show that $2^k|n\Longleftrightarrow 2^k|a_{n}$

Question

Let sequence $\{a_{n}\}$ such $a_{0}=0,a_{1}=1,a_{n}=2a_{n-1}+a_{n-2}$. show that $$2^k|n\Longleftrightarrow 2^k|a_{n}$$

I try to find the $\{a_{n}\}$ closed form $$a_{n}=\dfrac{(1+\sqrt{2})^n-(1-\sqrt{2})^n}{2\sqrt{2}},$$Any help would be greatly appreciated

Find a relation between $a_n$ and $a_{2n}$ and use induction. — Daniel Fischer, Mar 07 '16 at 13:58
How to find $a_{n}$ and $a_{2n}$ relation,can you explain detail? — math110, Mar 07 '16 at 14:02
Do you know a relation between $F_n$ and $F_{2n}$ for the Fibonacci numbers? It's very similar here. — Daniel Fischer, Mar 07 '16 at 14:04

xpaul · Answer 1 · 2016-03-08T13:17:40.850

1

Note that \begin{eqnarray} a_n&=&\frac{\sum_{i=0}^n\binom{n}{i}2^{i/2}-\sum_{i=0}^n(-1)^i\binom{n}{i}2^{i/2}}{2^{3/2}}\\ &=&\frac{2\sum_{i=0}^{[n/2]}\binom{n}{2i+1}2^{(2i+1)/2}}{2^{3/2}}\\ &=&\sum_{i=0}^{[n/2]}\binom{n}{2i+1}2^{i}. \end{eqnarray} Since $n|\binom{n}{2i+1}$ for $i=0,1,\cdots [n/2]$, we have if $2^k|n$, then $2^k|a_n$.

On the other hand, first note that $$ a_n=n(1+m) $$ for some natural number $m\ge 1$ and hence if $2^k|a_n$, then $2^k|n$.

edited Mar 08 '16 at 13:17

answered Mar 07 '16 at 14:04

xpaul

44,000

How to prove ：if $2^k|a_{n}$ then we have $2^k|n$ – math110 Mar 08 '16 at 03:17
@geromty, I modified the answer. – xpaul Mar 08 '16 at 13:18

score 1 · Answer 2 · edited Apr 13 '17 at 12:20

1

HINT

Use induction (such as seen here) or your closed form to show

$$a_{2n}=2a_{n}^2+2a_{n}a_{n-1} \Rightarrow 2a_{n}|a_{2n} $$

Thanks to @DanielFischer♦ .

edited Apr 13 '17 at 12:20

Community

1

answered Mar 07 '16 at 14:22

S.C.B.

22,768

can you prove $a_{2n}=2a^2_{n}+2a_{n}a_{n-1}?$ – math110 Mar 08 '16 at 03:13
and How to prove this case: if $2^k|a_{n}$ then $2^k|n$ – math110 Mar 08 '16 at 03:18
@geromty Use your closed Form. For your second question, as that in a seperate question. – S.C.B. Mar 08 '16 at 07:40

show that $2^k|n\Longleftrightarrow 2^k|a_{n}$

2 Answers2