## another number sequence

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### another number sequence

Continue this sequence:

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

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Meanwhile, those who know the solution could discuss the semantics of whether it really can be called a number sequence or not.
Last edited by Dashiva on Fri Oct 13, 2006 10:36 pm UTC, edited 2 times in total.
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Dashiva

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### not a number sequence

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

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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.

Torn Apart By Dingos

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

Gelsamel
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*groans*

THIS one!

For the record:

1113213211
31131211131221
13211311123113112211

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

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

>>Never, 3 is the highest.

Gelsamel
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Torn Apart By Dingos wrote: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.
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GreedyAlgorithm

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Gelsamel wrote: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...
Air Gear

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

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### Re:

Gelsamel wrote: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?

tricky77puzzle
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### Re: Re:

tricky77puzzle wrote:But how does one prove that?

What's wrong with a necro? It's a reasonable question. Anyway:
Spoiler:
The 2ith digit is always different from the 2i+2nd 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.
Buttons

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Location: Somerville

### Re: another number sequence

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.

quintopia

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### Re: another number sequence

quintopia wrote: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+1st digit in the number), which isn't quite right.
Buttons

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Location: Somerville

### Re: another number sequence

Buttons wrote:
quintopia wrote: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+1st 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.
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Macbi

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### Re: another number sequence

Buttons wrote:
quintopia wrote: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+1st 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.

quintopia

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### Re: another number sequence

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)
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hassellhoff

Posts: 30
Joined: Mon Jul 28, 2008 8:40 am UTC

### Re: another number sequence

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?
crzftx

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Location: Rockford, IL

### Re: another number sequence

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

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Joined: Thu Jul 03, 2008 3:07 am UTC

### Re: another number sequence

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.
crzftx

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Location: Rockford, IL

### Re: another number sequence

crzftx wrote: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?

mbrownmx

Posts: 24
Joined: Wed May 14, 2008 4:27 am UTC

### Re: another number sequence

mbrownmx wrote:
crzftx wrote: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.
crzftx

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Location: Rockford, IL

### Re: another number sequence

crzftx wrote:
mbrownmx wrote:
crzftx wrote: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.

mbrownmx

Posts: 24
Joined: Wed May 14, 2008 4:27 am UTC

### Re: another number sequence

mbrownmx wrote: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.
Buttons

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Location: Somerville

### Re: another number sequence

mbrownmx wrote:
crzftx wrote:
mbrownmx wrote:
crzftx wrote: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
crzftx

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### Re: another number sequence

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

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### Re: another number sequence

{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]"
ARP

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