I am having trouble solving this puzzle... See if you can figure it out...

1
11
21
1211
111221
312211
13112221
1113213211
31131211131221
13211311123113112211
???

What is the next number in the sequence?

11131221133112132113212221



im pretty sure thats it unless i made a stupid mistake...

Well if that answer is correct then you are amazing! I tip my hat to you.

this is how you do it: For example...


If the first number is 12


then for the next number, you basically say what is there... there is one one and one two, therefore the next number is 1112

the next is done the same way... there is three ones and one two... thus 312

get it?

you just keep going like that until you get what i got. Hope you understand

I sort of get it.

I probably could have stared at it for an hour before figuring that out. Good thinking.

Nice One Ryan!!!


Haha i was thinking if it was similar to that binary game that was going around a bit ago...


so, how long does the sequence have to be to get a number '4' in there?

This is one of my favourites:
0, 1, 8, 11, 69, 88, 96, 101, 111, ?

It took me many hours to solve, and that felt great!

Are the next two 181 and 609????

Yep. It's really simple, but it just took me a long time to see the idea. I like that sequence, because the idea is so unique.

Well, by now we know the answer, but what is the method *to tired to think*?

I have one for you:

3 1 4 1 5 ?

A tip is that it is not 1....

Good luck

I now around 300 first digits in that sequence so I saw it immediately

4 1 4 2 1 3 5 ?

This one is also very easy, at least for me... I always memorize useless things.

Oh, come on, quit bragging *lol*
I admire your speed in seeing that one...
(or is it that I admire your abillity to remember as many numbers?)



Well, I have to admit I didn't knew that "by heart", but I figured a couple of sequenses it could be, and found it when looking at the secound one (just a coinsident, huh?).
Would have been realy easy if you had put another number in there , like in the one I posted...

For something like this I admit perl is cool...

perl -e 'for($_="1\n";;s/(\d)\1*/length($&).$1/eg){print}'



Will it ever? Not in the first 50 elements, and their lengths (numbers of digits) are:

1
2
2
4
6
6
8
10
14
20
26
34
46
62
78
102
134
176
226
302
408
528
678
904
1182
1540
2012
2606
3410
4462
5808
7586
9898
12884
16774
21890
28528
37158
48410
63138
82350
107312
139984
182376
237746
310036
403966
526646
686646

Uh... huh? What's that? Is that for working out the first number sequence or something? L8r.

Well, 300 is not good at all . It just tells that sometimes I'm really bored...

Also this one... 6 1 8 0 3 3 9 8 8 7 ?
I only know 20 digits from that sequence though... lol

What, you don't speak perl? Yeah that prints the above sequence. Until it runs out of memory.

wow thanks Stefan...

Yes, the number he wrote down was how many digits were in each sequence (not necessarily the numbers in that sequence).

Actually... it's easy to prove there won't ever be a digit other than 1, 2, 3. Give it a try.

Well, that seems to be correct since as far as we have printed the maximum is 3 and every set of the same numbers generate 2 numbers in the following sequense...

This means that to recieve a digit larger than 3 you would have to have a sequence that spelled like this:
y number of x's, x number of x's and x number of z's
this isn't possible since it would be read:
(y+x) number of x's and x number of z's
this can never result in a sequence of equal numbers larger than 3, and since no digit larger than 3 has been presented in the sequenses yet, x+y can not exeed that...

I hope the above makes sense to you (I get it quite easily (would have been odd in other cases ), but I wonder if I have expressed myself clear enough...)

Hmm, I'm sure it's ok, but I don't quite get it Anyway, here's my proof, I've got a feeling the idea is similar:

The only way to get 4 in the sequence is to have ...b aaaa c... on the previous step. (This would then become ...d 4a e...) But you can't have ...b aaaa c..., it should have been written as ...b (a+a)a c...

I am quite sure we mean the same thing, but you explained it much easier!

Hrm, simpler?:

In order to have a 4, you must either have x 4's, which would mean that the line before it had a 4. (okay, that was the easy half)

Or you have 4 x's, which, for example:

1111

Is not possible, because if that existed, the row before that would be "11", and the row after that would be "21".

Yup, that's it, you just used numbers to illustrate what Haara and I were trying to say