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Problem 11828

(American Mathematical Monthly, Vol.122, March 2015) Proposed by R. Tauraso (Italy).

Let n be a positive integer, and let z be a complex number that is not a kth root of unity for any k with 1 ≤ k ≤ n. Let S be the set of all lists (a1, . . . , an) of n nonnegative integers such that Pn

k=1kak= n. Prove that X

a∈S n

Y

k=1

1

ak!kak(1 − zk)ak =

n

Y

k=1

1 1 − zk. For example, for n = 3 we have

1

6(1 − z)3 + 1

2(1 − z)(1 − z2)+ 1

3(1 − z3) = 1

(1 − z)(1 − z2)(1 − z3).

Solution proposed by Roberto Tauraso, Dipartimento di Matematica, Universit`a di Roma “Tor Vergata”, via della Ricerca Scientifica, 00133 Roma, Italy.

First solution. We have that

F (z, t) :=X

n≥0

tn X

α1+2α2+···=n n

Y

k=1

1

αk!kαk(1 − zk)αk

=

n

Y

k=1

X

αk≥0

tk αk!kαk(1 − zk)αk

= Y

k≥1

exp

 tk k(1 − zk)



= exp

 X

k≥1

tk k(1 − zk)

= exp

 X

j≥0

X

k≥1

(tzj)k k

=Y

j≥0

exp

 X

k≥1

(tzj)k k

=Y

j≥0

exp

 ln

 1

1 − tzj



=Y

j≥0

1

1 − tzj = 1 1 − t

Y

j≥1

1 1 − tzj

Note that [tnzN]F (z, t) is the number of integer partitions of N with at most n parts because it is known that

[tnzN]Y

j≥1

1 1 − tzj

is the number of integer partitions of N with exactly n parts. On the other side,

G(z, t) :=X

n≥0

tn

n

Y

k=1

1 1 − zk

(2)

and [tnzN]G(z, t) is the number of integer partitions of N with at most n parts.

Hence F and G are equal and the identity is proved. 

Second solution. It suffices to prove that

X

σ∈π(In) n

Y

k=1

1 1 −Qk

j=1xσ(j) = X

σ∈π(In) lσ

Y

k=1

1 1 −Q

j∈Ckxj

where σ ∈ π(In) is a permutation of In= {1, 2, . . . , n} which can be written as a product of disjoint cycles C1, C2, . . . , Clσ. The required identity follows by letting xj= z for j = 1, . . . , n.

We note that

n

Y

k=1

1 1 −Qk

j=1xσ(j) = X

a1≥a2≥···≥an≥0 n

Y

j=1

xaσ(j)j ,

and

lσ

Y

k=1

1 1 −Q

j∈Ckxj = X

a1≥0,a2≥0,··· ,a≥0 lσ

Y

k=1

 Y

j∈Ck

xj

ak

.

Hence, if a1= a2= · · · = an then

[xa11xa22· · · xann] X

σ∈π(In) n

Y

k=1

1 1 −Qk

j=1xσ(j) = n! = [xa11x2a2· · · xann] X

σ∈π(In) lσ

Y

k=1

1 1 −Q

j∈Ckxj. In general,

[xa11xa22· · · xann] X

σ∈π(In) n

Y

k=1

1 1 −Qk

j=1xσ(j) =Y

i≥0

(|Ai|!)

where Ai= {j : aj= i} for i ≥ 0. By the above remark,

|Ai|! = [Y

j∈Ai

xajj] X

σi∈π(Ai) lσi

Y

k=1

1 1 −Q

j∈Ck,ixj

.

Therefore

Y

i≥0

(|Ai|!) =Y

i≥0

[Y

j∈Ai

xajj] X

σi∈π(Ai) lσi

Y

k=1

1 1 −Q

j∈Ck,ixj

= [xa11xa22· · · xann] X

σ∈π(In) lσ

Y

k=1

1 1 −Q

j∈Ckxj

where σ is the product of σi ∈ π(Ai) for i ≥ 0. 

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