2

Find $ a(n) $ in function of n, $ n\geq 2 $ where $$ a(2n+1)=(2n+1)a(n+1) $$ $$ a(2n)=2a(n), $$ and $ a(2)=1, $ $ a(3)=1. $

I have showed that $ a(n)=\frac{na(n+1)a(2n)}{a(2n+1)-a(n+1)} $, but it seems it doesn't help much, can anyone help me with this please?

neo
  • 21
  • Note that according to $a(2n+1)=(2n+1)a(n+1)$, we must have $a(3)=3a(2)$ in order to get a reasonable closed form formula – polfosol Sep 17 '16 at 14:13

1 Answers1

0

Hint: Note that if $n$ is odd, then $$\frac{a(n+1)}{a(n)}=\frac{2}{n}$$ and if $n$ is even $$\frac{a(n+1)}{a(n)}=\left(\frac{n+1}{2}\right)\frac{a(n/2+1)}{a(n/2)}$$ now set: $n/2=k$.

Edit: A little more explanation

Set $n=2^{\alpha} m$ where $\alpha\ge 0$ is an integer and $m$ is odd. Hence $\alpha>0$ iff $n$ is even and $\alpha=0,n=m$ iff $n$ is odd. So we can write: $$\frac{a(2^\alpha m+1)}{a(2^\alpha m)}=\left(\frac{2^\alpha m+1}{2}\right)\frac{a(2^{\alpha-1} m+1)}{a(2^{\alpha-1} m)}\tag{*}\label{*}$$ for $\alpha>0$. And: $$\frac{a(n+1)}{a(n)}=\frac{a(m+1)}{a(m)}=\frac{2}{m}$$ for $\alpha=0$. Therefore, you may repeat the $\eqref{*}$ relationship $\alpha$ times to finally get: $$\frac{a(2^\alpha m+1)}{a(2^\alpha m)}=\left(\prod_{i=1}^{\alpha}\frac{2^i m+1}{2}\right)\frac{a(m+1)}{a(m)}=\left(\prod_{i=\color{red}1}^{\alpha}\frac{2^i m+1}{2}\right)\frac{2}{m}$$ Or more generally: $$\frac{a(n+1)}{a(n)}=\left(\prod_{\color{red}{i=0}}^{\alpha}\frac{2^i m+1}{2}\right)\frac{2}{m}\frac{2}{m+1}$$ for all $n$.

polfosol
  • 9,245
  • I'm sorry, but I still haven't found way to solve it with current information, could you explain me your idea? – neo Sep 17 '16 at 11:33
  • from two relations I wrote in a task I have f(2)=1,f(3)=1,f(4)=2,f(5)=5,f(6)=2,f(7)=14,f(8)=4,f(9)=45,f(10)=10,f(11)=22,f(12)=4,f(13)=182,f(14)=28 and so on. I tried to get what you wrote and apply your formula to get ratio of these first few values but unsuccessfully-results don't match. Can you help me? – neo Sep 17 '16 at 13:15
  • @neo: The problem is, the above formula doesn't apply to $a(2),a(3)$. In other words, if $a(3)=3$ then this recursive formula would have been valid for all $a(n)$. With $a(3)=1$ it doesn't look like we would ever be able to get a closed form formula – polfosol Sep 17 '16 at 13:55
  • ok. I'm interested how to use the recursion to get the formula above because I didn't make it yet, so would you explain me step by step? – neo Sep 17 '16 at 14:24
  • Yes, it is. With the assumption that a(3)=3, is it possible to find closed form – neo Sep 17 '16 at 15:41
  • for a(n), not just the ratio? – neo Sep 17 '16 at 15:42
  • @neo: Replace $m$ with $n/2^\alpha$ and simplify the ratio to get: $$a_{n+1}=\frac{2}{n}f(n)a_n$$ where $f(n)=\prod_{i=1}^\alpha (1+n/2^i)$ for even $n$ and $f(n)=1$ for odd $n$. Now you get the idea: $$a_{n+1}=\frac{2^n}{n!}a_1\prod_{j\text{ even}}^n f(j)$$ – polfosol Sep 17 '16 at 17:09