MATH IN MOSCOW
11. PARTITIONS.
COMBINATORICS, SPRINT 2016
1. Synopsis
A partition is dened as a decreasing sequence of positive integers: 1 ; : : : ; n . Notation: # def
= n, || =
1 + · · · + n . By p(n) denote the number of partitions with || = n (so that p(1) = 1, p(2) = 2, p(3) = 3,
p(4) = 5, etc.). Put also p(0) = 1 by denition. Also denote by ak () is the number of terms i equal to k; then it
is sometimes written = 1a1 () 2a2 () : : : (exponential notation).
P
1
m
2
2
4
3
6
Theorem 1. ∞
m=0 p(m)t = (1 + t + t + : : : )(1 + t + t + : : : )(1 + t + t + : : : ) · · · = (1−t)(1−t2 )(1−t3 )::: .
Proof. For any sequence of nonnegative integers a1 ; : : : ; an (with an > 0; or, equivalently, for any innite sequence
b1 ; b2 ; : : : for which there exists an N such that bk = 0 for k ≥ N ) there is a unique partition with ai () = bi
for all i = 1; : : : ; n). One has || = b1 + 2b2 + · · · + nbn . On the other hand, opening brackets in the (left) innite
product of the theorem, one obtains a term tb1 · t2b2 · : : : · tnbn = tb1 +2b2 +···+nbn for any sequence b1 ; : : : ; bn . Thus,
the number of terms tm in the sum is equal to the number of sequences b1 ; : : : ; bn with b1 + 2b2 + · · · + nbn = m,
that is, to the number of partitions with || = m.
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Partition = (1 ; : : : ; n ) is often represented graphically as a Young diagram: a sequence of n rectangles of
sizes (length × height) 1 × 1, 2 × 1, . . . , n × 1, situated one under the other and aligned on the left side. Then
|| is the total area of the diagram (the number of boxes in it), and # is the number of rows of the diagram.
P
k+1 (t(3k2 −k)=2 + t(3k2 +k)=2 ).
Theorem 2 (pentagonal identity). (1 − t)(1 − t2 )(1 − t3 ) · · · = ∞
k=1 (−1)
Corollary 1. p(n) = p(n − 1) + p(n − 2) − p(n − 5) − p(n − 7) + · · · + (−1)k p(n − (3k2 − k)=2) + (−1)k p(n − (3k2 +
k)=2) − : : : .
P
m
Proof. Note rst that if (1 − t)(1 − t2 )(1 − t3 ) · · · = ∞
m=0 q (m)t then q (m) = q+ (m) − q− (m) where q+ (m) is the
number of partitions with 1 > 2 > : : : (such partitions are called strict) such that || = m and # even, and
q− (m) is the number of strict partitions with || = m and # odd. It is proved in the same way as Theorem 1.
Thus, the theorem is equivalent to the statement that if m = (3k2 ± k)=2 with k even then q+ (m) = q− (m) + 1,
if m = (3k2 ± k)=2 with k odd then q+ (m) = q− (m) − 1, and q+ (m) = q− (m) for all other m.
Draw a strict partition as a Young diagram (it will have no rows of equal length) and dene for it two
new diagrams of strict partitions: W () and W −1 (). To obtain W () let k be the maximal integer such that
2 = 1 − 1, 3 = 2 − 1, . . . , k = k−1 − 1; graphically, k is the length of the \rightmost diagonal boundary"
of the Young diagram. Then cut this diagonal o (so that i will become smaller by 1 for all i = 1; : : : ; k), and
put the k boxes obtained to the bottom of the diagram, as a new row. If what obtained is a Young diagram of a
strict partition then denote it by W (); else W () is not dened. W −1 is the inverse operation: take the bottom
row of and distribute its boxes among rst rows of the diagram, provided these boxes all come to the rightmost
diagonal (that is, all the rows that grew by 1 as a result of the operation, satised the equality i = i−1 − 1). If
one obtains a diagram of a strict partition in this way then call it W −1 (), else W −1 () is not dened.
The operations W and W −1 have the following properties, evident from the denition:
• |W ()| = || and |W −1 ()| = ||.
• #W () = # + 1, #W −1 () = # − 1.
• W and W −1 are indeed inverse: if W () is dened then W −1 (W ()) is dened and equal to , and the
same with W and W −1 exchanged.
• If W () is dened then W −1 () is not dened (and vice versa, automatically).
Thus, all strict partitions with || = m are split into three pariwise nonintersecting classes: those with W ()
dened (class A1 ), those with W −1 () dened (class A2 ), and those where neither operation is dened (class A3 ).
Partitions from the A1 and A2 come in pairs: ∈ A1 is paired with W () ∈ A2 . By property 1 the contributions
of the two members of any pair into q+ (m) − q− (m) cancel. Therefore q+ (m) − q− (m) is equal to the number of
∈ A3 with || = m and # even, minus the number of ∈ A3 with || = m and # odd.
A partition may belong to A3 if and only if its bottom row and its rightmost diagonal intersect; also, the
length of the rightmost diagonal of should be equal either to the length of the bottom row or to the length
of the bottom row minus 1. In the rst case if k is the length of the bottom row, then # = k, and the
= 2k − 1; 2k − 2; : : : ; k, so that || = (3k − 1)k=2; in the second case if k + 1 is the length of the bottom row then
# = k and = (2k; 2k − 1; : : : ; k + 1), so that || = (3k + 1)k=2.
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1
A symmetric group Sn (or permutation group) is the group of all one-to-one maps x : {1; : : : ; n} → {1; : : : ; n}
called permutation; the group operation is the composition of maps. In other words, Sn consists of all sequences
[x(1); : : : ; x(n)] such that every number from 1 to n is met exactly once among x(i), 1 ≤ i ≤ n; the composition of
the permutations x = [x(1); : : : ; x(n)] and y = [y(1); : : : ; y(n)] is the permutation xy = [x(y(1)); : : : ; x(y(n))].
Sn is a group: x(yz ) = (xy)z for all permutations x; y; z ; there exists an identity permutation e = [1; 2; : : : ; n]
such that ex = xe = x for any x, and for any permutation x there exists its inverse x−1 (since x is a one-to-one
map). Sn is not commutative: generally xy 6= yx.
A cyclic permutation (q1 ; : : : ; qk ) (k ≤ n is called a length) where q1 ; : : : ; qk ∈ {1; 2; : : : ; n} are all distinct is
denes as: x(q1 ) = q2 , x(q2 ) = q3 , . . . , x(qk−1 ) = qk , x(qk ) = q1 , and x(i) = i if i ∈= {q1 ; : : : ; qk } (the set is called
a domain of the permutation). Cyclic permutations of length 2 are called transpositions, e.g. (23) ∈ S4 means
[1; 3; 2; 4].
Theorem 3. Cyclic permutations with nonintersecting domains commute: xy = yx if x = (q1 ; : : : ; qk ) and y =
(p1 ; : : : ; p` ) where {q1 ; : : : ; qk } ∩ {p1 ; : : : ; p` } = ∅. Every permutation x ∈ Sn can be represented as a product
x = c1 : : : cs of cyclic permutations with nonintersecting domains; this product is unique up to the order of factors.
Proof. Existence is proved by downward induction by the number of xed points of the permutation x (a number
i ∈ {1; : : : ; n} is a xed point if x(i) = i). If the number of xed points of x is n then x = e (the identity
permutation) is a product of an empty sequence of cyclic permutations. Otherwise, let q1 be not a xed point:
x(q1 ) = q2 6= q1 . Dene a sequence q3 ; q4 ; : : : recursively as qi = x(qi−1 ), i = 3; 4 : : : . All qi are elements of a nite
set {1; : : : ; n}, so the sequence must have repetitions. Let k be the rst repetition: qk = qi for some i < k. If i > 1
then x(qk−1 ) = qk = qi = x(qi−1 ). The map x is one-to-one, so this implies qk−1 = qi−1 contrary to the assumtion
that there are no repetitions in the sequence before qk . Thus i = 1, and therefore the sequence q1 ; q2 ; : : : is periodic:
qk = q1 , qk+1 = q2 , etc. Thus, x = c1 y where c1 = (q1 ; : : : ; qk−1 ) and the permutation y is as follows: y(qi ) = qi
for all i = 1; : : : ; k − 1 and y(j ) = x(j ) for all other j . The permutation y has more xed points than x and by the
induction hypothesis one has y = c2 : : : cs where c2 ; : : : ; cs are cyclic permutations with nonintersecting domains.
Since all qi are xed points of y, they do not belong to the domain of any ci , i = 2; : : : ; s. So x = c1 c2 : : : cs is the
desired presentation.
To prove the uniqueness note that if x = c1 : : : cs then a and b lie in a domain of some cycle ci if and only if
xk (a) = b for some k ≥ 0 (xk is the product of k factors all equal to x). Thus if x = c1 : : : cs = c01 : : : c0t then the
domain of evey c0i should be a domain of some cj . Since the cycles commute, one may suppose without loss of
generality that domains of c1 and c01 are the same, so that c1 = (q1 ; q2 ; : : : ; qk ) and c01 = (q1 ; q20 ; : : : ; qk0 ). But then
q20 = x(q1 ) = q2 , hence q30 = x(q20 ) = x(q2 ) = q3 , . . . , qk0 = x(qk0 −1 ) = x(qk−1 ) = qk by induction, so that c1 = c01 .
Then let c02 be the cyclic permutation with the same domain as c2 , so prove c2 = c02 in the same manner, and so
on.
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Theorem 4. Let x = c1 : : : cs where ci = (q1;i ; : : : ; qki ;i ) are cyclic permutations with nonintersecting domains,
and let y be any permutation. Then y−1 xy = c1 ` : : : c0s where c0i = (y(q1;i ); : : : ; y(qki ;i )) for all i.
Proof. If x = c1 is a cyclic permutation then the theorem is proved by a direct calculation. In the general case one
has y−1 xy = y−1 c1 y · y−1 c2 y · : : : · y−1 cs y.
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So, for any permutation x ∈ Sn one can dene a partition (x) = (1 ; : : : ; s ) with |x | = n where i is the
length of the i-th cycle in the representation of x as a product of cyclic permutations with nonintersecting domains.
Corollary 2. (x) = (z ) if and only if x and z are conjugate: there exists a permutation y such that z = y−1 xy.
Proof. Equality (x) = (y−1 xy) follows immediately from Theorem 4. Conversely, let (x) = (z ) def
= =
(1 ; : : : ; s ). Represent x and z as products of cyclic permutations with nonintersecting domains: x = c1 : : : cs and
y = c01 : : : c0s where ci = (qi;1 ; : : : ; qi ;i ) and ci = (qi;0 1 ; : : : ; q0 i ;i ); include there for convenience also cycles of length
1, i.e. xed points of x and z . All the numbers qji are pairwise distinct (the domains do not intersect), and so are
0
qji
. Thus there exists a permutation y sending every qji to qji0 . One has z = y−1 xy by Theorem 4.
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Thus, partitions with || = n are in one-to-one correspondence with conjugacy classes of Sn .
2. Exercises
Exercise 1. (a)PLet v(n) be the number of strict partitions such that || = n. Write down the generating
n
function f (q) = ∞
of partitions (not necessarily
n=0 v (n)q as an innite product. (b) Let w(n) be a number
P
n
strict) such that all the parts are odd. Write down the generating g(q) = ∞
n=0 w(n)q as an innite product.
(c) Prove that f (q) = g(q). (d) Prove directly that v(n) = w(n) putting strict partitions and partitions with odd
parts into one-to-one correspondence.
Exercise 2. Prove L. Euler's formula (in the left-hand side there is a summation over all positive integers, in the
right-hand side, a multiplication over the set of all prime numbers):
1
1
1
1
1
·
:::
1 + s + s + ··· =
2
3
1 − 1=2s 1 − 1=3s 1 − 1=5s
3. Homework
The problems marked with an asterisk do not inuence the homework score. Such problem require more thinking
than usual: they may be more dicult, deeper, or just less accurately formulated.
h i
h i
n is dened as a sum P
qa1 +···+ak . (a) Prove that n = (q)n =(q)k (q)n−k
0
≤
a
≤···≤
a
≤
n
−
k
1
k
k q
k q·
¸
h i
h i
¡
¢
n
2
`
n
n
n
where (q)` = (1 − q)(1 − q ) : : : (1 − q ). (b) Prove that limq→1
= k . (c) How are
and
related?
k q
k q
n−k q
·
¸
·
¸
h i
h
i
h i
h
i
(d) Prove that n = n−1 + qn−k n−1 and n = n−1 + qk n−1 .
k q
k q
k−1 q
k q
k −1 q
k q
h i
P
Problem 2. (a) Prove the q-binomial formula (1 − z )(1 − zq) : : : (1 − zqn−1 ) = nk=0 nk (−1)k z k qk(k−1)=2 .
q
P
(1−a)(1−aq):::(1−aqn−1 ) tn = Q∞ 1−atqm .
(b) Prove Cauchy formula 1 + ∞
n
m
m=0 1−tq
n=1
(1−q):::(1−q )
Problem 1. The number
Hint (for the question 2(b)). Prove that the left-hand side and the right-hand side of the identity satisfy the
equality F (t)(1 − t) = F (tq)(1 − at).