another number sequence

[enfolder]

Continue this sequence:

1, 11, 21, 1211, 111221, 312211, 13112221, ?

[Dashiva]

Meanwhile, those who know the solution could discuss the semantics of whether it really can be called a number sequence or not.

[enfolder]

Care to elaborate, or at least point to a definition of the term?

[Torn Apart By Dingos]

Of course it's a number sequence. But it's of the lateral thinking variety, and I'd argue that those aren't very interesting.

[Gelsamel]

I actually wrote out this sequence for ages a while ago. I think I wrote out like 20 numbers then I got bored.

[Air Gear]

*groans*

THIS one!

For the record:

1113213211
31131211131221
13211311123113112211

So when's the first time we get a 4 in there? A 5?

[Gelsamel]

So when's the first time we get a 4 in there? A 5?


>>Never, 3 is the highest.

[GreedyAlgorithm]

Of course it's a number sequence. But it's of the lateral thinking variety, and I'd argue that those aren't very interesting.

Clicking this link is a *spoiler*: A005150 has been studied quite a bit by Conway, so I'd argue that they're probably interesting.

[Air Gear]

So when's the first time we get a 4 in there? A 5?


>>Never, 3 is the highest.

Aw, I was hoping that it would take a bit longer than that...

[Zeph]

A005151 is also quite neat, since it gets a 4, and repeats after 13 steps.

[tricky77puzzle]

So when's the first time we get a 4 in there? A 5?

>>Never, 3 is the highest.

But how does one prove that?

Oops, necro-alert! Um, bump it back down, please...

[Buttons]

But how does one prove that?

Oops, necro-alert! Um, bump it back down, please...

What's wrong with a necro? It's a reasonable question. Anyway:

Spoiler:

The 2i^th digit is always different from the 2i+2^nd digit. (Did I use the right suffix there? I dunno.) Therefore, the longest possible substring with no two different letters ("monogrammatic substring?") is of length three.

[quintopia]

A more intuitive proof:

Spoiler:

It would be silly to write "1111" because why would you write it twice. The transformation specifies you should write "21" instead.

[Buttons]

A more intuitive proof:

Spoiler:

It would be silly to write "1111" because why would you write it twice. The transformation specifies you should write "21" instead.

Spoiler:

Yeah, but that assumes the first 1 is an odd digit (that is, it's the first, third, fifth, ..., or 2i+1^st digit in the number), which isn't quite right.

[Macbi]

A more intuitive proof:

Spoiler:

It would be silly to write "1111" because why would you write it twice. The transformation specifies you should write "21" instead.

Spoiler:

Yeah, but that assumes the first 1 is an odd digit (that is, it's the first, third, fifth, ..., or 2i+1^st digit in the number), which isn't quite right.

Spoiler:

If the first 1is an even digit then we still have the same problem: two consecutive even digits being 1s.

[quintopia]

A more intuitive proof:

Spoiler:

It would be silly to write "1111" because why would you write it twice. The transformation specifies you should write "21" instead.

Spoiler:

Yeah, but that assumes the first 1 is an odd digit (that is, it's the first, third, fifth, ..., or 2i+1^st digit in the number), which isn't quite right.

Spoiler:

The other case is something like "211113" wherein the rules say you should have written "3113" instead. I was just intuitivizing(?) what you said earlier, and so didn't feel the need to be completely general.

[hassellhoff]

its 4 in the morning, and i have nothing better to do..

i filled out about 4 pages of this in a 6th grade notebook...


1
11
21
1211
111221
312211
12112221
1112213211
312211121221
1211223112112211
11122122122112212221
3122112211222122113211
13112221222132112221131221
1113213211321113122132213112211
311312111312211312311311221113221113212221 ( its about here my eyes start to bleed)
132113111231131122211311121221132122311322311312113211 (now I've got a nose bleed)
111312211331121321132132211331121122211312112213211322132113111221131221 ( cardiac arrest, dont worry, just look at my siggy)
(FUCK THIS)

[crzftx]

What if, somewhere along the line you had 1 two, then 2 twos, then 2 threes. Wouldn't that be written 122223? Making the next line 114213?
Or can someone prove a similar situation is impossible?
Somebody within this thread may already have, and I didn't understand...

[Haragorn]

one two, two twos, two threes would really be three twos, two threes.in other words, 122223 is actually 3223.

[crzftx]

Nice one Haragorn, I don't know why I didn't see the huge flaw.
Here's the first 50 terms of a related series:
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, 894810
Try to find the 51st.

[mbrownmx]

Nice one Haragorn, I don't know why I didn't see the huge flaw.
Here's the first 50 terms of a related series:
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, 894810
Try to find the 51st.

Quick guess on the generating function; too lazy to follow it through to the value here:

Spoiler:

value of ith element of this sequence = length of ith element of the previous sequence?

[crzftx]

Nice one Haragorn, I don't know why I didn't see the huge flaw.
Here's the first 50 terms of a related series:
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, 894810
Try to find the 51st.

Quick guess on the generating function; too lazy to follow it through to the value here:

Spoiler:

value of ith element of this sequence = length of ith element of the previous sequence?

Yeah, you got it. OEIS only has up to the 49th element, so I thought I'd post through 50 and see if anyone could get the 51st.

[mbrownmx]

Nice one Haragorn, I don't know why I didn't see the huge flaw.
Here's the first 50 terms of a related series:
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, 894810
Try to find the 51st.

Quick guess on the generating function; too lazy to follow it through to the value here:

Spoiler:

value of ith element of this sequence = length of ith element of the previous sequence?

Yeah, you got it. OEIS only has up to the 49th element, so I thought I'd post through 50 and see if anyone could get the 51st.

I don't think it'd've been *hard* to do so, just more time consuming that I really have time to spare right now. Maybe if I had the appropriate 50th element, it wouldn't be so bad.

[Buttons]

Quick guess on the generating function; too lazy to follow it through to the value here:

Spoiler:

value of ith element of this sequence = length of ith element of the previous sequence?

See, and here I thought someone actually found the generating function for the sequence. That would have been... surprising.

[crzftx]

Nice one Haragorn, I don't know why I didn't see the huge flaw.
Here's the first 50 terms of a related series:
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, 894810
Try to find the 51st.

Quick guess on the generating function; too lazy to follow it through to the value here:

Spoiler:

value of ith element of this sequence = length of ith element of the previous sequence?

Yeah, you got it. OEIS only has up to the 49th element, so I thought I'd post through 50 and see if anyone could get the 51st.

I don't think it'd've been *hard* to do so, just more time consuming that I really have time to spare right now. Maybe if I had the appropriate 50th element, it wouldn't be so bad.

Hmmmm, well the 50th element is pretty big... but if you really want it: 50.txt

[antonfire]

One of my favorite numbers is Conway's constant. "Not very interesting", my ass.

[ARP]

{1, 11, 21, 1211, 111221, 312211, 13112221,...}
Look and Say sequence
each term describes the previous term
Continuation:
1, 11, 21, 1211, 111221, 312211, 13112221, 1113213211, 31131211131221, 13211311123113112211, 11131221133112132113212221, 3113112221232112111312211312113211, ...
Mathematica:
"FromDigits /@ NestList[Flatten[Reverse /@ (Through[{First, Length}[#1]] &) /@ Split[#]] &, {1}, n-1]"