Math Is Fun Forum

  Discussion about math, puzzles, games and fun.   Useful symbols: ÷ × ½ √ ∞ ≠ ≤ ≥ ≈ ⇒ ± ∈ Δ θ ∴ ∑ ∫ • π ƒ -¹ ² ³ °

You are not logged in.

#1 2007-04-24 22:18:56

litik
Member
Registered: 2007-04-24
Posts: 2

Series 000111222...

Hello everyone!

I have found the general formula for:
0011223344...

n - (n - (n - (3 + (-1)^n)/2)/2)

Can any one help me for:
000111222333...
or
0000111122223333....
and so

Offline

#2 2007-04-24 22:30:10

JaneFairfax
Member
Registered: 2007-02-23
Posts: 6,868

Re: Series 000111222...

0.0011223344…

Er, what happens after 99?

0.00112233445566778899… ???dunno

Or does it stop there (so it’s actually a rational number with a terminating decimal)?

Offline

#3 2007-04-24 22:40:09

mathsyperson
Moderator
Registered: 2005-06-22
Posts: 4,900

Re: Series 000111222...

Wow that's a complicated formula. I think you can simplify it a bit by chopping the front off, if you wanted.
n - (3 + (-1)^n)/2)/2 should work as well.

For the other sequences, I'm not sure. You've made two consecutive terms give the same answer by using (-1)^n, but for more than two that won't work. I'll have a good think about it though, this seems like a fun challenge.

An easy solution would be something like rounddown ((n-1)/3), but presumably you're not allowed to do that kind of thing.


Why did the vector cross the road?
It wanted to be normal.

Offline

#4 2007-04-24 23:21:00

litik
Member
Registered: 2007-04-24
Posts: 2

Re: Series 000111222...

Thanks for your replay.
Jane
I think it should work for numbers:
1 2 3 4 5 6 7 8 ........................... n
0 0 1 1 2 2 3 3.......
Correct me if I'm wrong

mathsyperson:
you are right: n - (3 + (-1)^n)/2)/2 works

The formula I found is working for:
001122334455...... only
it is based on finding if the n is odd or even (-1)^n

if I go for 000111222.... or 000011112222....
then I thing the (-1)^n is not valid

Let see for 000111222333...
1 2 3 4 5 6 7 8 9 10 ........n

n
1->0  3/3-1=0 ?
2->0  3/3-1=0 ?
3->0  3/3-1=0

4->1  6/3-1=1 ?
5->1  6/3-1=1 ?
6->1  6/3-1=1

7->2  9/3-1=2 ?
8->2  9/3-1=2 ?
9->2  9/3-1=2
.
.

So it works for n that complitely divides by 3 (no remaining)
So how do I know (by a formula) that a number n complitely divides by 3
or it is: n<3   for 1 and 2
or it is: n<6   for 4 and 5

In fact I did an algorithm for that problem and it works fine:
get the n
see if it complitely devides by 3 then n/3-1
if not n=n+1
then see again if it complitely devides by 3 then n/3-1
and so on

this works but I wanted something beter
like the formula I found for generating 001122334455...
for 1 2 3 4 5 ......n

I'm not that good in maths but I like chalanges,
Maby that information I gave you will do some help.

Thank you again for your replays

Offline

Board footer

Powered by FluxBB