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Wizard
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Wizard
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Thanks for this post from:
rsactuarytringlomane
August 15th, 2019 at 12:20:40 PM permalink
To respond to my last post, I noticed that quite often the number of partitions was evenly divisible by 11. Here is a count of how often the total partitions are divisible by 2, 3, 5, 7, and 11, starting with total equal to that divisor and up to as far as Excel would let me.

Divisor Yes No Ratio Expected
2 44 57 43.6% 50.0%
3 37 66 35.9% 33.3%
5 35 72 32.7% 20.0%
7 33 76 30.3% 14.3%
11 52 60 46.4% 9.1%


If the number of partitions were random, you would expect 1 in 11 totals to be divisible by 11, or 9.1%. Instead we have 46.4%.

It's not whether you win or lose; it's whether or not you had a good bet.
TomG
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August 15th, 2019 at 5:41:53 PM permalink
5, 7, 9, 11, 12, 15, 16, 18, 20, 21, 24, 25, 26, 29, 30, 32, 33, 35 -- that's how many integers there are for each number of digits in each partition (if that makes sense). There are five integers with one digit, seven with two, nine with three. The rate of growth is fairly predictable. Good estimates for the number of partitions of n can be done with very little computing power, if someone wants to try,
Wizard
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Wizard
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August 15th, 2019 at 9:05:39 PM permalink
Quote: TomG

5, 7, 9, 11, 12, 15, 16, 18, 20, 21, 24, 25, 26, 29, 30, 32, 33, 35 -- that's how many integers there are for each number of digits in each partition (if that makes sense). There are five integers with one digit, seven with two, nine with three. The rate of growth is fairly predictable. Good estimates for the number of partitions of n can be done with very little computing power, if someone wants to try,



There is a formula, involving pi and e, of course, that is a very good predictor of number of partitions. However, there is nothing like finding exact patterns.
It's not whether you win or lose; it's whether or not you had a good bet.
FleaStiff
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August 15th, 2019 at 10:10:28 PM permalink
the math is, of course, beyond me but this seems to be what dice dealers do. How many chips in what denominations will pay the various bets with sufficient speed and the fewest trips to his stacks to obtain different colors or to deplete the general supply of the most commonly used chips. When there are pumpkins, Barneys, blacks, greens, reds, and whites in play he has to pay different players in sequence with different hands and yet not delay the game.
Wizard
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Wizard
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August 16th, 2019 at 2:31:19 AM permalink
Here are the Ramanujan identities. Let P(x) = number of partitions of x.

P(5x + 4) is always divisible by 5
P(7x + 5) is always divisible by 7
P(11x + 6) is always divisible by 11

Here are a couple more that were discovered later:

P(17303x + 237) is always divisible by 13
P(206839x + 839) is always divisible by 17
P(1977147619x + 815655) is always divisible by 19

Makes you suspect there is another identity for the next prime, 23, but we haven't found it yet.
It's not whether you win or lose; it's whether or not you had a good bet.
gordonm888
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August 16th, 2019 at 7:25:13 AM permalink
I do have a question about terminology.

Consider the 7 partitions of 5:

1-1-1-1-1
2-1-1-1
2-2-1, 3-1-1
4-1, 3-2
5

What is the terminology to describe the number of clumps or elements of a partition? I have been using "length," as in: the partitions listed above start with partitions of length 5, then length 4 and so on down to length 1. But I realize that is probably not the accepted term.
So many better men, a few of them friends, were dead. And a thousand thousand slimy things lived on, and so did I.
ThatDonGuy
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August 16th, 2019 at 7:39:00 AM permalink
Quote: Wizard

* To coin my own term, meaning to have rather few factors, compared to the size of the number. Is there a more technical term for this, and please don't say "relatively prime."


Especially as "relatively prime" refers to something completely different. (Two numbers are relatively prime if they do not share any prime factors, even if neither is prime. For example, 16 and 25 are relatively prime to each other. For that matter, any power of 2 and any odd number are relatively prime.)

More information on numbers of partitions than you probably want to know
RS
RS
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August 16th, 2019 at 7:53:32 AM permalink
If I'm understanding this correctly, your goal (or the "mystery of mathematics") is to find a simple formula for finding the number of partitions of x, without having to know the number of partitions of x-1?

Also, you wrote "partions" instead of "partitions" in your OP linking to wikipedia.


Speaking of numberphile, have you considered making videos like that? The only one I can think of is the video where you made BJ basic strategy in excel.
gordonm888
gordonm888
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August 16th, 2019 at 8:48:37 AM permalink
Quote: Wizard

To respond to my last post, I noticed that quite often the number of partitions was evenly divisible by 11. Here is a count of how often the total partitions are divisible by 2, 3, 5, 7, and 11, starting with total equal to that divisor and up to as far as Excel would let me.

Divisor Yes No Ratio Expected
2 44 57 43.6% 50.0%
3 37 66 35.9% 33.3%
5 35 72 32.7% 20.0%
7 33 76 30.3% 14.3%
11 52 60 46.4% 9.1%


If the number of partitions were random, you would expect 1 in 11 totals to be divisible by 11, or 9.1%. Instead we have 46.4%.



That is interesting. What are the results for a divisor of 13?

I have taken all the primes from 2-1,000,000 and converted them into different bases/radices from 2-100+ and looked for frequency of integers in the digits (ignoring the first and last digit of each prime.) It is extraordinarily regular, with a digit frequency that has very little variance around the expected number. Given that as a background, I am surprised to see your large discrepancy in the frequency of divisor 11. It seems like it might be significant.
So many better men, a few of them friends, were dead. And a thousand thousand slimy things lived on, and so did I.
Wizard
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Wizard
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August 16th, 2019 at 9:30:09 AM permalink
Quote: gordonm888

What is the terminology to describe the number of clumps or elements of a partition? I have been using "length," as in: the partitions listed above start with partitions of length 5, then length 4 and so on down to length 1. But I realize that is probably not the accepted term.



If there's a term for it, I don't know what it is. The way my math works is I keep track of the maximum height in any given partition. For example, the number of partitions in 100 equals:

P(100) = P(1) + P(2) + ... + P(50) (these are for the cases where the height of the longest stack is 50 to 100) + P(51,49) + P(52,48) + ... P (100,1),
where P(x,y) = Number of partitions where x is the total items and y is the maximum height of a stack.

For example if the height of the largest stack is 30, then there are 70 left, but you can't have a stack higher than 30.

Just one way of getting the answer by brute force.
It's not whether you win or lose; it's whether or not you had a good bet.

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