UJRefugees.net

Animal wrote: ↑Tue Apr 19, 2022 2:25 pm

Deathproof wrote: ↑Tue Apr 19, 2022 11:40 am

Animal wrote: ↑Mon Apr 18, 2022 6:43 pm

Deathproof wrote: ↑Mon Apr 18, 2022 5:22 pm

Animal wrote: ↑Mon Apr 18, 2022 4:02 pm

Deathproof wrote: ↑Mon Apr 18, 2022 3:46 am

Good job.

How high would you have to continue this sequence before one of the numbers became a "4"?

Hmmm... good question. Let's try it out. The next number in the sequence would be 13211321322113. Your turn.

1113122113121113222113. your turn.

31131122211311123113322113

132113213221133112132123222113

11131221131211132221232112111312111213322113

Holy fuck. I'm beginning to suspect you'd need to start with a 4 to ever get to a 4. But I'll keep going as long as you do.

Deathproof wrote: ↑Tue Apr 19, 2022 7:20 pm

Animal wrote: ↑Tue Apr 19, 2022 2:25 pm

Deathproof wrote: ↑Tue Apr 19, 2022 11:40 am

Animal wrote: ↑Mon Apr 18, 2022 6:43 pm

Deathproof wrote: ↑Mon Apr 18, 2022 5:22 pm

Animal wrote: ↑Mon Apr 18, 2022 4:02 pm

How high would you have to continue this sequence before one of the numbers became a "4"?

Hmmm... good question. Let's try it out. The next number in the sequence would be 13211321322113. Your turn.

1113122113121113222113. your turn.

31131122211311123113322113

132113213221133112132123222113

11131221131211132221232112111312111213322113

Holy fuck. I'm beginning to suspect you'd need to start with a 4 to ever get to a 4. But I'll keep going as long as you do.

I don't think its possible to get to a 4.

31131122211311123113321112131221123113111231121123222113

Animal wrote: ↑Tue Apr 19, 2022 7:23 pm

Deathproof wrote: ↑Tue Apr 19, 2022 7:20 pm

Animal wrote: ↑Tue Apr 19, 2022 2:25 pm

Deathproof wrote: ↑Tue Apr 19, 2022 11:40 am

Animal wrote: ↑Mon Apr 18, 2022 6:43 pm

Deathproof wrote: ↑Mon Apr 18, 2022 5:22 pm

Hmmm... good question. Let's try it out. The next number in the sequence would be 13211321322113. Your turn.

1113122113121113222113. your turn.

31131122211311123113322113

132113213221133112132123222113

11131221131211132221232112111312111213322113

Holy fuck. I'm beginning to suspect you'd need to start with a 4 to ever get to a 4. But I'll keep going as long as you do.

I don't think its possible to get to a 4.

31131122211311123113321112131221123113111231121123222113

1321132132211331121221231231121311222112132113311213211221121332211213

Are you seeing any patterns forming? I'm not.
Also, can you imagine what some of the more sponge-brained UJRers are thinking when they see this? Like CTC or Who, for instance? Utter confusion.

Deathproof wrote: ↑Wed Apr 20, 2022 12:12 am
1321132132211331121221231231121311222112132113311213211221121332211213

Are you seeing any patterns forming? I'm not.
Also, can you imagine what some of the more sponge-brained UJRers are thinking when they see this? Like CTC or Who, for instance? Utter confusion.

I have not noticed any patterns other than there never seems to be a 4.

1113122113121113222123211211221112131112132112111321322112111312212321121113122122211211232221121113

Animal wrote: ↑Wed Apr 20, 2022 12:19 am

Deathproof wrote: ↑Wed Apr 20, 2022 12:12 am
1321132132211331121221231231121311222112132113311213211221121332211213

Are you seeing any patterns forming? I'm not.
Also, can you imagine what some of the more sponge-brained UJRers are thinking when they see this? Like CTC or Who, for instance? Utter confusion.

I have not noticed any patterns other than there never seems to be a 4.

1113122113121113222123211211221112131112132112111321322112111312212321121113122122211211232221121113

31131122211311123113321112131221122122311211133112111312211231131211132221123113112211121312211231131122113221122112133221123113

Deathproof wrote: ↑Wed Apr 20, 2022 12:27 am

Animal wrote: ↑Wed Apr 20, 2022 12:19 am

Deathproof wrote: ↑Wed Apr 20, 2022 12:12 am
1321132132211331121221231231121311222112132113311213211221121332211213

Are you seeing any patterns forming? I'm not.
Also, can you imagine what some of the more sponge-brained UJRers are thinking when they see this? Like CTC or Who, for instance? Utter confusion.

I have not noticed any patterns other than there never seems to be a 4.

1113122113121113222123211211221112131112132112111321322112111312212321121113122122211211232221121113

31131122211311123113321112131221122122311211133112111312211231131211132221123113112211121312211231131122113221122112133221123113

13211321322113311213212312311211131122212211221321123123211231131122211213211311123113322112132113212231121113112221121321132122211322212221121123222112132113

fuck

Animal wrote: ↑Wed Apr 20, 2022 12:40 am

Deathproof wrote: ↑Wed Apr 20, 2022 12:27 am

Animal wrote: ↑Wed Apr 20, 2022 12:19 am

Deathproof wrote: ↑Wed Apr 20, 2022 12:12 am
1321132132211331121221231231121311222112132113311213211221121332211213

Are you seeing any patterns forming? I'm not.
Also, can you imagine what some of the more sponge-brained UJRers are thinking when they see this? Like CTC or Who, for instance? Utter confusion.

I have not noticed any patterns other than there never seems to be a 4.

1113122113121113222123211211221112131112132112111321322112111312212321121113122122211211232221121113

31131122211311123113321112131221122122311211133112111312211231131211132221123113112211121312211231131122113221122112133221123113

13211321322113311213212312311211131122212211221321123123211231131122211213211311123113322112132113212231121113112221121321132122211322212221121123222112132113

fuck

I'm ready to call it and just assume it'll never hit a 4 if you are.

11131221131213222123211211131211121311121321123113213211222122111312211213111213122112132113213221121113122113311213212322211211131221131211221321123113213221121113122113121132211332113221122112133221121113122113

Deathproof wrote: ↑Wed Apr 20, 2022 12:47 am
I'm ready to call it and just assume it'll never hit a 4 if you are.

11131221131213222123211211131211121311121321123113213211222122111312211213111213122112132113213221121113122113311213212322211211131221131211221321123113213221121113122113121132211332113221122112133221121113122113

sure. now explain to me why there can never be a 4.

Animal wrote: ↑Wed Apr 20, 2022 12:49 am

Deathproof wrote: ↑Wed Apr 20, 2022 12:47 am
I'm ready to call it and just assume it'll never hit a 4 if you are.

11131221131213222123211211131211121311121321123113213211222122111312211213111213122112132113213221121113122113311213212322211211131221131211221321123113213221121113122113121132211332113221122112133221121113122113

sure. now explain to me why there can never be a 4.

I assume because none of the individual digits ever reaches a string of 4. But I doubt that's the answer you're looking for.

I really just don't know.

first, let's work it backward.

we had lots of 111's. if you have 13blah blah blah then the next number is 1113. so that is 3 1's in a row. now how do you get another one in front or behind that? you can't get a 1 in front of it because the only string that ends in a 1 is one that has a 1 in it. which would turn 13blah blah into 113blah blah. which would become 2113. And you can't put another 1 in between the 111 and the 3. So there is no way to get a 4th 1 in a string.

now if you have 2211blah blah. then the next string is 2221. that gets 3 2's in a row. How do you get another 2 in that string? you can't get one in front or you completely change the result. And you can't squeeze one between the 2 and 1. Now, this logic becomes exactly the same as the logic used above. and it also becomes the same logic for why you can't have 4 3's in a row.

so, you have created an endless sequence that can only be made from 1's, 2's and 3's. No other numbers.

Animal wrote: ↑Wed Apr 20, 2022 1:07 am first, let's work it backward.

we had lots of 111's. if you have 13blah blah blah then the next number is 1113. so that is 3 1's in a row. now how do you get another one in front or behind that? you can't get a 1 in front of it because the only string that ends in a 1 is one that has a 1 in it. which would turn 13blah blah into 113blah blah. which would become 2113. And you can't put another 1 in between the 111 and the 3. So there is no way to get a 4th 1 in a string.

now if you have 2211blah blah. then the next string is 2221. that gets 3 2's in a row. How do you get another 2 in that string? you can't get one in front or you completely change the result. And you can't squeeze one between the 2 and 1. Now, this logic becomes exactly the same as the logic used above. and it also becomes the same logic for why you can't have 4 3's in a row.

so, you have created an endless sequence that can only be made from 1's, 2's and 3's. No other numbers.

What if your starting number had a 4 in it? Then you'd never get a 5?

Deathproof wrote: ↑Wed Apr 20, 2022 3:14 am

Animal wrote: ↑Wed Apr 20, 2022 1:07 am first, let's work it backward.

we had lots of 111's. if you have 13blah blah blah then the next number is 1113. so that is 3 1's in a row. now how do you get another one in front or behind that? you can't get a 1 in front of it because the only string that ends in a 1 is one that has a 1 in it. which would turn 13blah blah into 113blah blah. which would become 2113. And you can't put another 1 in between the 111 and the 3. So there is no way to get a 4th 1 in a string.

now if you have 2211blah blah. then the next string is 2221. that gets 3 2's in a row. How do you get another 2 in that string? you can't get one in front or you completely change the result. And you can't squeeze one between the 2 and 1. Now, this logic becomes exactly the same as the logic used above. and it also becomes the same logic for why you can't have 4 3's in a row.

so, you have created an endless sequence that can only be made from 1's, 2's and 3's. No other numbers.

What if your starting number had a 4 in it? Then you'd never get a 5?

yes. you just can't get more then 3 of any number in a sequence. And, if you had a 4, you would never have more than a single 4 in the series. You would never get a 44 or a 444. You would only get 14s.

Animal wrote: ↑Wed Apr 20, 2022 1:34 pm

Deathproof wrote: ↑Wed Apr 20, 2022 3:14 am

Animal wrote: ↑Wed Apr 20, 2022 1:07 am first, let's work it backward.

we had lots of 111's. if you have 13blah blah blah then the next number is 1113. so that is 3 1's in a row. now how do you get another one in front or behind that? you can't get a 1 in front of it because the only string that ends in a 1 is one that has a 1 in it. which would turn 13blah blah into 113blah blah. which would become 2113. And you can't put another 1 in between the 111 and the 3. So there is no way to get a 4th 1 in a string.

now if you have 2211blah blah. then the next string is 2221. that gets 3 2's in a row. How do you get another 2 in that string? you can't get one in front or you completely change the result. And you can't squeeze one between the 2 and 1. Now, this logic becomes exactly the same as the logic used above. and it also becomes the same logic for why you can't have 4 3's in a row.

so, you have created an endless sequence that can only be made from 1's, 2's and 3's. No other numbers.

What if your starting number had a 4 in it? Then you'd never get a 5?

yes. you just can't get more then 3 of any number in a sequence. And, if you had a 4, you would never have more than a single 4 in the series. You would never get a 44 or a 444. You would only get 14s.

Interesting. Have I stumbled onto a mathematical theorem or something?

Deathproof wrote: ↑Wed Apr 20, 2022 2:25 pm

Animal wrote: ↑Wed Apr 20, 2022 1:34 pm

Deathproof wrote: ↑Wed Apr 20, 2022 3:14 am

Animal wrote: ↑Wed Apr 20, 2022 1:07 am first, let's work it backward.

we had lots of 111's. if you have 13blah blah blah then the next number is 1113. so that is 3 1's in a row. now how do you get another one in front or behind that? you can't get a 1 in front of it because the only string that ends in a 1 is one that has a 1 in it. which would turn 13blah blah into 113blah blah. which would become 2113. And you can't put another 1 in between the 111 and the 3. So there is no way to get a 4th 1 in a string.

now if you have 2211blah blah. then the next string is 2221. that gets 3 2's in a row. How do you get another 2 in that string? you can't get one in front or you completely change the result. And you can't squeeze one between the 2 and 1. Now, this logic becomes exactly the same as the logic used above. and it also becomes the same logic for why you can't have 4 3's in a row.

so, you have created an endless sequence that can only be made from 1's, 2's and 3's. No other numbers.

What if your starting number had a 4 in it? Then you'd never get a 5?

yes. you just can't get more then 3 of any number in a sequence. And, if you had a 4, you would never have more than a single 4 in the series. You would never get a 44 or a 444. You would only get 14s.

Interesting. Have I stumbled onto a mathematical theorem or something?

the Deathproof Conjecture

Animal wrote: ↑Thu Apr 21, 2022 1:21 am

Deathproof wrote: ↑Wed Apr 20, 2022 2:25 pm

Animal wrote: ↑Wed Apr 20, 2022 1:34 pm

Deathproof wrote: ↑Wed Apr 20, 2022 3:14 am

Animal wrote: ↑Wed Apr 20, 2022 1:07 am first, let's work it backward.

we had lots of 111's. if you have 13blah blah blah then the next number is 1113. so that is 3 1's in a row. now how do you get another one in front or behind that? you can't get a 1 in front of it because the only string that ends in a 1 is one that has a 1 in it. which would turn 13blah blah into 113blah blah. which would become 2113. And you can't put another 1 in between the 111 and the 3. So there is no way to get a 4th 1 in a string.

now if you have 2211blah blah. then the next string is 2221. that gets 3 2's in a row. How do you get another 2 in that string? you can't get one in front or you completely change the result. And you can't squeeze one between the 2 and 1. Now, this logic becomes exactly the same as the logic used above. and it also becomes the same logic for why you can't have 4 3's in a row.

so, you have created an endless sequence that can only be made from 1's, 2's and 3's. No other numbers.

What if your starting number had a 4 in it? Then you'd never get a 5?

yes. you just can't get more then 3 of any number in a sequence. And, if you had a 4, you would never have more than a single 4 in the series. You would never get a 44 or a 444. You would only get 14s.

Interesting. Have I stumbled onto a mathematical theorem or something?

the Deathproof Conjecture

Fuck yeah. I'm making it official.

Bluespruce1964 wrote: ↑Tue May 10, 2022 10:52 am

2 Number of circles in the numbers

Antknot wrote: ↑Tue May 10, 2022 11:59 am

Bluespruce1964 wrote: ↑Tue May 10, 2022 10:52 am

2 Number of circles in the numbers

ok, it wasn't that interesting.

There is a quantity (11¹⁰⁰⁰ - 1). How many trailing zeros are at the end of this quantity?

QillerDaemon wrote: ↑Sun Oct 09, 2022 4:24 pm There is a quantity (11¹⁰⁰⁰ - 1). How many trailing zeros are at the end of this quantity?

I am going to guess that there is 1 trailing 0.

I'm coming to you guys when my kid needs help in high school math.

1 + 2 = 3
4 + 5 + 6 = 7 + 8
9 + 10 + 11 + 12 = 13 + 14 + 15
...

3² * 4² = 5²
10² * 11² * 12² = 13² * 14²
21² * 22² * 23² * 24² = 25² * 26² * 27²
...

Questions:
1) What is the pattern that produces these series?
2) What is the next line or two of each series?
3) Does this hold for even higher exponents?
4) If not, how can it be made to hold true?

You will have to give Animal some time. It's been awhile since he powered up his calculator.

Well since Animal's calculator apparently needs new C batteries ( ), here's a few hints:

1) the pattern is *not* that the first series' lines are all the positive integers, with the beginning digit of the next line being 1 plus the digit ending the previous line. It's much more subtle than that. The squared series is actually related to the first series.

2) nope, figure it out yourself.

3) obviously not, that's the gist of Fermat's Last Theorem, proved a while back finally.

4) The first two series hold because it's possible to cut up a number on the left side of each equation, then "wrap" the cut parts around the uncut numbers until they come out equal to the right side. That only holds for equations of exponent 1 and 2. The wrapping does not work on higher exponents because it's not possible to cut up any of the numbers on the left side to make equal equations. The cut parts don't add up enough to wrap.

But it is possible to "complete the square" (essentially add some bits and pieces) for higher powers that cause Fermat's Last Theorem to hold for non-integer higher powers, just not with integers. However, there is no pattern that holds for every higher power. The non-integer working power will be either a little bigger or a little smaller than each non-working integer higher exponent. But if there is any rhyme or reason for every power series, it's not at all apparent. (fixed wording a bit)