4. Integer partitions and q -binomial coefficients
4.1. Partition basics
The following definition is built in analogy to Definition 3.9.1:
Definition 4.1.1. (a) An (integer) partition means a (finite) weakly decreas-ing tuple of positive integers – i.e., a finite tuple (λ1,λ2, . . . ,λm) of positive integers such thatλ1 ≥λ2 ≥ · · · ≥λm.
Thus, partitions are the same as weakly decreasing compositions. Hence, the notions of size and length of a partition are automatically defined, since we have defined them for compositions (in Definition 3.9.1).
(b) The parts of a partition (λ1,λ2, . . . ,λm) are simply its entries λ1,λ2, . . . ,λm.
(c)Let n∈ Z. Apartition of n means a partition whose size isn.
(d) Let n ∈ Z and k ∈ N. A partition of n into k parts is a partition whose size is nand whose length isk.
Example 4.1.2. The partitions of 5 are
(5), (4, 1), (3, 2), (3, 1, 1), (2, 2, 1), (2, 1, 1, 1), (1, 1, 1, 1, 1).
Definition 4.1.3. (a)Let n∈ Zand k∈ N. Then, we set pk(n) := (# of partitions of ninto kparts). (b)Let n∈ Z. Then, we set
p(n) := (# of partitions of n). This is called then-th partition number.
Example 4.1.4. Our above list of partitions of 5 reveals that p0(5) = 0;
p1(5) = 1;
p2(5) = 2;
p3(5) = 2;
p4(5) = 1;
p5(5) = 1;
pk(5) = 0 for any k>5;
and finally p(5) =7.
Here are the values of p(n) for the first 15 nonnegative integers n:
n 0 1 2 3 4 5 6 7 8 9 10 11 12 13 14
p(n) 1 1 2 3 5 7 11 15 22 30 42 56 77 101 135 . The sequence (p(0),p(1),p(2), . . .) is remarkable for being an integer se-quence that grows faster than polynomially, but still considerably slower than exponentially. (See (154) for an asymptotic expansion.) This not-too-fast growth (for instance, p(100) =190 569 292 is far smaller than 2100) makes integer par-titions rather convenient for computer experiments.
We will next state some elementary properties of pk(n) and p(n), but first we introduce a few very basic notations:
Definition 4.1.5. We will use the Iverson bracket notation: If A is a log-ical statement, then [A] means the truth value of A; this is the integer (1, if Ais true;
0, if Ais false.
For example,[2+2=4] = 1 and[2+2=5] =0.
Note that the Kronecker delta notation is a particular case of the Iverson bracket: We have
δi,j = [i =j] for any objectsi and j.
Definition 4.1.6. Let abe a real number.
Then,bac (called thefloorofa) means the largest integer that is ≤a.
Likewise,dae(called theceilingofa) means the smallest integer that is≥a.
For example, the number π ≈ 3.14 satisfies bπc = 3 and dπe = 4 and b−πc = −4 and d−πe = −3. For another example, bnc = dne = n for each n∈ Z.
The following proposition collects various basic properties of the numbers introduced in Definition 4.1.3:
Proposition 4.1.7. Let n∈ Zandk ∈ N.
(a)We have pk(n) =0 whenever n<0 andk∈ N.
(b)We have pk(n) =0 whenever k>n.
(c)We have p0(n) = [n=0]. (d)We have p1(n) = [n>0].
(e)We have pk(n) = pk(n−k) +pk−1(n−1) wheneverk>0.
(f)We have p2(n) = bn/2c whenevern ∈ N.
(g)We have p(n) = p0(n) +p1(n) +· · ·+pn(n)whenever n∈ N.
(h)We have p(n) =0 whenever n<0.
Proof of Proposition 4.1.7 (sketched). (a)The size of a partition is always nonneg-ative (being a sum of positive integers). Thus, a negnonneg-ative number n has no partitions whatsoever. Thus, pk(n) =0 whenever n<0 andk∈ N.
(b) If (λ1,λ2, . . . ,λk) is a partition, then λi ≥ 1 for each i ∈ {1, 2, . . . ,k} (because a partition is a tuple of positive integers, i.e., of integers≥1). Hence, if(λ1,λ2, . . . ,λk) is a partition of ninto kparts, then
n =λ1+λ2+· · ·+λk (since (λ1,λ2, . . . ,λk) is a partition of n)
≥1+1+· · ·+1
| {z }
ktimes
(sinceλi ≥1 for each i∈ {1, 2, . . . ,k})
=k.
Thus, a partition ofnintokparts cannot satisfyk>n. Thus, no such partitions exist ifk>n. In other words, pk(n) = 0 ifk >n.
(c)The integer 0 has a unique partition into 0 parts, namely the empty tuple (). A nonzero integerncannot have any partitions into 0 parts, since the empty tuple has size 06=n. Thus, p0(n)equals 1 for n=0 and equals 0 forn 6=0. In other words, p0(n) = [n =0].
(d) Any positive integer n has a unique partition into 1 part – namely, the 1-tuple (n). On the other hand, if n is not positive, then this 1-tuple is not a partition, so in this casen has no partition into 1 part. Thus, p1(n) equals 1 if nis positive and equals 0 otherwise. In other words, p1(n) = [n>0].
(e)Assume thatk>0. We must prove thatpk(n) = pk(n−k) +pk−1(n−1). We consider all partitions of n into k parts. We classify these partitions into two types:
• Type 1consists of all partitions that have 1 as a part.
• Type 2consists of all partitions that don’t.
For example, here are the partitions of 5 along with their types:
(4, 1), (3, 1, 1), (2, 2, 1), (2, 1, 1, 1), (1, 1, 1, 1, 1)
| {z }
Type 1
, (5), (3, 2)
| {z }
Type 2
.
Let us count the type-1 partitions and the type-2 partitions separately.
Any type-1 partition has 1 as a part, therefore as its last part (because it is weakly decreasing). Hence, any type-1 partition has the form(λ1,λ2, . . . ,λk−1, 1). If(λ1,λ2, . . . ,λk−1, 1)is a type-1 partition (ofnintokparts), then(λ1,λ2, . . . ,λk−1) is a partition ofn−1 into k−1 parts. Thus, we have a map
{type-1 partitions of ninto k parts} → {partitions ofn−1 into k−1 parts}, (λ1,λ2, . . . ,λk−1, 1) 7→ (λ1,λ2, . . . ,λk−1).
This map is a bijection (since it has an inverse map, which simply inserts a 1 at the end of a partition). Thus, the bijection principle shows that
(# of type-1 partitions ofn into kparts)
= (# of partitions ofn−1 into k−1 parts) = pk−1(n−1) (by the definition of pk−1(n−1)).
Now let us count type-2 partitions. A type-2 partition does not have 1 as a part; hence, all its parts are larger than 1 (because all its parts are positive inte-gers), and therefore we can subtract 1 from each part and still have a partition in front of us. To be more specific: If (λ1,λ2, . . . ,λk) is a type-2 partition of n into k parts, then subtracting 1 from each of its parts produces the k-tuple (λ1−1,λ2−1, . . . ,λk−1), which is a partition ofn−kinto kparts. Hence, we have a map
{type-2 partitions of ninto kparts} → {partitions of n−kinto k parts}, (λ1,λ2, . . . ,λk) 7→(λ1−1,λ2−1, . . . ,λk−1).
This map is a bijection (since it has an inverse map, which simply adds 1 to each entry of a partition). Thus, the bijection principle shows that
(# of type-2 partitions ofn intok parts)
= (# of partitions ofn−k intok parts) = pk(n−k) (by the definition of pk(n−k)).
Since any partition ofnintok parts is either type-1 or type-2 (but not both at the same time), we now have
(# of partitions of ninto kparts)
= (# of type-1 partitions of ninto k parts)
| {z }
=pk−1(n−1)
+ (# of type-2 partitions of ninto k parts)
| {z }
=pk(n−k)
= pk−1(n−1) +pk(n−k) = pk(n−k) +pk−1(n−1).
Since the left hand side of this equality is pk(n), we thus have proved that pk(n) = pk(n−k) +pk−1(n−1).
(f)Let n∈ N. The partitions ofn into 2 parts are
(n−1, 1), (n−2, 2), (n−3, 3), . . . ,
n− bn/2c
| {z }
=dn/2e
,bn/2c
. Thus there arebn/2c of them. In other words, p2(n) = bn/2c.
(g)Let n∈ N. Any partition ofn must havek parts for somek ∈ N. Thus, p(n) =
∑
k∈N
pk(n) =
∑
n k=0pk(n) +
∑
∞ k=n+1pk(n)
| {z }
=0
(by Proposition 4.1.7(b))
=
∑
n k=0pk(n) = p0(n) +p1(n) +· · ·+pn(n). (h)Same argument as for(a).
Proposition 4.1.7 (e) is a recursive formula that makes it not too hard to computepk(n)for reasonably small values ofnand k. Then, using Proposition 4.1.7(g), we can compute p(n) from these pk(n)’s. However, one might want a better, faster method.
To get there, let me first express the generating function of the numbersp(n): Theorem 4.1.8. In the FPS ringZ[[x]], we have
n
∑
∈Np(n)xn =
∏
∞ k=11 1−xk.
(The product on the right hand side is well-defined, since multiplying a FPS by 1
1−xk does not affect its firstk coefficients.)
Example 4.1.9. Let us check the above equality “up tox5”, i.e., let us compare the coefficients of xi for i < 5. (In doing so, we can ignore all powers of x higher thanx4.) We have
∏
∞ k=11
1−xk = 1
1−x1 · 1
1−x2 · 1
1−x3 · 1
1−x4 · · · ·
= 1+x+x2+x3+x4+· · ·
·1+x2+x4+· · ·
·1+x3+· · ·
·1+x4+· · ·
·(1+· · ·)
·(1+· · ·)
· · · ·
=1+x+2x2+3x3+5x4+· · ·
= p(0) +p(1)x+p(2)x2+p(3)x3+p(4)x4+· · · .
Proof of Theorem 4.1.8. We have
∏
∞ k=11 1−xk
| {z }
=1+xk+x2k+x3k+···
=
∏
∞ k=1
1+xk+x2k+x3k+· · ·
| {z }
= ∑
u∈Nxku
=
∏
∞ k=1∑
u∈N
xku
=
∑
(u1,u2,u3,...)∈N∞is essentially finite
x1u1x2u2x3u3· · ·
here, we expanded the product using Proposition 3.11.25
=
∑
(u1,u2,u3,...)∈N∞is essentially finite
x1u1+2u2+3u3+···
=
∑
n∈N
|Qn|xn, where
Qn ={(u1,u2,u3, . . .) ∈N∞ essentially finite | 1u1+2u2+3u3+· · · =n}. Thus, it will suffice to show that
|Qn| = p(n) for eachn ∈N.
Let us fixn∈ N. We want to construct a bijection fromQn to{partitions ofn}. Here is how to do this: For any(u1,u2,u3, . . .) ∈ Qn, define a partition
π(u1,u2,u3, . . .) := (the partition that contains each iexactlyui times)
=
. . . , 3, 3, . . . , 3
| {z }
u3times
, 2, 2, . . . , 2
| {z }
u2times
, 1, 1, . . . , 1
| {z }
u1times
.
This is a partition of n, since its size is 1u1+2u2+3u3+· · · = n (because (u1,u2,u3, . . .) ∈ Qn). Thus, we have defined a partition π(u1,u2,u3, . . .) of n for each(u1,u2,u3, . . .)∈ Qn. In other words, we have defined a map
π : Qn → {partitions ofn}.
It remains to show that this mapπ is a bijection. We define a map ρ: {partitions ofn} → Qn
that sends each partitionλof nto the sequence
(# of 1’s inλ, # of 2’s inλ, # of 3’s inλ, . . .) ∈ Qn
(this is indeed a sequence inQn, since
∑
∞ i=1i(# ofi’s in λ) = (the sum of all entries ofλ) = |λ|=n
becauseλis a partition ofn). It is now easy to check that the maps π andρare mutually inverse, so thatπis a bijection. The bijection principle therefore yields
|Qn| = (# of partitions ofn) = p(n); but this is precisely what we wanted to show. The proof of Theorem 4.1.8 is thus complete.
Theorem 4.1.8 has a “finite” analogue (finite in the sense that the product
∏∞ k=1
1
1−xk is replaced by a finite product; the FPSs are still infinite):
Theorem 4.1.10. Let m ∈ N. For each n ∈ N, let pparts≤m(n) be the # of partitionsλ ofnsuch that all parts of λare≤m. Then,
n
∑
∈Npparts≤m(n)xn =
∏
m k=11 1−xk.
Proof of Theorem 4.1.10. This proof is mostly analogous to the above proof of Theorem 4.1.8, and to some extent even simpler because it usesm-tuples instead of infinite sequences.
We have
∏
m k=11 1−xk
| {z }
=1+xk+x2k+x3k+···
=
∏
m k=1
1+xk+x2k+x3k+· · ·
| {z }
= ∑
u∈Nxku
=
∏
m k=1∑
u∈N
xku
=
∑
(u1,u2,...,um)∈Nm
x1u1x2u2· · ·xmum
here, we expanded the product using Proposition 3.11.23
=
∑
(u1,u2,...,um)∈Nm
x1u1+2u2+···+mum =
∑
n∈N
|Qn|xn, where
Qn ={(u1,u2, . . . ,um)∈ Nm | 1u1+2u2+· · ·+mum =n}. Thus, it will suffice to show that
|Qn|= pparts≤m(n) for each n∈ N.
Let us fix n ∈ N. We want to construct a bijection from Qn to the set {partitions λofn such that all parts ofλ are ≤m}.
Here is how to do this: For any(u1,u2, . . . ,um) ∈ Qn, define a partition π(u1,u2, . . . ,um) := (the partition that contains eachi exactlyui times)
=
m,m, . . . ,m
| {z }
umtimes
, . . . , 2, 2, . . . , 2
| {z }
u2times
, 1, 1, . . . , 1
| {z }
u1times
. Thus, we have defined a map
π : Qn → {partitionsλofn such that all parts ofλ are ≤m}.
It is easy to see that this mapπis a bijection43. The bijection principle therefore yields
|Qn| = (# of partitionsλ ofnsuch that all parts of λare ≤m) = pparts≤m(n); but this is precisely what we wanted to show. The proof of Theorem 4.1.10 is thus complete.
Next, we shall state a result of Euler that we have already discovered in a different language.
Definition 4.1.11. Let n∈ Z.
(a)A partition of n into odd partsmeans a partition of nwhose all parts are odd.
(b)Apartition of n into distinct partsmeans a partition of nwhose parts are distinct.
(c)Let
podd(n) := (# of partitions of ninto odd parts) and pdist(n) := (# of partitions of ninto distinct parts).
Example 4.1.12. We have
podd(7) =|{(7), (5, 1, 1), (3, 3, 1), (3, 1, 1, 1, 1), (1, 1, 1, 1, 1, 1, 1)}| =5;
pdist(7) =|{(7), (6, 1), (5, 2), (4, 3), (4, 2, 1)}| =5.
Theorem 4.1.13 (Euler’s odd-distinct identity). We have podd(n) = pdist(n) for each n∈ N.
43The argument is analogous to the one used in the proof of Theorem 4.1.8.
We have already encountered this theorem before (as Theorem 3.11.30, albeit in less precise language), and we have proved it using the generating function identity
∏
i>0
1−x2i−1−1
=
∏
k>0
1+xk . Let me outline a different, bijective proof.
Second proof of Theorem 4.1.13 (sketched). Let n ∈ N. We want to construct a bi-jection
A: {partitions ofninto odd parts} → {partitions ofninto distinct parts}. We shall do this as follows: Given a partitionλ ofn into odd parts, we repeat-edly merge pairs of equal parts inλuntil no more equal parts appear. The final result will be A(λ). Here are two examples:
• To compute A(5, 5, 3, 1, 1, 1), we compute44 5, 5, 3, 1, 1, 1
→ 10, 3, 1, 1, 1
→(10, 3, 2, 1). Thus, A(5, 5, 3, 1, 1, 1) = (10, 3, 2, 1).
• To compute A(5, 3, 1, 1, 1, 1), we compute 5, 3, 1, 1, 1, 1
→ 5, 3, 2, 1, 1
→ 5, 3, 2, 2
→(5, 4, 3). Thus, A(5, 3, 1, 1, 1, 1) = (5, 4, 3).
Why is this mapAwell-defined? We only specified the sort of steps we are al-lowed to take when computing A(λ); however, there is often a choice involved in taking these steps (since there are often several pairs of equal parts).45 So we have specified a non-deterministic algorithm. Why is the resulting partition independent of the choices we make?
One way to prove this is using thediamond lemma, which is a general tool for proving that certain non-deterministic algorithms have unique final outcomes (independent of the choices taken). We will hopefully learn more about it later on.
For the map A, we can also proceed differently, by analyzing the algorithm that we used to define A. Namely, we observe what is really going on when we are merging equal parts. Let us say our original partitionλ has pmany 1s.
44The two entries underlined are the two equal entries that are going to get merged in the next step. Note that there are usually several candidates, and we just pick one pair at will.
45For example, we could have also computedA(5, 5, 3, 1, 1, 1)as follows:
(5, 5, 3, 1, 1, 1)→ 5, 5, 3, 2, 1
→(10, 3, 2, 1).
Let us first merge them in pairs, so that we getbp/2c many 2s and maybe one single 1. Then, let us merge the 2s in pairs, so that we getbbp/2c/2cmany 4s, maybe a single 2, and maybe a single 1. Proceed until no more than one 1, no more than one 2, no more than one 4, no more than one 8, and so on remain.
This clears out any duplicate parts of the form 2k. Next do the same with parts of the form 3·2k (that is, with parts equal to 3, 6, 12, 24, and so on), then with parts of the form 5·2k, and so on.
The nice thing about this way of proceeding is that we can explicitly describe the final outcome. Indeed, if the original partition λ (a partition of n into odd parts) contains an odd part k precisely m many times, and if the binary representation ofmism = (mimi−1· · ·m1m0)2(that is, ifm0,m1, . . . ,mi ∈ {0, 1} satisfy m = ∑i
j=0
mj2j), then the partition A(λ) will contain the number 20k exactlym0 times, the number 21k exactlym1 times, the number 22k exactly m2
times, and so on. Since the binary digitsm0,m1, . . . ,miare all≤1, this partition A(λ) will therefore not contain any number more than once, i.e., it will be a partition into distinct parts.
It is not hard to check that this map A is indeed a bijection. Indeed, in order to see this, we construct a map B that will turn out to be its inverse. Here, we start with a partition λ of n into distinct parts. Let us represent each part of this partition in the form k·2i for some odd k ≥ 1 and some integer i ≥ 0.
(Recall that any positive integer can be represented uniquely in this form.) Now, replace this part k·2i by 2i many k’s. The resulting partition (once all parts have been replaced) will usually have many equal parts, but all its parts are odd. We define B(λ) to be this resulting partition. Alternatively, B(λ) can also be constructed step-by-step by a non-deterministic algorithm: Starting with λ, keep “breaking even parts into halves” (i.e., whenever you see an even partm, replace it by two parts m
2 and m
2), until no even parts remain any more.
The result is B(λ). It is not hard to see that both descriptions of B(λ) describe the same partition. It is furthermore easy to see that this map B is indeed an inverse ofA, so that Ais indeed a bijection. Thus, the bijection principle yields
|{partitions ofninto odd parts}| =|{partitions ofninto distinct parts}|. In other words, podd(n) = pdist(n). This proves Theorem 4.1.13.
Here is another situation in which two kinds of partitions are equinumerous:
Proposition 4.1.14. Let n∈ Nand k ∈N. Then,
pk(n) = (# of partitions of nwhose largest part isk). Here and in the following, we use the following convention:
Convention 4.1.15. We agree to say that the largest part of the empty parti-tion ()is 0 (even though this partition has no parts).
Example 4.1.16. Forn=4 andk =3, we have
pk(n) = p3(4) =1 (due to the partition (2, 1, 1)) and
(# of partitions of nwhose largest part isk)
= (# of partitions of 4 whose largest part is 3)
=1 (due to the partition (3, 1)). Thus, Proposition 4.1.14 holds for n=4 andk=3.
Proof of Proposition 4.1.14 (sketched). We do a “proof by picture” (it can be made rigorous – see Exercise A.3.1.1 for this). We pickn =14 and k =4 for example, and we start with the partitionλ= (5, 4, 4, 1) ofn intok parts.
We draw a table of k left-aligned rows, where the length of each row equals the corresponding part of λ (that is, the i-th row from the top has λi boxes, whereλ= (λ1,λ2, . . . ,λk)):
5→ 4→ 4→ 1→
Now, let us flip this table across the “main diagonal” (i.e., the diagonal that goes from the top-left corner to the bottom-right corner)46:
5 4 4 1
↓ ↓ ↓ ↓
The lengths of the rows of the resulting table again form a partition ofn. (In our case, this new partition is (4, 3, 3, 3, 1).) Moreover, the largest part of this new partition is k (because the original table had k rows, so the flipped table
46This kind of flip is precisely how you would transpose a matrix.
has k columns, and this means that its top row has k boxes). This procedure (i.e., turning a partition into a table, then flipping the table across the “main diagonal”, and then reading the lengths of the rows of the resulting table again as a partition) therefore gives a map from
{partitions ofninto k parts} to
{partitions ofnwhose largest part isk}.
Moreover, this map is a bijection (indeed, its inverse can be effected in the exact same way, by flipping the table). This bijection is calledconjugationof partitions, and will be studied in more detail later.
Here are some pointers to how this proof can be formalized (see Exercise A.3.1.1 for much more): For any partition λ = (λ1,λ2, . . . ,λk), we define the Young diagramof λto be the set
Y(λ):=n(i,j) ∈ Z2 | 1≤i ≤k and 1≤j ≤λio.
This Young diagram is precisely the table that we drew above, as long as we agree to identify each pair (i,j) ∈ Y(λ) with the box in row i and column j.
Now, theconjugate of the partitionλ is the partitionλt uniquely determined by Y λt
=flip(Y(λ)) = {(j,i) | (i,j) ∈ Y(λ)}.
Explicitly,λt can be defined byλt = µ1,µ2, . . . ,µp, where pis the largest part ofλand where
µi = (# of parts of λthat are ≥i) for eachi ∈ {1, 2, . . . ,p}.
(This conjugate λt is also often called λ0.) Now, it is not hard to show that
λt
= |λ| and λtt
= λ for each partition λ, and that the largest part of λt equals the length of λ. Using these observations (which are proved in Exercise A.3.1.1), we see that the map
{partitions ofninto k parts} → {partitions of nwhose largest part isk}, λ7→λt
is well-defined and is a bijection; thus, the above proof of Proposition 4.1.14 becomes fully rigorous.
The word “Young” in “Young diagram” (and, later, “Young tableau”) does not imply any novelty (Young diagrams have been around in some form or another since the 19th century – if often in the superficially different guise of
“Ferrers diagrams”), but rather honors Alfred Young, who built up the rep-resentation theory of symmetric groups (and significantly forwarded invariant theory) using these objects.
Corollary 4.1.17. Let n ∈Nand k∈ N. Then, p0(n) +p1(n) +· · ·+pk(n)
= (# of partitions ofnwhose largest part is ≤k). Proof of Corollary 4.1.17. We have
p0(n) +p1(n) +· · ·+pk(n)
=
∑
k i=0pi(n)
| {z }
=(# of partitions ofnwhose largest part isi) (by Proposition 4.1.14,
applied toiinstead ofk)
=
∑
k i=0(# of partitions ofn whose largest part isi)
= (# of partitions ofn whose largest part is ≤k). This proves Corollary 4.1.17.
Corollary 4.1.17 leads to yet another FPS identity:
Theorem 4.1.18. Let m∈ N. Then,
n
∑
∈N(p0(n) +p1(n) +· · ·+pm(n))xn =
∏
m k=11 1−xk.
Proof of Theorem 4.1.18. For eachn∈ N, we have p0(n) +p1(n) +· · ·+pm(n)
= (# of partitions of nwhose largest part is ≤m) (by Corollary 4.1.17)
= (# of partitions of nwhose all parts are ≤m)
because the condition “the largest part is ≤m” for a partition is clearly equivalent to “all parts are ≤m”
= pparts≤m(n),
where pparts≤m(n) is defined as in Theorem 4.1.10. Hence,
n
∑
∈N(p0(n) +p1(n) +· · ·+pm(n))
| {z }
=pparts≤m(n)
xn =
∑
n∈N
pparts≤m(n)xn =
∏
m k=11 1−xk (by Theorem 4.1.10). This proves Theorem 4.1.18.