Our chief weapon is surprise...
surprise and fear...
fear and surprise...
Our two weapons are fear and surprise...
and ruthless efficiency...
The Spanish Inquisition Monty Python For reasons we shall see soon, we divert in a surprising way.
A number is pentagonal if it is of the form k(3k − 1)/2 or generalized pentag-onal OEIS A001318 if it is of the form k(3k + 1)/2 for some positive integer k.
Alternatively, one can simply use k(3k − 1)/2 for arbitrary integers k. The first few pentagonal numbers for positive index are 1(3 ⋅ 1 − 1)/2 = 1, 2(3 ⋅ 2 − 1)/2 = 5, 3(3 ⋅ 3 − 1)/2 = 12, 4(3 ⋅ 4 − 1)/2 = 22 as depicted in Figure V.2, respectively 0(3⋅0−1)/2 = 0, −1(3⋅(−1)−1)/2 = 2, −2(3⋅(−2)−1)/2 = 7, −3(3⋅(−3)−1)/2 = 15 for other indices.
The pentagonal numbers give the following extraordinary2theorem.
Theorem V.4 (Euler’s Pentagonal Numbers Theorem): a) If n is not a pentagonal number, the number of partitions of n into an even and an odd number of distinct parts are equal.
2as in the title: une loi tout extraordinaire, [Eul15]
V.3. PENTAGONAL NUMBERS 71
Figure V.2: The first few pentagonal numbers: 1,5,12,22
b) If n = k(3k −1)/2 for k ∈ Z even, then the number of partitions of n into an even number of distinct parts exceeds the number of partitions into an odd number of distinct parts by one.
c) If n = k(3k − 1)/2 for k ∈ Z odd, then the number of partitions of n into an even number of distinct parts is one less than the number of partitions into an odd number of distinct parts.
Example: There are 2 partitions of 6 into an even number of distinct parts and 2 into an odd number of distinct parts. There are 3 partitions of 7 into an even number of distinct parts, and 2 into an odd number of distinct parts. There are 7 partitions of 12 into even parts, and 8 into odd parts. Proof: The proof will consist of constructing a bijection between partitions with an even number of distinct parts and partitions with an odd number of distinct parts. In the case that n is pentagonal this will have leave out exactly one partition.
For a partition λ ⊢ n into different parts, we consider two subsets of the cells, as depicted in Figure V.3:
The base is the bottom row of cells (the shortest row), the slope is the cells start-ing at the end of the top row and descendstart-ing straight down-left as far as possible (this might be only of length 1).
Figure V.3: Base (dashed) and Slope of two partitions
Base and slope might intersect in one cell, we call those cells of the base that are not in the slope the pure base; similarly the pure slope are those slope cells that are not also in the base.
We now divide the partitions of n with distinct parts into three classes:
Class 1 are those partitions for which the pure base contains more cells than the slope.
Class 2 are those partitions for which the pure slope is at least as long as the base.
Class 3 are all remaining partitions3.
For a partition λ in Class 1, we create a new partition µ by removing the slope (possibly including the base point) and placing these cells in a new row at the bot-tom of the diagram. As the pure base is larger than the slope this results in a parti-tion into different parts. Furthermore, the slope of µ is at least as long as the slope of λ (which is the base of µ) if base and slope of µ are distinct; if they intersect the slope of µ is strictly larger. This means µ is in Class 2.
Vice versa, if µ is in Class 2, we remove the base (possibly including the low-est point of the slope) and attach it from top right as new slope. This results in a partition λ into different parts which lies in Class 1.
For example, the left partition in Figure V.3 is in Class 2 and will be transformed into the partition on the right side in Class 1, and vice versa.
It is easily seen that these two operations are mutually inverse, that is Class 1 and Class 2 contain the same number of partitions. Furthermore the operations change the number of parts by exactly one. That means that in Class 1 ∪ Class 2, there are exactly as many partitions with an even number of parts, as with an odd number of parts.
Finally consider Class 3: A partition in this class must have base and slope in-tersecting, and the base contains the same number of cells, or exactly one more, than the slope, see Figure V.4.
Figure V.4: Possibilities for partitions in Class 3
This means that, if there are k parts, it consists of a k × k rectangle and a 1, 2, . . . , k triangle that might or might not overlap. In the first case there are k2+ k(k − 1)/2 = k(3k − 1)/2 parts, respectively k2+ k(k + 1)/2 = k(3k + 1)/2 = (−k)(3(−k) − 1)/2 parts. For a given n only one of these two options is possible, and it implies that n is a pentagonal number. This proves the theorem. ◻
3Classifications become easy if we allow an “all the rest” category
V.3. PENTAGONAL NUMBERS 73 To simplify notation we write ω(k) = k(3k−1)/2 for k ∈ Z to state the following consequence for the inverse of the generating function of partitions.
Corollary V.5:
Proof: The second equality simply relies on the now familiar two ways to describe generalized pentagonal numbers.
Let even(n) be the number of partitions of n into an even number of dis-tinct parts, odd(n) similarly for an odd number of parts. By the Pentagonal Num-bers Theorem, the right hand side of the equation is the generating function for even(n) − odd(n). We aim to show that this is true also for the left hand side:
The coefficient for tnwill be made up by contributions from terms of the form tni, all distinct, such that n = n1+ ⋯ + nl. This partition contributes (−1)l. That is, every partition into an even number of distinct parts contributes 1, every partition into an odd number of distinct parts contributes −1, which was to be shown. ◻
This now permits us to formulate a recurrence formula for p(n):
Corollary V.6:
We now consider the coefficient of tnis this product (which is zero for n > 0). This gives
0 = p(n) + ∑
k>0
(−1)k(p(n − ω(k)) + p(n − ω(−k))) .
By using that p(n) = 0 for n < 0 and substituting values for the pentagonal numbers
we get the explicit recursion. ◻
This formula can be used for example to calculate values for p(n) effectively, or to estimate the growth of the function p(n).
While it might not seems so we have by now wandered deep into Number the-ory. In this area we get for example Ramanujan’s famous congruence identities
p(5n + 4) ≡ 0 (mod 5) p(7n + 5) ≡ 0 (mod 7) p(11n + 6) ≡ 0 (mod 11)
that recently have been generalized by Ono and collaborators.
Euler’s theorem and its consequence are a special case of the Jacobi triple product identity
∏∞ n=1
(1 − q2n)(1 + q2n−1t)(1 + q2n−1t−1) = ∑∞
r=−∞
qr
2
tr
which has applications in the theory of theta-functions – the territory of Number Theory – and Physics.