\}+\left\{\log_2\left(\frac{a_j}{a_i}\right)\right\}=1$ if $\log_2\left(\frac{a_i}{a_j}\right)\not\in\mathbb Z$ and 0 otherwise. This means that at least one of $B_1,B_2,\ldots,B_n$ as a product of at most $\left(2^{\binom n2}(a_1a_2\ldots a_n)^n\right)^\frac 1n=2^{\frac{n-1}2}a_1a_2\ldots a_n$, as desired.

Just to clarify, ${\left\{\log_2\left(\frac{a_i}{a_j}\right)\right\}}$ represents fractional part?

By permuting the indices we may assume $a_1<a_2<...<a_n$. Now for each $1\leq i\leq j\leq n$ let $c_{i,j}$ be the smallest integer such that $$2^{c_{i,j}}a_j>a_i$$Claim. If $i>j$ then $c_{i,j}+c_{j,i}=1$. Proof. Notice that $$2^{c_{i,j}-1}a_j<a_i<a_j$$Hence $$2^{-c_{i,j}+1}a_i>a_j>2^{-c_{i,j}}$$$\blacksquare$ Moreover, $c_{i,i}=0$ obviously. Now notice that for each $1\leq j\leq n$ $$(b^j_1,...,b^j_n)=(2^{c_{1,j}}a_1,2^{c_{2,j}}a_2,...,2^{c_{n,j}}a_n)$$is a choice for the sequence $\{b_n\}$. Meanwhile $$\prod_{1\leq i\leq n}\prod_{1\leq j\leq n}b_j^i=\prod_{1\leq i<j\leq n}2^{c_{i,j}+c_{j,i}}\prod_{1\leq i\leq n}\prod_{1\leq j\leq n}a_j^i=2^{\binom{n}{2}}(a_1a_2...a_n)^n$$Therefore by pigeonhole principle there exists a some $j$ such that $$\prod_{1\leq j\leq n}b_j^i>(2^{\binom{n}{2}}(a_1a_2...a_n)^n)^{\frac{1}{n}}=2^{\frac{n-1}{2}}a_1...a_n$$as desired.