COMBINATORICS. FINAL EXAM
∞ (k + 1)2
P
. Find the similar expansion of F (u + 1) as the sum over negative
uk
k=0
∞ c
P
k
.
powers of u, i.e., find the ck ’s in F (u + 1) =
k
u
k=0
Problem 1. Let F (u) =
Solution 1. We have to sum the power series for F :
∞
∞
X
X
2 −k
F (u) =
(k + 1) u =
(k + 1)2 tk ,
k=0
k=0
−1
where t = u . We know that
∞
X
2
1
1+t
d
=
,
(k + 1) t = t + 1
3
dt
1
−
t
(1
−
t)
k=0
2 k
so substituting back t = u−1 , we get (multiplying both parts of the fraction by u3 ):
F (u) =
u2 (u + 1)
.
(u − 1)3
Now F (u + 1) = (u + 1)2 (u + 2)/u3 , so the desired decomposition is actually finite:
5
2
4
F (u + 1) = 1 + + 2 + 3
u u
u
(c0 = 1, c1 = 4, c2 = 5, c3 = 2, all other ck ’s are zero).
Problem 2. Solve the recurrence relation (and check yourself by substituting your answer back in the
relation):
an+2 = 4an + 2n (n ≥ 0),
a0 = 0, a1 = 14 .
P
n
Solution 2. Using ordinary GF’s: Let f (x) be the ordinary generating function for an : f (x) = ∞
n=0 an x .
n
Multiply both sides of the recurrence relation by x and sum over all n ≥ 0:
∞
∞
∞
X
X
X
n
n
(2x)n ,
an x +
an+2 x = 4
n=0
n=0
n=0
this clearly gives
1
1
(f (x) − 41 x) = 4f (x) +
,
2
x
1 − 2x
so after a simple calculation one gets:
x
1
f (x) =
.
4 (1 − 2x)2
To expand this, write this as
∞
x
2x
1
1
1X
f (x) =
=
=
n(2x)n ,
4 (1 − 2x)2
8 (1 − 2x)2
8 n=0
so an = n2n−3 .
P
Using exponential GF’s: Let F (x) = ∞
n=0
an n
x
n!
be the exponential GF for an ’s. Then
X an+2 xn
F 00 (x) =
.
n!
n≥0
1
2
COMBINATORICS. FINAL EXAM
Multiply both sides of the recurrence relation by xn /n! and sum over all n ≥ 0. One gets the differential
equation with initial conditions:
F 00 (x) = 4F (x) + e2x ,
F (0) = 0,
F 0 (0) = 14 .
The general linear solution is c1 e2x + c2 e−2x , and a partial solution is 41 xe2x . In fact, this partial solution
also satisfies that initial conditions, so one has
∞
∞
1 2x 1 X 2n xn+1
1 X n2n−1 xn
xe =
=
,
4
4 n=0 n!
4 n=0
n!
which gives the same answer for an ’s.
Problem 3. Recall that the number of standard Young tableaux (= the ways to grow a Young diagram
by adding one box at a time) of shape (n, n) is Cn , the nth Catalan number. E.g., for n = 3 we have 5 such
tableaux:
1 2 3
1 2 4
1 2 5
1 3 4
1 3 5
4 5 6
3 5 6
3 4 6
2 5 6
2 4 6
Compute the number of standard Young tableaux of shape (n + 2, n) (you can use the hook formula only
to check your answer). E.g., for n = 2 one has 9 such tableaux:
1 2 3 4
5 6
1 2 3 5
4 6
1 2 5 6
3 4
1 2 3 6
4 5
1 2 4 6
3 5
1 3 4 5
2 6
1 3 5 6
2 4
1 3 4 6
2 5
1 2 4 5
3 6
Solution 3. One solution: See problem set 4, where we count up-right paths on the lattice which do not
go above the diagonal. This problem clearly corresponds to paths from (0, 0) to (n + 2, n), and by the
reflection principle one learns that the number of paths here is
2n + 2
2n + 2
−
.
n
n−1
3
2n + 2
By the hook formula, this is
, and it is clear that these two numbers are equal to each other.
n+3
n
Another solution: Consider standard Young tableaux of shape (n + 2, n + 2), there are Cn+2 such
tableaux. In each of the tableaux the number 2n + 4 is in the lower right corner (see also the next problem).
Where is the number 2n + 3? There are two possibilities:
(1) If 2n + 3 is in the second row, it must be in the (n + 1)-th column. Deleting 2n + 4 and 2n + 3, we
readily get a standard Young tableau of shape (n + 2, n), and every such tableau can be obtained
as explained here.
(2) If 2n + 3 is in the first row, it must be in the last column. Deleting 2n + 3 and 2n + 4, we get a
standard Young tableau of shape (n + 1, n + 1).
Thus, one observes that the number we need is equal to
1
2n + 4
1
2n + 2
Cn+2 − Cn+1 =
−
,
n+3 n+2
n+2 n+1
and it can be readily verified that this is the desired answer.
Problem 4. What is the number of standard Young tableaux of shape (n + 1, n) ?
Solution 4. Observe that if one takes standard Young tableaux of shape (n + 1, n + 1), then the number
2n + 2 must always be in the lower right corner. Deleting this number, one clearly gets a bijection between
standard Young tableaux of shape (n + 1, n) and standard Young tableaux of shape (n + 1, n + 1). The
number of the latter tableaux is Cn+1 .
COMBINATORICS. FINAL EXAM
3
Problem 5. Express the number of partitions of 100 into even parts which are ≥ 3 and ≤ 10 as a suitable
coefficient in a certain generating function.
Solution 5. Since one can take arbitrary number of parts equal to 4, 6, 8, 10, the generating function for
partitions into these parts is
1
.
(1+q 4 +q 8 +. . . )(1+q 6 +q 12 +. . . )(1+q 8 +q 16 +. . . )(1+q 10 +q 20 +. . . ) =
(1 − q 4 )(1 − q 6 )(1 − q 8 )(1 − q 10 )
The desired number of partitions is the coefficient by q 100 in the above generating function.
Problem 6. An arithmetic function f is called strongly multiplicative if f (mn) = f (m)f (n) for all m, n (not
necessaryPrelatively prime). Find a strongly multiplicative function a(n) such that the arithmetic function
b(n) =
a(n) is not strongly multiplicative. Will b(n) be just multiplicative (i.e., b(xy) = b(x)b(y) for
d : d|n
relatively prime x, y)?
Solution 6. Indeed, take a(n) = 1 for all n, then b(n) is the number of divisors of n. We have b(2) = 2, and
b(4) = 3 6= 2 · 2. Of course, b(n) will be multiplicative by the known properties of arithmetic functions.
Problem 7. Let µ(n) be the Möbius arithmetic function. Find (in terms of the zeta function ζ(s)) the
Dirichlet GF for the sequence {|µ(n)|}.
Solution 7. Since |µ(n)| is a multiplicative arithmetic function, and, moreover,
(
1,
if k ≤ 1,
k
|µ(p )| =
0,
otherwise
(for prime p), we can write
∞
X
|µ(n)|
n=1
ns
=
Y
Y 1 − p−2s
ζ(s)
(1 + p−s ) =
=
.
−s
1
−
p
ζ(2s)
p
p
Problem 8. Let d(n) be the number of divisors of n, and let


X
X

a(n) :=
d(m) .
k : k|n
m : m|k
Find the Dirichlet GF for the sequence a(n).
Solution 8. The Dirichlet GF for d(n) is ζ 2 (s). From the formula for coefficients of a product of two Dirichlet
GF’s, we see that each of the summations in the definition of a(n) corresponds to multiplication by ζ(s).
Thus, the Dirichlet GF for a(n) is just ζ 4 (s).
Problem 9. Consider the complete (labeled) graph K1,1,5 which has two distinguished vertices A, B connected to each other, and 5 other vertices C1 , . . . , C5 of degree 2 which are connected to both A and B.
Compute the number of spanning trees for this graph (in the labeled setting, as usual).
Solution 9. One solution: One can write the corresponding Laplace matrix, cut the first and the last row
and get


−6 1
1
1
1
1
0
0
0 
 1 −2 0


0 −2 0
0
0 
 1


0
0 −2 0
0 
 1
 1
0
0
0 −2 0 
1
0
0
0
0 −2
The determinant of this matrix is 112.
Another solution: The spanning trees are of two types:
4
COMBINATORICS. FINAL EXAM
(1) A and B are connected, then A must be connected to some of Ci ’s, and also B. There are 25 = 32
ways to chose such a tree.
(2) A and B are not connected, so they are both connected to some of Ci ’s, and each other vertex of
Ci ’s is attached to A or to B. In this situation we have 5 · 24 such trees.
Summing, we get 112.
Problem 10. Consider htheimodel of gluing of edges of a 2n-polygon. Show that the genus of the resulting
n
surface does not exceed
(integer part).
2
Solution 10. Consider the 2n-polygon. Connect by a line (curve) the edges which you are gluing. A handle
in the surface arises when there is an “intersection” between two curves. I.e., geometrically the “maximum
genus configuration”
arises in the gluing of the form ababcdcd . . . . Thus, the number of handles cannot be
hni
bigger than
.
2
Alternatively, this follows from the Euler graph theorem V − E + F = 2 − 2g, where E = n. For maximal
genus one should have F = 1 and the minimal possible V , which is 1 or 2 depending on the parity of n.
Problem 11. Compute the number of gluings of the hexagon (n = 3) which give the torus.
Solution 11. The total number of gluings of the hexagon is 15 = 3!!, and 5 = C3 of them give the sphere
(the Catalan number). Thus, 10 other give the torus, because by the previous problem, g = 2 cannot occur.
Alternatively, one can use the Harer-Zagier theorem as in the solution of Problem ??.
Supplementary problems
Problem 12. Express the number of integer solutions of the equation
a + b + c = 100
with constraints 60 ≥ a ≥ b ≥ c ≥ 4 as a suitable coefficient in a certain generating function. (Hint: you
will need a two-variable generating function.)
Solution 12. The number of solutions in the problem is clearly the number of partitions of 100 with cor60
Q
1
responding constraints. Consider the generating function
. It is clear that q enumerates the
k
k=4 1 − tq
number of boxes in a partition in the usual way, and t enumerates the number of parts in a partition. This
means that the coefficient by tx q y is the number of partitions of y with x parts,
60 and each
of the parts is
Q
1
between 4 and 60. Thus, the answer to the problem is the coefficient [t3 q 100 ]
.
k
k=4 1 − tq
Problem 13. Let λ be the multiplicative function such that λ(pk ) = (−1)k (p prime). Prove that
(
X
1,
n is a square,
λ(d) =
0,
otherwise.
d : d|n
Solution 13. The Dirichlet GF for λ(n) is
∞
Y
Y
YX
−s
−2s
(1 − p + p − . . . ) =
(−p−s )k =
p
p
Thus, the Dirichlet GF for the sequence
k=0
P
p
Y 1 − p−s
1
ζ(2s)
=
=
.
−s
−2s
1+p
1−p
ζ(s)
p
λ(d) is ζ(2s), and it has the form
d : d|n
∞
X
X 1
1
ζ(2s) =
=
.
n2s n square ns
n=1
Problem 14. Solve Problem ?? for K1,1,n .
COMBINATORICS. FINAL EXAM
5
Solution 14. As in the second solution of Problem ??, we have 2n + n2n−1 spanning trees.
Problem 15. Compute the number of gluings of the octagon (n = 4) which give the torus.
Solution 15. Use Harer-Zagier theorem: find the coefficient by s5 in the expansion
N
1+s
= (1 + 2s + 2s2 + 2s3 + 2s4 + 2s5 + . . . )N .
1−s
To get s5 , one should take some factors from some of N braces. There are 7 ways to do this according to 7
partitions of 5:
(5),
(4, 1),
(3, 2),
(3, 1, 1),
(2, 2, 1),
(2, 1, 1, 1),
(1, 1, 1, 1, 1)
Thus, the term with s5 will be
N −1
N −1
3
N (2s ) + N (2s )(2s) + N (2s )(2s ) + N
(2s )(2s)(2s) +
N (2s2 )(2s2 )(2s)
2
2
N −1
N
+N
(2s2 )(2s)(2s)(2s) +
(2s)(2s)(2s)(2s)(2s).
3
5
5
2
4
2
3
2
We only need N 3 term, because the number of vertices should be 3. This term can only come from the last
four summands, and it is equal to
1
1
6
35
4
· 8 + · 8 − · 16 +
· 32 = .
2
2
6
120
3
Dividing this by 2 and multiplying by 7!! = 105 (as in the Harer-Zagier theorem), one gets 70.