# Thread: Sequence numbers

1. ## Sequence numbers

Hello again (I’m stuck again),

Given the following E sequence: 1,2,1,3,2,4,2,6,3,8,3,12.....

I’ve been trying this for a bit and can’t seem to come up with a recurrence relation that makes sense so I can find the following numbers in the sequence.

the numbers increases then decreases. Thats what is making me stuck.

Help is much appreciated!

2. ## Re: Sequence numbers

Look at the critters this way:
1,2
1,3

2,4
2,6

3,8
3,12
Then pretty obvious that next is:
4,16
4,24

3. ## Re: Sequence numbers

Originally Posted by DenisB
Look at the critters this way:
1,2
1,3

2,4
2,6

3,8
3,12
Then pretty obvious that next is:
4,16
4,24
Wow. That makes so much sense. Thank you! After looking at that I was trying to come up with a recurrence relation for generating the sequence and nothing I came up with made sense. So I’m puzzled.

for every second number I understand it’s multiplied by two, and the first one is and addition of one after two numbers.

4. ## Re: Sequence numbers

@DenisB

Can you elaborate on your sequence. I'm confused.

A sequence is a function. I always thought for each integer in the domain P^+ there can only be assigned one unique value of the sequence hence 1-1 and onto. Your sequence is many-one.

Edit: Actually the sequence <(-1)^n> has proved me wrong about 1-1. But it's not many-one.

5. ## Re: Sequence numbers

It's two sequences running together.

n,<2>,n,<3> for n=1

n,<2n>,n,<3n> for n>1

6. ## Re: Sequence numbers

Originally Posted by PilgrimsPath
@DenisB
Can you elaborate on your sequence. I'm confused.
It's not MY sequence!
I simply showed a way to "look" at the mess!!

7. ## Re: Sequence numbers

So what is your "look" on the problem to formalise this mess?

8. ## Re: Sequence numbers

WhassaMatta you, Pilgrim?!

1,2
1,3

2,4
2,6

3,8
3,12

4,16*
4,24**

2*2 = 4; 4*2 = 8; 8*2 = 16*
3*2 = 6; 6*2 = 12; 12*2 = 24**

10. ## Re: Sequence numbers

Let's call the sequence $a_n$ where $a_0$ is the first number.
$a_{n+4} = \left(\dfrac{3+(-1)^{n+1}}{2}\right)a_n+\dfrac{1+(-1)^n}{2}, a_0=a_2=1, a_1=2, a_3=3$