Partitions

I've been thinking about partions off and on lately. If the term sounds familiar, it was at the center of the movie The Man Who Knew Infinity.

If you didn't see the movie, it is the number of ways you can break down an integer into other integers of lesser or equal size. For example, let's say a blackjack dealer has chips in denominations from $1 to $5. You ask to break down a $5 chip and she asks "how?" How many ways can it be done?

The answer is 7:

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

Granted, getting another $5 chip back isn't breaking it down, but to be consistent, it counts as a partition.

I wrote both a spreadsheet and a program to get at answers for large numbers. Here they are for 1 to 405. At 406, I exceed the size of a 64-bit unsigned integer.



I'll put my code in spoiler tags, if anyone wants to tackle this as a programming exercise.



All that said, I'm not sure what my question is. Has anyone else messed with this? Please don't just plop in a Numberfile video and call it a day, let's try to have a discussion.

One of the mysteries of mathematics is a simple formula to the number of partitions of x, that doesn't just sum partitions of smaller numbers (as I did). As the movie I linked to goes into, Srinivasa Ramanujan found some relationships between a number and its number of partitions.

The question for the poll is how do you feel about partitions?

A similar concept is sometimes used as an interview question for software development roles. "Given standard USA coins, write a function that will take a dollar amount and return the number of different ways you can make that dollar amount in coins." I think I tried to solve it with recursion, but iteration would work too. Your function is also printing out the combinations which wasn't a requirement the one time I was asked this.

Because I hate interviewing, I don't feel good about partitions :P

I've been working on math behind the Grand Jackpot for the Lightning Link / Dragon Link games and I had to research and use partitions in that. Crazy that you mention it right now, because I'm about ready to share my work if anyone is interested.

Not sure how helpful this is, but here is a link to an article on Discovery.com that indicates something is there waiting to be described.

Quote: rsactuary

I'm about ready to share my work if anyone is interested.

I'm interested.

PM me an email address? We might need to get on the phone for me to walk you through it. It's a LOT of math.

Quote: Wizard

I'm interested.

I would also be interested.

I have recently been doing some analytical/theoretical work on partitions, although going in a completely different direction than you have. I'm writing it up in a way suitable for WOV and I'll try to post some of it in a day or two and hope for some feedback.

Quote: tringlomane

I would also be interested.

Let me run it by Wiz and see what he thinks. I stress that it is strictly a theoretical exercise as there are too many unknowns without the PAR sheets to be able to apply it.

I noticed the total number of partitions is often prime or primey* . Just look at the first five total partitions: 2,3,5,7,11.

That caused me to factor the number of partitions to see how primey they were.



* 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."

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%.

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,

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.

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.

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.

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.

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

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.

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.

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.

Quote: gordonm888

What are the results for a divisor of 13?

Damn Excel doesn't handle big numbers very well, so my sample size is rather small:

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%
13	4	109	3.5%	7.7%

So 13's are underrepresented. I think it is indeed significant how often 11 pops up and how often 13 does not.

I think I'll add more code to check the factorization for larger numbers. I can handle numbers up to 2^64 in C++. Projects like this make me jealous of the people who have Mathematica.

Quote:

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.

Can you show me a table to illustrate what you mean?

I took the analysis of how often total partitions are divisible by a prime number up to an initial quantity of 405. Here are the results.

Prime	Count in 2 to 405	Ratio	Expected
2	181	44.8%	50.0%
3	139	34.4%	33.3%
5	139	34.4%	20.0%
7	101	25.0%	14.3%
11	108	26.7%	9.1%
13	25	6.2%	7.7%
17	17	4.2%	5.9%
23	18	4.5%	4.3%
29	11	2.7%	3.4%

This shows a huge surplus of total partitions divisible by 5, 7, and 11, especially 11. You may recall these are the same primes in the Ramanujan identities. Very interesting...

« Primey »

- Semi-prime
- k-almost prime
- Sphenic number

https://en.wikipedia.org/wiki/Table_of_prime_factors?wprov=sfti1

A favor of the Wiz:

Can you run your code for partitions of 10, 11, 12, 13, 14 and 15 and PM or email the lists to me please? I don't have the software to run the code myself.

Quote: Wizard

Can you show me a table to illustrate what you mean?

Sorry, I spent some time looking through my many spreadsheets and I didn't find the work that I had mentioned. I did find some spreadsheets with the primes from 3-19,997 converted into prime number radices (bases) so I used them to produce the (modest) results shown in the table below.

Explanation: This table is the frequency of digits (as shown in the first column) in prime numbers converted into base 2, 5, 7,and 11 - in which I only count the digits as they appear in the "2nd to last digit" and "3rd to last digit" of each prime number.

So in the 2nd row I am counting the frequency of digits 1-4 as they appear in all the primes from 29-19997, when those primes are written in Base 5 -and only counting the digits in the 5^3 and 5^2 columns of the prime numbers.

As an example the prime number 19,917 is written in base 5 as 1114202, so I count one "2" and one "0" as shown in bold in the number.

Digits	Base 3: 11-19997	Base 5: 29-19997	Base 7: 53-19997	Base 11: 127-19997
0	1492	885	661	410
1	1503	917	627	409
2	1517	916	633	394
3		898	634	412
4		885	649	396
5			641	404
6			645	414
7				417
8				423
9				388
10				393

This was just a quick effort to show the kind of work that I had done. The original work was on every prime less than 1,000,000 and the variance in the calculated frequencies of various digits seemed pretty damn small, as I remember it.

Quote: rsactuary

A favor of the Wiz:

Can you run your code for partitions of 10, 11, 12, 13, 14 and 15 and PM or email the lists to me please? I don't have the software to run the code myself.

Does this table not answer your question?

Quote: gordonm888

Sorry, I spent some time looking through my many spreadsheets and I didn't find the work that I had mentioned. I did find some spreadsheets with the primes from 3-19,997 converted into prime number radices (bases) so I used them to produce the (modest) results shown in the table below...

Thank you. I can't think of a good comment offhand.

In this and the following posts I briefly describe some of the research I am doing on partitions

Partitions in a linear sequence
It is well known that if you have 5 objects that you can partition them into these seven arrangements:
1-1-1-1-1; 2-1-1-1; 2-2-1; 3-1-1; 3-2; 4-1; 5

Now let’s consider the 13 ranks in a standard deck of cards, and for the moment let’s define an Ace to be a high card only. (Equivalently, we could define an ace as a low card only with identical results)
Given: All the various combinations of 5 cards of different rank, and ignoring the suits of the cards
Defining: a 5 card straight as 5 consecutive ranks, a 4 card straight as 4 consecutive ranks, etc. ranging down to a 1-card straight which is a card that has a rank with both adjacent ranks empty, let’s look at some combinations of 5 different ranks

Ex: QJ962. Graphically, this hand looks like this: __QJ_9__6___2. It is easy to see that it has one 2- card straight and three 1-card straights. An obvious way to label this hand (from a connect-ness or straightness point of view) is 2-1-1-1

Ex: T7654. Graphically, this hand is ____T__7654__. It is one 4-card straight and one 1-card straight, which we label as 4-1.
Clearly, all the “straight patterns” are equivalent to the partitions of 5, because we are literally filling 5 of 13 ‘slots’ and noting how the 5 objects are partitioned.

Again, in the cards analogy I am defining aces as either high or low, but not both. However, we can define a mathematical relationship that is more general than cards:

Define a linear string(or array) of 13 slots such that each of the 11 interior slots is each connected to two adjacent slots and such that the two end slots are only connected to one adjacent slot each. Now consider every possible combination of ways to populate the 13 slots with 5 objects. There will be c(13,5) =1287 different combinations of ways to populate the 13 slots with 5 objects.

Surprisingly, given this definition, we can calculate the number of combinations that correspond to the various partitions of 5, equivalently we say that we can calculate the frequencies or probability densities of the partitions of 5 when in a linear string (without loops) of 13 spaces.

Case	1-1-1-1-1	2-1-1-1	2-2-1	3-1-1	3-2	4-1	5
5 into 13, line	126	504	252	252	72	72	9

Of course, there is nothing fundamental about having a string length of 13. Here are some combination frequencies for partitions of 5 when the length of the string is 8-13 available slots.

Case	1-1-1-1-1	2-1-1-1	2-2-1	3-1-1	3-2	4-1	5
5 into 13, line	126	504	252	252	72	72	9
5 into 12, line	56	280	168	168	56	56	8
5 into 11, line	21	140	105	105	42	42	7
5 into 10, line	6	62	58	60	30	30	6
5 into 9, line	1	20	30	30	20	20	5
5 into 8, line	0	4	12	12	12	12	4

A couple of observations based on the above table:
1. The partitions 4-1 and 3-2 appear to be equally likely when 5 objects are placed in a linear string.
2. The “5 into 8” case shows an uncanny symmetry in the number of combinations for the partitions. I have yet to see anything that matches it in the various cases I have analyzed.

Partitions in a Closed Linear String
Consider the game Clock Solitaire where all 13 ranks are arrayed a circle. Further, define the existence of a “straight” in this circular configuration to include KA2 as a 3-card straight, QKA23 as a 5-card straight, etc. More generally consider a closed linear string such that the ends of the string are adjacent (as in a loop or circle) and such that every grid spot in the string has two connections; i.e. no ends.

Given this configuration, the partition frequencies are different as is shown in the table below.

Case	1-1-1-1-1	2-1-1-1	2-2-1	3-1-1	3-2	4-1	5
5 into 13, open line	141	497	244	252	72	72	9
5 into 13, closed (loop) line	92	455	272	274	91	90	13

The difference in these two cases has been illustrated by geometric differences in the configuration of the linear array, but the essential differences in the case arise not from geometry but from the definitions of adjacency or connect-ness.

Partitions of 3
The number 3 has the following three partitions: 1-1-1, 2-1, and 3. It’s a very simple set of partitions, but a good starting point for becoming familiar with this kind of analysis For game analysts, remembering that Ace is either high only or low only, this is “You are dealt 3 unpaired cards, how often do you have a 3-card straight, a 2-card straight, or no connected ranks at all (1-1-1).”

Case	1-1-1	2-1	3	Total
3 of 13, line	165	110	11	286
3 of 12, line	120	90	10	220
3 of 11, line	84	72	9	165
3 of 10, line	56	56	8	120
3 of 9, line	35	42	7	84
3 of 8, line	20	30	6	56
3 of 7, line	10	20	5	35
3 of 6, line	4	12	4	20

Now, let’s look at the those same results for partitions of three when the line is a closed loop (no end points)

Case	1-1-1	2-1	3	Total
3 of 13, closed loop	157	116	13	286
3 of 12, closed loop	112	97	12	220
3 of 11, closed loop	77	77	11	165
3 of 10, closed loop	50	60	10	120
3 of 9, closed loop	30	45	9	84
3 of 8, closed loop	16	32	8	56
3 of 7, closed loop	7	21	7	35
3 of 6, closed loop	2	12	6	20

There appears to be a lot of symmetry in this closed loop version of partitions of 3. At this point, I have not analyzed closed loop systems very much because I have become intrigued by some aspects of possible applications of the “open line” partition frequencies.

Partitions of 4.
The number 4 has the following five partitions: 1-1-1-1, 2-1-1, 2-2, 3-1, and 4. Just for chuckles, let’s see the frequencies of these partitions for open linear arrays of length 8-13.

Case	1-1-1-1	2-1-1	2-2	3-1	4	Total
4 of 13, line	210	360	45	90	10	715
4 of 12, line	126	252	36	72	9	495
4 of 11, line	70	168	28	56	8	330
4 of 10, line	35	105	21	42	7	210
4 of 9, line	15	60	15	30	6	126
4 of 8, line	5	30	10	20	5	70

I always like to look for patterns in numbers, look for primes and special numbers. But when I first calculated these partition frequencies/combinations they frustrated me. The only prime numbers are trivial. I learned to dig into them deeper, which I’ll discuss later in the next few posts.

Calculating Partition Frequency

I will explain an efficient way of calculating the frequency of any partition of the number m when chosen from an open linear array of n objects.

Let me define an example based on card games. Example: You are designing a bonus payout for a card game in which a player is dealt 7 cards and in which the Ace counts only as a high or low card and in which you are ignoring flushes. You want to evaluate a bonus payout for the player when his 7-card hand has no pairs and has two straights that are exactly 3-cards long. There are c(13,7) = 1716 possible combinations of 7 ranks, but how often will 7 cards of different ranks be partitioned into a 3-3-1 partition?

In order to implement the methodology I will describe, we must define for any given partition a TOTAL, a LENGTH and the NUMBER OF PERMUTATIONS

For 3-3-1, The total is 3+3+1=7. The length is 3, because this partition has 3 elements or clumps. And the number of permutations is 3, because it can be sequenced as followed: 3-3-1; 3-1-3; and 1-3-3.
Of specific importance to this calculation is the partition 3-3-1 has 3 permutations.

The next step is to examine the partition of the spaces between the 3 substrings that make up 3-3-1. I call this “partitioning the void.” When selecting 7 objects from a total of 13, you can define the 7 objects that were selected equally well by defining the 6 objects that you have not selected. When picking 7 card ranks from 13, we can equally well think of this as defining 6 card ranks that are NOT SELECTED. This is the reason that c(13,7) =c(13,6) and more generally, that


Now, given 7 selected from 13, and given the 7 objects will be partitioned as 3-3-1, we know that there must be 6 objects that were not selected and the 6 unselected objects must have a partition that has a length in the range of 2-4. Clearly, the unselected objects must have at least two ‘clumps ‘ so as to divide the 7 objects into three clumps. Similarly, if the 6 unselected objects were partitioned into 5 or more clumps (2-1-1-1, or 1-1-1-1-1-1) they would have to divide the 7 selected objects into more than 3 clumps.

So given an open linear string of n objects and selecting m objects with a partition of length l, the unselected objects must total (n-m) and be arrayed as a partition of (n-m) with length in the rangel -1 to l+1.

For our example, which involves 7 ranks selected from 13 and arrayed as a partition of 3-3-1, let’s look at the partitions of 6. We’ll order them by their length:



For our example of 7 ranks selected from 13 and arrayed as a partition of 3-3-1, the 6 unselected ranks must be partitioned as either 3-1-1-1; 2-2-1-1, 4-1-1; 3-2-1; 2-2-2, 5-1; 4-2; 3-3 because these are all the possible partitions of 6 that have a length of 2-4.

Now, I define a parameter that will be useful in some applications.
p(n,l) = the sum of the number of permutations for all partitions of n with length l.

Example p(6,3) is the sum of number of permutations for all partitions of 6 with length =3. Referring to the list above we have:
4-1-1 which has 3 permutations
3-2-1 which has 6 permutations
2-2-2 which has 1 permutation

So, p(6,3) = 3+6+1 =10.

Similarly, p(6,6)=1; p(6,5)=5; p(6,4)=10: p(6,2)=5 and p(6,1)=1.

The number of combinations of 7 ranks picked out of 13 that have a partition of 3-3-1 I is the product:




So, when selecting 7 ranks out of 13 (and restricting the ace to be either high or low) the 7 ranks will be partitioned as 3-3-1 in 105 of 1716 possible combinations.

It is straightforward to calculate values of p(n,l) for small values of n; I have found that it is useful to have precalculated tables of this parameter when working certain kinds of problems.

Permutations and Binary Integers
The binary form of integers is an obvious opportunity to investigate partitions in open linear strings. I have just started to consider this, but here are some preliminary examples and observations.

The decimal integer 137 may be written in binary notation as 10001001. Now notice that 10001001 is a string of 8 digits: 3 ones and 5 zeros. The ones are partitioned as 1-1-1 and the 5 zeros are partitioned as 3-2.

The partitions 1-1-1|3-2 do not uniquely connote 197, there are 2 binary numbers with those partitions. We know it is 2 because 1-1-1 has 1 permutation and 3-2 has 2 permutations and 1x2=2.

The set of binary integers with partitions 1-1-1|3-1 is (145, 197).

As another example, the set of binary integers with partitions 2-1-1|2-1 is (75, 77, 83, 89, 101, 105). Here are some other sets, as denoted by combinations of simple partitions.
2-1|1 = (11, 13)
3-1|1 = (23, 29)
4-1|1 = (47, 61)
5-1|1 = (95, 125)
6-1|1 = (191, 253)
2-1|2 = (19, 25)
3-1|3 = (71, 113)
4-1|3 = (143, 241)
2-1-1|1-1 = (43, 45, 53)

Here are some rules for interpreting the partition nomenclature:
1. The first partition is the partition of the 1’s, the second partition refers to the partition of the 0’s.
2. The number of binary integers that have that specific set of partitions for the 1s and 0s will be equal to the product of the permutations of the two partitions.
3. In order to be a valid binary integer, the length of the zeros partition must be equal to or one less than the length of the ones partition.
4. If the length of the partition of zeros is one less than the length of the partition of the 1s, than all the integers in the set will be ODD. If the length of the two partitions are equal, then all the integers in the set will be EVEN.

So given this symbol: 4-3-1-1|2-2-1 it can be immediately deduced that 36 integers will have that configuration of partitions, and they will all be odd numbers in the range 2¹⁴ to 2^15-1.

Now let’s look at partitions when writing binary numbers with some of the leading zeroes. To do this it is necessary to define a number “space” or region. Let’s consider all the integers from 0 to (n¹³-1) and write them all as having 13 digits.


Notice that for 7,621 the ones partition has a length l=4, while the zeroes partition has a length that is 3, or l-1. Here again, we see that if the ones partition has a length n then the length of the ones partition is constrained to be n-1, n, or n+1.

Let’s make a definition that in the space 0 to (2¹³-1) that a number is:
SMALL if it is 0 to (2¹²-1)
LARGE if it is 2¹² to (2¹²-1)

Now if the ones partition of an integer in the space 0 to (2¹³-1) is of length l, then it is easily proven that:
- The integer will be SMALL and ODD if the zeros partition is of length ( l - 1 )
- The integer will be LARGE and EVEN if the zeros partition is of length ( l + 1 )
- The integer will be either SMALL and EVEN or LARGE and ODD if the zeros partition is of length l

Combinations and Factor of Two
Let me write down one particular mathematical identity that came to me as I was working with binary numbers.

_{m=1 to n}Σ c(n,m) = 2ⁿ – 1

where, again, c(n,m) is the classic formula for combinations when selecting m from n.

Example: Take n=13
c(13,13) =1
c(13,1) = c(13,12) = 13
c(13,2) = c(13,11) =78
c(13,3) = c(13,10) =286
c(13,4) = c(13,9) =715
c(13,5) = c(13,8) =1287
c(13,6) = c(13,7) =1716

and:
1+ 2 x (13 + 78 + 286 + 715 + 1287 + 1716) = 8,191 = 2¹³ - 1

This simple formula potentially links combination math to other areas of number theory such as Mersenne Primes.

The identity may also be written as:

_{m=0 to n}Σc(n,m) = 2ⁿ

Where now the summation starts at m=0.

Interesting, huh?

Quote: Wizard

Does this table not answer your question?

No, I'm looking for an actual listing of the partitions, not the number of them.

ie: here's the listing for 7....

1,1,1,1,1,1,1
2,1,1,1,1,1
2,2,1,1,1
3,1,1,1
2,2,2,1
3,2,1,1
4,1,1,1
3,2,2
3,3,1
4,2,1
5,1,1,
4,3
5,2
6,1
7

I'm looking for this list for 10, 11, 12, 13, 14 and 15.

I could manually do it myself, but worried about missing something. I thought that's what your program did, but maybe not?

Quote: rsactuary

No, I'm looking for an actual listing of the partitions, not the number of them.

"I see," said the flea (referring to myself). Yes, I think I can get my program to do that. I've got a lot on my plate today, but hopefully I can provide it for any reasonable number soon.

Quote: rsactuary

No, I'm looking for an actual listing of the partitions, not the number of them.

Thank you!!! yikes it gets big fast!

Quote: Wizard

Exact patterns for 4

Partitions for 4
4
3,1
2,2
2,1,1
1,1,1,1

Exact patterns for 5

Partitions for 5
5
4,1
3,2
3,1,1
2,2,1
2,1,1,1
1,1,1,1,1

Exact patterns for 6

Partitions for 6
6
5,1
4,2
4,1,1
3,3
3,2,1
3,1,1,1
2,2,2
2,2,1,1
2,1,1,1,1
1,1,1,1,1,1

Exact patterns for 7

Partitions for 7
7
6,1
5,2
5,1,1
4,3
4,2,1
4,1,1,1
3,3,1
3,2,2
3,2,1,1
3,1,1,1,1
2,2,2,1
2,2,1,1,1
2,1,1,1,1,1
1,1,1,1,1,1,1

Exact patterns for 8

Partitions for 8
8
7,1
6,2
6,1,1
5,3
5,2,1
5,1,1,1
4,4
4,3,1
4,2,2
4,2,1,1
4,1,1,1,1
3,3,2
3,3,1,1
3,2,2,1
3,2,1,1,1
3,1,1,1,1,1
2,2,2,2
2,2,2,1,1
2,2,1,1,1,1
2,1,1,1,1,1,1
1,1,1,1,1,1,1,1

Exact patterns for 9

Partitions for 9
9
8,1
7,2
7,1,1
6,3
6,2,1
6,1,1,1
5,4
5,3,1
5,2,2
5,2,1,1
5,1,1,1,1
4,4,1
4,3,2
4,3,1,1
4,2,2,1
4,2,1,1,1
4,1,1,1,1,1
3,3,3
3,3,2,1
3,3,1,1,1
3,2,2,2
3,2,2,1,1
3,2,1,1,1,1
3,1,1,1,1,1,1
2,2,2,2,1
2,2,2,1,1,1
2,2,1,1,1,1,1
2,1,1,1,1,1,1,1
1,1,1,1,1,1,1,1,1

Exact patterns for 10

Partitions for 10
10
9,1
8,2
8,1,1
7,3
7,2,1
7,1,1,1
6,4
6,3,1
6,2,2
6,2,1,1
6,1,1,1,1
5,5
5,4,1
5,3,2
5,3,1,1
5,2,2,1
5,2,1,1,1
5,1,1,1,1,1
4,4,2
4,4,1,1
4,3,3
4,3,2,1
4,3,1,1,1
4,2,2,2
4,2,2,1,1
4,2,1,1,1,1
4,1,1,1,1,1,1
3,3,3,1
3,3,2,2
3,3,2,1,1
3,3,1,1,1,1
3,2,2,2,1
3,2,2,1,1,1
3,2,1,1,1,1,1
3,1,1,1,1,1,1,1
2,2,2,2,2
2,2,2,2,1,1
2,2,2,1,1,1,1
2,2,1,1,1,1,1,1
2,1,1,1,1,1,1,1,1
1,1,1,1,1,1,1,1,1,1

Exact patterns for 11

Partitions for 11
11
10,1
9,2
9,1,1
8,3
8,2,1
8,1,1,1
7,4
7,3,1
7,2,2
7,2,1,1
7,1,1,1,1
6,5
6,4,1
6,3,2
6,3,1,1
6,2,2,1
6,2,1,1,1
6,1,1,1,1,1
5,5,1
5,4,2
5,4,1,1
5,3,3
5,3,2,1
5,3,1,1,1
5,2,2,2
5,2,2,1,1
5,2,1,1,1,1
5,1,1,1,1,1,1
4,4,3
4,4,2,1
4,4,1,1,1
4,3,3,1
4,3,2,2
4,3,2,1,1
4,3,1,1,1,1
4,2,2,2,1
4,2,2,1,1,1
4,2,1,1,1,1,1
4,1,1,1,1,1,1,1
3,3,3,2
3,3,3,1,1
3,3,2,2,1
3,3,2,1,1,1
3,3,1,1,1,1,1
3,2,2,2,2
3,2,2,2,1,1
3,2,2,1,1,1,1
3,2,1,1,1,1,1,1
3,1,1,1,1,1,1,1,1
2,2,2,2,2,1
2,2,2,2,1,1,1
2,2,2,1,1,1,1,1
2,2,1,1,1,1,1,1,1
2,1,1,1,1,1,1,1,1,1
1,1,1,1,1,1,1,1,1,1,1

Exact patterns for 12

Partitions for 12
12
11,1
10,2
10,1,1
9,3
9,2,1
9,1,1,1
8,4
8,3,1
8,2,2
8,2,1,1
8,1,1,1,1
7,5
7,4,1
7,3,2
7,3,1,1
7,2,2,1
7,2,1,1,1
7,1,1,1,1,1
6,6
6,5,1
6,4,2
6,4,1,1
6,3,3
6,3,2,1
6,3,1,1,1
6,2,2,2
6,2,2,1,1
6,2,1,1,1,1
6,1,1,1,1,1,1
5,5,2
5,5,1,1
5,4,3
5,4,2,1
5,4,1,1,1
5,3,3,1
5,3,2,2
5,3,2,1,1
5,3,1,1,1,1
5,2,2,2,1
5,2,2,1,1,1
5,2,1,1,1,1,1
5,1,1,1,1,1,1,1
4,4,4
4,4,3,1
4,4,2,2
4,4,2,1,1
4,4,1,1,1,1
4,3,3,2
4,3,3,1,1
4,3,2,2,1
4,3,2,1,1,1
4,3,1,1,1,1,1
4,2,2,2,2
4,2,2,2,1,1
4,2,2,1,1,1,1
4,2,1,1,1,1,1,1
4,1,1,1,1,1,1,1,1
3,3,3,3
3,3,3,2,1
3,3,3,1,1,1
3,3,2,2,2
3,3,2,2,1,1
3,3,2,1,1,1,1
3,3,1,1,1,1,1,1
3,2,2,2,2,1
3,2,2,2,1,1,1
3,2,2,1,1,1,1,1
3,2,1,1,1,1,1,1,1
3,1,1,1,1,1,1,1,1,1
2,2,2,2,2,2
2,2,2,2,2,1,1
2,2,2,2,1,1,1,1
2,2,2,1,1,1,1,1,1
2,2,1,1,1,1,1,1,1,1
2,1,1,1,1,1,1,1,1,1,1
1,1,1,1,1,1,1,1,1,1,1,1

Exact patterns for 13

Partitions for 13
13
12,1
11,2
11,1,1
10,3
10,2,1
10,1,1,1
9,4
9,3,1
9,2,2
9,2,1,1
9,1,1,1,1
8,5
8,4,1
8,3,2
8,3,1,1
8,2,2,1
8,2,1,1,1
8,1,1,1,1,1
7,6
7,5,1
7,4,2
7,4,1,1
7,3,3
7,3,2,1
7,3,1,1,1
7,2,2,2
7,2,2,1,1
7,2,1,1,1,1
7,1,1,1,1,1,1
6,6,1
6,5,2
6,5,1,1
6,4,3
6,4,2,1
6,4,1,1,1
6,3,3,1
6,3,2,2
6,3,2,1,1
6,3,1,1,1,1
6,2,2,2,1
6,2,2,1,1,1
6,2,1,1,1,1,1
6,1,1,1,1,1,1,1
5,5,3
5,5,2,1
5,5,1,1,1
5,4,4
5,4,3,1
5,4,2,2
5,4,2,1,1
5,4,1,1,1,1
5,3,3,2
5,3,3,1,1
5,3,2,2,1
5,3,2,1,1,1
5,3,1,1,1,1,1
5,2,2,2,2
5,2,2,2,1,1
5,2,2,1,1,1,1
5,2,1,1,1,1,1,1
5,1,1,1,1,1,1,1,1
4,4,4,1
4,4,3,2
4,4,3,1,1
4,4,2,2,1
4,4,2,1,1,1
4,4,1,1,1,1,1
4,3,3,3
4,3,3,2,1
4,3,3,1,1,1
4,3,2,2,2
4,3,2,2,1,1
4,3,2,1,1,1,1
4,3,1,1,1,1,1,1
4,2,2,2,2,1
4,2,2,2,1,1,1
4,2,2,1,1,1,1,1
4,2,1,1,1,1,1,1,1
4,1,1,1,1,1,1,1,1,1
3,3,3,3,1
3,3,3,2,2
3,3,3,2,1,1
3,3,3,1,1,1,1
3,3,2,2,2,1
3,3,2,2,1,1,1
3,3,2,1,1,1,1,1
3,3,1,1,1,1,1,1,1
3,2,2,2,2,2
3,2,2,2,2,1,1
3,2,2,2,1,1,1,1
3,2,2,1,1,1,1,1,1
3,2,1,1,1,1,1,1,1,1
3,1,1,1,1,1,1,1,1,1,1
2,2,2,2,2,2,1
2,2,2,2,2,1,1,1
2,2,2,2,1,1,1,1,1
2,2,2,1,1,1,1,1,1,1
2,2,1,1,1,1,1,1,1,1,1
2,1,1,1,1,1,1,1,1,1,1,1
1,1,1,1,1,1,1,1,1,1,1,1,1

Exact patterns for 14

Partitions for 14
14
13,1
12,2
12,1,1
11,3
11,2,1
11,1,1,1
10,4
10,3,1
10,2,2
10,2,1,1
10,1,1,1,1
9,5
9,4,1
9,3,2
9,3,1,1
9,2,2,1
9,2,1,1,1
9,1,1,1,1,1
8,6
8,5,1
8,4,2
8,4,1,1
8,3,3
8,3,2,1
8,3,1,1,1
8,2,2,2
8,2,2,1,1
8,2,1,1,1,1
8,1,1,1,1,1,1
7,7
7,6,1
7,5,2
7,5,1,1
7,4,3
7,4,2,1
7,4,1,1,1
7,3,3,1
7,3,2,2
7,3,2,1,1
7,3,1,1,1,1
7,2,2,2,1
7,2,2,1,1,1
7,2,1,1,1,1,1
7,1,1,1,1,1,1,1
6,6,2
6,6,1,1
6,5,3
6,5,2,1
6,5,1,1,1
6,4,4
6,4,3,1
6,4,2,2
6,4,2,1,1
6,4,1,1,1,1
6,3,3,2
6,3,3,1,1
6,3,2,2,1
6,3,2,1,1,1
6,3,1,1,1,1,1
6,2,2,2,2
6,2,2,2,1,1
6,2,2,1,1,1,1
6,2,1,1,1,1,1,1
6,1,1,1,1,1,1,1,1
5,5,4
5,5,3,1
5,5,2,2
5,5,2,1,1
5,5,1,1,1,1
5,4,4,1
5,4,3,2
5,4,3,1,1
5,4,2,2,1
5,4,2,1,1,1
5,4,1,1,1,1,1
5,3,3,3
5,3,3,2,1
5,3,3,1,1,1
5,3,2,2,2
5,3,2,2,1,1
5,3,2,1,1,1,1
5,3,1,1,1,1,1,1
5,2,2,2,2,1
5,2,2,2,1,1,1
5,2,2,1,1,1,1,1
5,2,1,1,1,1,1,1,1
5,1,1,1,1,1,1,1,1,1
4,4,4,2
4,4,4,1,1
4,4,3,3
4,4,3,2,1
4,4,3,1,1,1
4,4,2,2,2
4,4,2,2,1,1
4,4,2,1,1,1,1
4,4,1,1,1,1,1,1
4,3,3,3,1
4,3,3,2,2
4,3,3,2,1,1
4,3,3,1,1,1,1
4,3,2,2,2,1
4,3,2,2,1,1,1
4,3,2,1,1,1,1,1
4,3,1,1,1,1,1,1,1
4,2,2,2,2,2
4,2,2,2,2,1,1
4,2,2,2,1,1,1,1
4,2,2,1,1,1,1,1,1
4,2,1,1,1,1,1,1,1,1
4,1,1,1,1,1,1,1,1,1,1
3,3,3,3,2
3,3,3,3,1,1
3,3,3,2,2,1
3,3,3,2,1,1,1
3,3,3,1,1,1,1,1
3,3,2,2,2,2
3,3,2,2,2,1,1
3,3,2,2,1,1,1,1
3,3,2,1,1,1,1,1,1
3,3,1,1,1,1,1,1,1,1
3,2,2,2,2,2,1
3,2,2,2,2,1,1,1
3,2,2,2,1,1,1,1,1
3,2,2,1,1,1,1,1,1,1
3,2,1,1,1,1,1,1,1,1,1
3,1,1,1,1,1,1,1,1,1,1,1
2,2,2,2,2,2,2
2,2,2,2,2,2,1,1
2,2,2,2,2,1,1,1,1
2,2,2,2,1,1,1,1,1,1
2,2,2,1,1,1,1,1,1,1,1
2,2,1,1,1,1,1,1,1,1,1,1
2,1,1,1,1,1,1,1,1,1,1,1,1
1,1,1,1,1,1,1,1,1,1,1,1,1,1

Exact patterns for 15

Partitions for 15
15
14,1
13,2
13,1,1
12,3
12,2,1
12,1,1,1
11,4
11,3,1
11,2,2
11,2,1,1
11,1,1,1,1
10,5
10,4,1
10,3,2
10,3,1,1
10,2,2,1
10,2,1,1,1
10,1,1,1,1,1
9,6
9,5,1
9,4,2
9,4,1,1
9,3,3
9,3,2,1
9,3,1,1,1
9,2,2,2
9,2,2,1,1
9,2,1,1,1,1
9,1,1,1,1,1,1
8,7
8,6,1
8,5,2
8,5,1,1
8,4,3
8,4,2,1
8,4,1,1,1
8,3,3,1
8,3,2,2
8,3,2,1,1
8,3,1,1,1,1
8,2,2,2,1
8,2,2,1,1,1
8,2,1,1,1,1,1
8,1,1,1,1,1,1,1
7,7,1
7,6,2
7,6,1,1
7,5,3
7,5,2,1
7,5,1,1,1
7,4,4
7,4,3,1
7,4,2,2
7,4,2,1,1
7,4,1,1,1,1
7,3,3,2
7,3,3,1,1
7,3,2,2,1
7,3,2,1,1,1
7,3,1,1,1,1,1
7,2,2,2,2
7,2,2,2,1,1
7,2,2,1,1,1,1
7,2,1,1,1,1,1,1
7,1,1,1,1,1,1,1,1
6,6,3
6,6,2,1
6,6,1,1,1
6,5,4
6,5,3,1
6,5,2,2
6,5,2,1,1
6,5,1,1,1,1
6,4,4,1
6,4,3,2
6,4,3,1,1
6,4,2,2,1
6,4,2,1,1,1
6,4,1,1,1,1,1
6,3,3,3
6,3,3,2,1
6,3,3,1,1,1
6,3,2,2,2
6,3,2,2,1,1
6,3,2,1,1,1,1
6,3,1,1,1,1,1,1
6,2,2,2,2,1
6,2,2,2,1,1,1
6,2,2,1,1,1,1,1
6,2,1,1,1,1,1,1,1
6,1,1,1,1,1,1,1,1,1
5,5,5
5,5,4,1
5,5,3,2
5,5,3,1,1
5,5,2,2,1
5,5,2,1,1,1
5,5,1,1,1,1,1
5,4,4,2
5,4,4,1,1
5,4,3,3
5,4,3,2,1
5,4,3,1,1,1
5,4,2,2,2
5,4,2,2,1,1
5,4,2,1,1,1,1
5,4,1,1,1,1,1,1
5,3,3,3,1
5,3,3,2,2
5,3,3,2,1,1
5,3,3,1,1,1,1
5,3,2,2,2,1
5,3,2,2,1,1,1
5,3,2,1,1,1,1,1
5,3,1,1,1,1,1,1,1
5,2,2,2,2,2
5,2,2,2,2,1,1
5,2,2,2,1,1,1,1
5,2,2,1,1,1,1,1,1
5,2,1,1,1,1,1,1,1,1
5,1,1,1,1,1,1,1,1,1,1
4,4,4,3
4,4,4,2,1
4,4,4,1,1,1
4,4,3,3,1
4,4,3,2,2
4,4,3,2,1,1
4,4,3,1,1,1,1
4,4,2,2,2,1
4,4,2,2,1,1,1
4,4,2,1,1,1,1,1
4,4,1,1,1,1,1,1,1
4,3,3,3,2
4,3,3,3,1,1
4,3,3,2,2,1
4,3,3,2,1,1,1
4,3,3,1,1,1,1,1
4,3,2,2,2,2
4,3,2,2,2,1,1
4,3,2,2,1,1,1,1
4,3,2,1,1,1,1,1,1
4,3,1,1,1,1,1,1,1,1
4,2,2,2,2,2,1
4,2,2,2,2,1,1,1
4,2,2,2,1,1,1,1,1
4,2,2,1,1,1,1,1,1,1
4,2,1,1,1,1,1,1,1,1,1
4,1,1,1,1,1,1,1,1,1,1,1
3,3,3,3,3
3,3,3,3,2,1
3,3,3,3,1,1,1
3,3,3,2,2,2
3,3,3,2,2,1,1
3,3,3,2,1,1,1,1
3,3,3,1,1,1,1,1,1
3,3,2,2,2,2,1
3,3,2,2,2,1,1,1
3,3,2,2,1,1,1,1,1
3,3,2,1,1,1,1,1,1,1
3,3,1,1,1,1,1,1,1,1,1
3,2,2,2,2,2,2
3,2,2,2,2,2,1,1
3,2,2,2,2,1,1,1,1
3,2,2,2,1,1,1,1,1,1
3,2,2,1,1,1,1,1,1,1,1
3,2,1,1,1,1,1,1,1,1,1,1
3,1,1,1,1,1,1,1,1,1,1,1,1
2,2,2,2,2,2,2,1
2,2,2,2,2,2,1,1,1
2,2,2,2,2,1,1,1,1,1
2,2,2,2,1,1,1,1,1,1,1
2,2,2,1,1,1,1,1,1,1,1,1
2,2,1,1,1,1,1,1,1,1,1,1,1
2,1,1,1,1,1,1,1,1,1,1,1,1,1
1,1,1,1,1,1,1,1,1,1,1,1,1,1,1

Seems like there’s should be an elegant recursive formula out there to delineate partitions.