Question

Let
\[ a_{1}=2,\; a_{n+1}=\left\lfloor\frac{3}{2}a_{n}\right\rfloor. \]
be a sequence.
Show that it has infinitely many elements that are even integers and infinitely many elements that are odd integers

Answer 1

Based on the information, \(a_2\) would be 1.5 * 2 i.e 3
Here we have an odd integer.
Thinking the rest :/

Answer 2

Here are the first 10 elements:
{2, 3, 4, 6, 9, 13, 19, 28, 42, 63}

Answer 3

I think we must get (3/2)^n as an even integer :/

Answer 4

Is there any other way to prove it? I'm not good at proving.

Answer 5

Suppose first the number of odd elements is finite and prove that there is a contradiction and then suppose the opposite and see what you can get.

Answer 6

Haha prof. spare me I'm just 12

Answer 7

could this be broken into two cases as follows:

Answer 8

case 1: \(a_n\) is even, say \(a_n=2m\). this implies:\[a_{n+1}=\left\lfloor\frac{3}{2}a_{n}\right\rfloor=\left\lfloor\frac{3}{\cancel{2}}*\cancel{2}m\right\rfloor=\left\lfloor3m\right\rfloor=3m\]which implies \(a_{n+1}\) is odd and a multiple of 3

Answer 9

case 2: \(a_n\) is odd, say \(a_n=2m+1\). this implies\[a_{n+1}=\left\lfloor\frac{3}{2}a_{n}\right\rfloor=\left\lfloor\frac{3}{2}*(2m+1)\right\rfloor=\left\lfloor3m+1+0.5\right\rfloor=3m+1\]which can be odd or even

Answer 10

subcases:
m is odd
m is even

Answer 11

I'm not sure if this proves there are an infinite number of odds/evens

Answer 12

I guess it does, since m=0, 1, 2, ...\(\infty\)

Answer 13

am I right professor elias?

Answer 14

you don't have free choice over what \(a_n\) is so it is not a proof

Answer 15

good point

Answer 16

but we can ascertain that all \(a_n\)'s must be of the form \(3m\) or \(3m+1\) correct?

Answer 17

i don't seem to get well along with numbers,
i think there are two kinds of even, and two kinds of odd ...
2n, where n is even and where n is odd
[2n+1], where ...
if we can obtain some kind of cyclic relation ... i guess we can say it is proof.

Answer 18

@asnaseer yes

Answer 19

Suppose first the number of odd elements is finite and prove that there is a contradiction and then suppose the opposite and see what you can get.

Answer 20

Summary of my findings so far:

Case 1. I assumed \(a_n=2m\) and that led to \(a_{n+1}=3m\implies a_{n+1} \text{ is odd or even depending on m and } a_{n+1}\gt a_n\)

Case 2. I assumed \(a_n=2m+1\) and that led to \(a_{n+1}=3m+1\implies a_{n+1} \text{ is odd or even depending on m and } a_{n+1}\gt a_n\)

In both cases, \(a_{n+1}\gt a_n\) so at least we know the sequence is infinite.

Now lets say the sequence only produces even numbers after a certain point, say n=e. This would imply that after that point all \(a_n\) must be of the form \(a_n=2^x\). This would only remain even after x applications of the function, since each application multiplies this by \(\frac{3}{2}\). Therefore we would start getting odd numbers again after x applications of this sequence generator. Which implies the number of odds cannot be finite.

Similarly, lets sy the sequence only produces odd numbers after a certain point, say n=o. This would imply that after that point all \(a_n\) must be of the form \(a_n=2m+1\) where m MUST be even. Therefore all \(a_n\) must be of the form \(a_n=2^x+1\) . But this immediately produces an even number on the next application of the function.

Therefore the sequence contains an infinite number of odd and even numbers.

Answer 21

@asnaseer, your steps for case 1 and case 2 are almost fine, but needs some tuning:
For case 1, if all \( a_n\) are even on and after n=e.
then
\[
e= 2^u s \text { where s is odd} \\
a_{e+1} = \frac 3 2 e\\
a_{e+2} = \left(\frac 3 2\right)^2 e\\
a_{e+u} = \left(\frac 3 2\right)^u 2^u s = 3^u s\\
\]
which is odd, a contradiction.

Answer 22

Case 2 can be reduced to case 1, by putting \( b_n= a_n -1 \) for \( n \ge o \)
\( b_n\) are all even integer. I f we can show that
\[ b_{n+1} = \frac 3 2 b_n \]
We will be done by the first case.

Answer 23

ah yes! I didn't spot that. that makes more sense now - thanks.

Answer 24

Suppose that
\[
a_n = 2 m +1 \] then\[
m= \frac 1 2(a_n -1)\\
\frac { 3 a_n} 2= \frac {6m +3} 2= 3m + \frac 3 2\\
a_{n+1}=\left \lfloor \frac { 3 a_n} 2 \right\rfloor= 3 m +1=\frac3 2(a_n -1) +1\\
a_{n+1} -1 = \frac3 2(a_n -1)\\
b_{n+1} = \frac 3 2 b_n
\]

Answer 25

neat!

Answer 26

Thanks

Answer 27

I look forward to seing more of your problems - they are a lot of fun to tackle. :)

Answer 28

I am glad you feel that way.

Answer 29

you must have been a great teacher in your time - you certainly are now!