Partitional Pulchritude

Wednesday, 17th Sep, 2025 by Krilling for Company in Arithmetic, Geometry, Integer Sequences, Mathematics and tagged animated gifs, count of partition lengths, partition lengths, partitions, recreational math, recreational mathematics, recreational maths | Leave a comment

If you want a good example of how, in math, something very simple can quickly get very deep, just look at partitions. Here are the partitions of 1 to 5, that is, the ways 1 to 5 can be expressed as a sum of integers smaller than or equal to themselves:

1 = 1

numbpart(1) = 1

2 = 2
1 + 1 = 2

numbpart(2) = 2

3 = 3
1 + 2 = 3
1 + 1 + 1 = 3

numbpart(3) = 3

4 = 4
1 + 3 = 4
2 + 2 = 4
1 + 1 + 2 = 4
1 + 1 + 1 + 1 = 4

numbpart(4) = 5

5 = 5
1 + 4 = 5
2 + 3 = 5
1 + 1 + 3 = 5
1 + 2 + 2 = 5
1 + 1 + 1 + 2 = 5
1 + 1 + 1 + 1 + 1 = 5

numbpart(5) = 7

It’s very easy to understand the concept of partitions, but very difficult to understand how partitions behave. For example, here is numbpart(n), the count of partitions for 1, 2, 3,…

1, 2, 3, 5, 7, 11, 15, 22, 30, 42, 56, 77, 101, 135, 176, 231, 297, 385, 490, 627, 792, 1002, 1255, 1575, 1958, 2436, 3010, 3718, 4565, 5604, 6842, 8349, 10143, 12310, 14883, 17977, 21637, 26015, 31185, 37338, 44583, 53174, 63261, 75175, 89134, 105558, 124754, 147273, 173525, 204226, … A000041 at the Online Encyclopedia of Integer Sequences, “a(n) is the number of partitions of n (the partition numbers)”

What’s the formula for numbpart(n)? That’s a tricky question. And what’s the formula for the curves produced by counting the various lengths of partitions(n)? That’s another tricky question, but one thing is easy to see. As n gets bigger, the graph of countlen(partitions(n)) acquires a strange, lopsided beauty. Here are the partitions of 8, with the count of how many partitions of a particular length there are:

8 = 8 (1 partition of length 1)
1 + 7 = 8
2 + 6 = 8
3 + 5 = 8
4 + 4 = 8 (4 partitions of length 2)
1 + 1 + 6 = 8
1 + 2 + 5 = 8
1 + 3 + 4 = 8
2 + 2 + 4 = 8
2 + 3 + 3 = 8 (5 of length 3)
1 + 1 + 1 + 5 = 8
1 + 1 + 2 + 4 = 8
1 + 1 + 3 + 3 = 8
1 + 2 + 2 + 3 = 8
2 + 2 + 2 + 2 = 8 (5 of length 4)
1 + 1 + 1 + 1 + 4 = 8
1 + 1 + 1 + 2 + 3 = 8
1 + 1 + 2 + 2 + 2 = 8 (3 of length 5)
1 + 1 + 1 + 1 + 1 + 3 = 8
1 + 1 + 1 + 1 + 2 + 2 = 8 (2 of length 6)
1 + 1 + 1 + 1 + 1 + 1 + 2 = 8 (1 of length 7)
1 + 1 + 1 + 1 + 1 + 1 + 1 + 1 = 8 (1 of length 8)

When counts like that are shown as a graph, the graphs look like this (maximum counts are normalized to the same height):

graph of countlen(partitions(2))

countlen(partitions(3))

countlen(partitions(4))

countlen(partitions(5))

countlen(partitions(6))

countlen(partitions(7))

countlen(partitions(8))

countlen(partitions(9))

countlen(partitions(10))

countlen(partitions(15))

countlen(partitions(20))

countlen(partitions(30))

countlen(partitions(40))

countlen(partitions(50))

countlen(partitions(60))

countlen(partitions(70))

countlen(partitions(80))

countlen(partitions(90))

countlen(partitions(100))

Animated gif of partlen graphs (courtesy EZgif)

The graphs have a long, low right tail because the counts rise to great heights very quick, then fall away again, as you can see with partitions(100):

1 = count(partitions(10),len=1)
50 = count(partitions(10),len=2)
833 = count(partitions(10),len=3)
7153 = count(partitions(10),len=4)
38225 = count(partitions(10),len=5)
143247 = count(partitions(10),len=6)

[…]

10643083 = count(partitions(10),len=16)
11022546 = count(partitions(10),len=17)
11087828 = count(partitions(10),len=18)
10885999 = count(partitions(10),len=19)
10474462 = count(partitions(10),len=20)

[…]

30 = count(partitions(10),len=91)
22 = count(partitions(10),len=92)
15 = count(partitions(10),len=93)
11 = count(partitions(10),len=94)
7 = count(partitions(10),len=95)
5 = count(partitions(10),len=96)
3 = count(partitions(10),len=97)
2 = count(partitions(10),len=98)
1 = count(partitions(10),len=99)
1 = count(partitions(10),len=100)

Graham’s Sumber

Wednesday, 8th Jan, 2025 by Krilling for Company in Mathematics, Reciprocals and tagged 77, 78, Journal of the Australian Mathematical Society, partitions, Penguin Dictionary of Curious and Interesting Mathematics, R.L. Graham, reciprocals, Ronald Graham, sum to 1 | Leave a comment

Every number greater than 77 is the sum of integers, the sum of whose reciprocals is 1.

For example, 78 = 2 + 6 + 8 + 10 + 12 + 40 and 1/2 + 1/6 + 1/8 + 1/10 + 1/12 + 1/40 = 1.

R.L. Graham, “A Theorem on Partitions”, Journal of the Australian Mathematical Society, 1963; quoted in Le Lionnais, 1983.

• From David Wells’ Penguin Dictionary of Curious and Interesting Mathematics (1986)

Post-Performative Post-Scriptum…

The title of this post is a pun on the gargantuan Graham’s number, described by the same American mathematician and famous among math-fans for its mindboggling size. “Le Lionnais, 1983” must refer to a book called Les Nombres remarquables by the French mathematician François Le Lionnais (1901-84).

Sliv and Let Tri

Tuesday, 13th Apr, 2021 by Krilling for Company in Integer Sequences, Mathematics and tagged Arithmetic, elephant, Gregory the Great, lamb, Latin quotations, math, mathematics, maths, partitions, partitions of n, Pope Gregory I, prime numbers, primes, products, recreational math, recreational mathematics, recreational maths, river, St Gregory, sums | Leave a comment

“Fluvius, planus et altus, in quo et agnus ambulet et elephas natet,” wrote Pope Gregory the Great (540-604). “There’s a river, wide and deep, where a lamb may wade and an elephant swim.” He was talking about the Word of God, but you can easily apply his words to mathematics. However, in the river of mathematics, the very shallow and the very deep are often a single step apart.

Here’s a good example. Take the integer 2. How many different ways can it be represented as an sum of separate integers? Easy. First of all it can be represented as itself: 2 = 2. Next, it can be represented as 2 = 1 + 1. And that’s it. There are two partitions of 2, as mathematicians say:

2 = 2 = 1+1 (p=2)

Now try 3, 4, 5, 6:

3 = 3 = 1+2 = 1+1+1 (p=3) 4 = 4 = 1+3 = 2+2 = 1+1+2 = 1+1+1+1 (p=5) 5 = 5 = 1+4 = 2+3 = 1+1+3 = 1+2+2 = 1+1+1+2 = 1+1+1+1+1 (p=7) 6 = 6 = 1+5 = 2+4 = 3+3 = 1+1+4 = 1+2+3 = 2+2+2 = 1+1+1+3 = 1+1+2+2 = 1+1+1+1+2 = 1+1+1+1+1+1 (p=11)

So the partitions of 2, 3, 4, 5, 6 are 2, 3, 5, 7, 11. That’s interesting — the partition-counts are the prime numbers in sequence. So you might conjecture that p(7) = 13 and p(8) = 17. Alas, you’d be wrong. Here are the partitions of n = 1..10:

1 = 1 (p=1) 2 = 2 = 1+1 (p=2) 3 = 3 = 1+2 = 1+1+1 (p=3) 4 = 4 = 1+3 = 2+2 = 1+1+2 = 1+1+1+1 (p=5) 5 = 5 = 1+4 = 2+3 = 1+1+3 = 1+2+2 = 1+1+1+2 = 1+1+1+1+1 (p=7) 6 = 6 = 1+5 = 2+4 = 3+3 = 1+1+4 = 1+2+3 = 2+2+2 = 1+1+1+3 = 1+1+2+2 = 1+1+1+1+2 = 1+1+1+1+1+1 (p=11) 7 = 7 = 1+6 = 2+5 = 3+4 = 1+1+5 = 1+2+4 = 1+3+3 = 2+2+3 = 1+1+1+4 = 1+1+2+3 = 1+2+2+2 = 1+1+1+1+3 = 1+1+1+2+2 = 1+1+1+1+1+2 = 1+1+1+1+1+1+1 (p=15) 8 = 8 = 1+7 = 2+6 = 3+5 = 4+4 = 1+1+6 = 1+2+5 = 1+3+4 = 2+2+4 = 2+3+3 = 1+1+1+5 = 1+1+2+4 = 1+1+3+3 = 1+2+2+3 = 2+2+2+2 = 1+1+1+1+4 = 1+1+1+2+3 = 1+1+2+2+2 = 1+1+1+1+1+3 = 1+1+1+1+2+2 = 1+1+1+1+1+1+2 = 1+1+1+1+1+1+1+1 (p=22) 9 = 9 = 1+8 = 2+7 = 3+6 = 4+5 = 1+1+7 = 1+2+6 = 1+3+5 = 1+4+4 = 2+2+5 = 2+3+4 = 3+3+3 = 1+1+1+6 = 1+1+2+5 = 1+1+3+4 = 1+2+2+4 = 1+2+3+3 = 2+2+2+3 = 1+1+1+1+5 = 1+1+1+2+4 = 1+1+1+3+3 = 1+1+2+2+3 = 1+2+2+2+2 = 1+1+1+1+1+4 = 1+1+1+1+2+3 = 1+1+1+2+2+2 = 1+1+1+1+1+1+3 = 1+1+1+1+1+2+2 = 1+1+1+1+1+1+1+2 = 1+1+1+1+1+1+1+1+1 (p=30) 10 = 10 = 1+9 = 2+8 = 3+7 = 4+6 = 5+5 = 1+1+8 = 1+2+7 = 1+3+6 = 1+4+5 = 2+2+6 = 2+3+5 = 2+4+4 = 3+3+4 = 1+1+1+7 = 1+1+2+6 = 1+1+3+5 = 1+1+4+4 = 1+2+2+5 = 1+2+3+4 = 1+3+3+3 = 2+2+2+4 = 2+2+3+3 = 1+1+1+1+6 = 1+1+1+2+5 = 1+1+1+3+4 = 1+1+2+2+4 = 1+1+2+3+3 = 1+2+2+2+3 = 2+2+2+2+2 = 1+1+1+1+1+5 = 1+1+1+1+2+4 = 1+1+1+1+3+3 = 1+1+1+2+2+3 = 1+1+2+2+2+2 = 1+1+1+1+1+1+4 = 1+1+1+1+1+2+3 = 1+1+1+1+2+2+2 = 1+1+1+1+1+1+1+3 = 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 (p=42)

It’s very simple to understand what a partition is, but very difficult to say how many partitions, p(n), a particular number will have. Here’s a partition: 11 = 4 + 3 + 2 + 2. But what is p(11)? Is there a formula for the sequence of p(n)?

1, 2, 3, 5, 7, 11, 15, 22, 30, 42, 56, 77, 101, 135, 176, 231, 297, 385, 490, 627, 792, 1002, 1255, 1575, 1958, 2436, 3010, 3718, 4565, 5604, 6842, 8349, 10143, 12310, 14883, 17977, 21637, 26015, 3118 5, 37338, 44583, 53174, 63261... (A000041 at the OEIS)

Yes, there is a formula, but it is very difficult to understand the Partition function that supplies it. So that part of the river of mathematics is very deep. But a step away the river of mathematics is very shallow. Here’s another question: If you multiply the numbers in a partition of n, what’s the largest possible product? Try using the partitions of 5:

4 = 1 * 4 6 = 2 * 3 3 = 1 * 1 * 3 4 = 1 * 2 * 2 2 = 1 * 1 * 1 * 2 1 = 1 * 1 * 1 * 1 * 1

The largest product is 6 = 2 * 3. So the answer is easy for n = 5, but I assumed that as n got bigger, the largest product got more interesting, using a subtler and subtler mix of prime factors. I was wrong. You don’t have to struggle to find a formula for what you might call the maximum multiplicity of the partitions of n:

1 = 1 (n=1) 2 = 2 (n=2) 3 = 3 (n=3) 4 = 2 * 2 (n=4) 6 = 2 * 3 (n=5) 9 = 3 * 3 (n=6) 12 = 2 * 2 * 3 (n=7) 18 = 2 * 3 * 3 (n=8) 27 = 3 * 3 * 3 (n=9) 36 = 2 * 2 * 3 * 3 (n=10) 54 = 2 * 3 * 3 * 3 (n=11) 81 = 3 * 3 * 3 * 3 (n=12) 108 = 2 * 2 * 3 * 3 * 3 (n=13) 162 = 2 * 3 * 3 * 3 * 3 162(n=14) 243 = 3 * 3 * 3 * 3 * 3 (n=15) 324 = 2 * 2 * 3 * 3 * 3 * 3 (n=16) 486 = 2 * 3 * 3 * 3 * 3 * 3 (n=17) 729 = 3 * 3 * 3 * 3 * 3 * 3 (n=18)

It’s easy to see why the greatest prime factor is always 3. If you use 5 or 7 as a factor, the product can always be beaten by splitting the 5 into 2*3 or the 7 into 2*2*3:

15 = 3 * 5 < 18 = 3 * 2*3 (n=8) 14 = 2 * 7 < 24 = 2 * 2*2*3 (n=9) 35 = 5 * 7 < 72 = 2*3 * 2*2*3 (n=12)

And if you’re using 7 → 2*2*3 as factors, you can convert them to 1*3*3, then add the 1 to another factor to make a bigger product still:

14 = 2 * 7 < 24 = 2 * 2*2*3 < 27 = 3 * 3 * 3 (n=9) 35 = 5 * 7 < 72 = 2*3 * 2*2*3 < 81 = 3 * 3 * 3 * 3 (n=12)

Post-Performative Post-Scriptum

The title of this post is, of course, a paronomasia on core Beatles album Live and Let Die (1954). But what does it mean? Well, if you think of the partitions of n as slivers of n, then you sliv n to find its partitions:

9 = 9 = 1+8 = 2+7 = 3+6 = 4+5 = 1+1+7 = 1+2+6 = 1+3+5 = 1+4+4 = 2+2+5 = 2+3+4 = 3+3+3 = 1+1+1+6 = 1+1+2+5 = 1+1+3+4 = 1+2+2+4 = 1+2+3+3 = 2+2+2+3 = 1+1+1+1+5 = 1+1+1+2+4 = 1+1+1+3+3 = 1+1+2+2+3 = 1+2+2+2+2 = 1+1+1+1+1+4 = 1+1+1+1+2+3 = 1+1+1+2+2+2 = 1+1+1+1+1+1+3 = 1+1+1+1+1+2+2 = 1+1+1+1+1+1+1+2 = 1+1+1+1+1+1+1+1+1 (p=30)

And when you find the greatest product among those partitions, you let 3 or “tri” work its multiplicative magic. So you “Sliv and Let Tri”:

8 = 1 * 8 14 = 2 * 7 18 = 3 * 6 20 = 4 * 5 7 = 1 * 1 * 7 12 = 1 * 2 * 6 15 = 1 * 3 * 5 16 = 1 * 4 * 4 20 = 2 * 2 * 5 24 = 2 * 3 * 4 27 = 3 * 3 * 3 ← 6 = 1 * 1 * 1 * 6 10 = 1 * 1 * 2 * 5 12 = 1 * 1 * 3 * 4 16 = 1 * 2 * 2 * 4 12 = 1 * 2 * 3 * 3 24 = 2 * 2 * 2 * 3 5 = 1 * 1 * 1 * 1 * 5 8 = 1 * 1 * 1 * 2 * 4 9 = 1 * 1 * 1 * 3 * 3 12 = 1 * 1 * 2 * 2 * 3 16 = 1 * 2 * 2 * 2 * 2 4 = 1 * 1 * 1 * 1 * 1 * 4 6 = 1 * 1 * 1 * 1 * 2 * 3 8 = 1 * 1 * 1 * 2 * 2 * 2 3 = 1 * 1 * 1 * 1 * 1 * 1 * 3 4 = 1 * 1 * 1 * 1 * 1 * 2 * 2 2 = 1 * 1 * 1 * 1 * 1 * 1 * 1 * 2 1 = 1 * 1 * 1 * 1 * 1 * 1 * 1 * 1 * 1

The Choice of the Circle

Saturday, 26th Sep, 2015 by Krilling for Company in Geometry, Mathematics and tagged animated gif, animated gifs, circle, Circles, combinations, factorial, factorial n, factorials, graphics, math, mathematics, maths, n!, partitions, points around a circle, Polygons, recreational math, recreational mathematics, recreational maths | Leave a comment

Here’s an elementary mathematical problem: how many ways are there to choose three numbers from a set of six numbers? If the set is (1, 2, 3, 4, 5, 6), these are the possible choices (or combinations):

(1, 2, 3), (1, 2, 4), (1, 2, 5), (1, 2, 6), (1, 3, 4), (1, 3, 5), (1, 3, 6), (1, 4, 5), (1, 4, 6), (1, 5, 6), (2, 3, 4), (2, 3, 5), (2, 3, 6), (2, 4, 5), (2, 4, 6), (2, 5, 6), (3, 4, 5), (3, 4, 6), (3, 5, 6), (4, 5, 6) (c = 20)

So 6C3 = 20 (C stands for “combination”). The general formula is nCr = (n! / (n-r)!) / r!, where n is the number to choose from, r is the number of choices and n! is factorial n, or n multiplied by all numbers less than itself. For example, 6! = 6 * 5 * 4 * 3 * 2 * 1 = 720. When n = 6 and c = 3, 6C3 = (6! / (6-3)!) / 3! = (720 / 6) / 6 = 20.

There isn’t much visual appeal in the choices above, but there’s a simple way to change that. Take the ways of choosing two numbers from a set of ten. They start like this:

(1, 2), (1, 3), (1, 4), (1, 5), (1, 6), (1, 7), (1, 8), (1, 9), (1, 10), (2, 3), (2, 4), (2, 5), (2, 6), (2, 7), (2, 8), (2, 9), (2, 10), (3, 4), (3, 5), (3, 6)…

Suppose each choice represents the midpoint of two points chosen from a set of ten points around a pentagon, so that (1, 2) is half-way between points 1 and 2, (3, 5) is half-way between points 3 and 5, and so on:

Now take the ways of choosing three numbers from a set of ten:

(1, 2, 3), (1, 2, 4), (1, 2, 5), (1, 2, 6), (1, 2, 7), (1, 2, 8), (1, 2, 9), (1, 2, 10), (1, 3, 4), (1, 3, 5), (1, 3, 6), (1, 3, 7), (1, 3, 8), (1, 3, 9), (1, 3, 10)…

Now the pentagon looks like this, with (1, 2, 3) representing the point midway between 1, 2 and 3, (1, 3, 9) representing the point midway between 1, 3 and 9, and so on:

Now here are 10C4, 10C5 and 10C6 for the pentagon:

You can also generate the points 5C4 = 5, then add them to the original five points and generate 10C4:

5C4

10C4

And here are 5C5, 6C5 and 12C5:

Here are 7C7 and 8C8, adding points as for 5C4:

And here is 12C6 using a dodecagon:

And various nCr for dodecagons and other polygons:

This method can also be used to represent the partitions of n, or the number of sets whose members sum to n. The partitions of 5 are these:

(5), (4, 1), (3, 2), (3, 1, 1), (2, 2, 1), (2, 1, 1, 1), (1, 1, 1, 1, 1)

There are seven partitions, so p(5) = 7. Partitions start small and get very large, starting with p(1), p(2), p(3) and so on:

1, 2, 3, 5, 7, 11, 15, 22, 30, 42, 56, 77, 101, 135, 176, 231, 297, 385, 490, 627, 792, 1002, 1255, 1575, 1958, 2436, 3010, 3718, 4565, 5604, 6842, 8349, 10143, 12310, 14883, 17977, 21637, 26015, 31185, 37338, 44583, 53174, 63261, 75175, 89134, 105558, 124754, 147273, 173525, 204226, 239943, 281589, 329931, 386155, 451276, 526823, 614154, 715220, 831820, 966467, 1121505, 1300156…

Suppose the partitions of n are treated as sets of points around a polygon with n vertices. Each set is then used to generate the point midway between its members. For example, (5, 4, 4, 2) is one partition of 15 and would represent the point midway between 5, 4, 4 and 2 of a pentadecagon. Here is a graphical representation of p(30):

Here are graphical representations for the partitions 5 to 15, then 15 to 60 in increments of 5 (15, 20, 25, etc):

And here are some close-ups for the partitions of 35 and 40:

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Tag Archives: partitions

Partitional Pulchritude

Graham’s Sumber

Sliv and Let Tri

The Choice of the Circle