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# We have that an= n X k=1 k gcd(n, k

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Problem 12114

(American Mathematical Monthly, Vol.126, May 2019) Proposed by Z. Franco (USA).

Letn be a positive integer, and let An = {1/n, 2/n, . . . , n/n}. Let anbe the sum of the numerators inAn when these fractions are expressed in lowest terms. FindP

n=1an/n4.

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

Solution. More generally, we show that for any s > 3

X

n=1

an

ns = ζ(s) 2



1 + ζ(s − 2) ζ(s − 1)

 .

We have that

an=

n

X

k=1

k

gcd(n, k) = 1 +1 2

n−1

X

k=1

k gcd(n, k)+

n−1

X

k=1

k gcd(n, n − k)

!

= 1 +1 2

n−1

X

k=1

k gcd(n, k)+

n−1

X

k=1

n − k gcd(n, k)

!

= 1 + 1 2

n−1

X

k=1

n

gcd(n, k) = 1 + bn

2 where

bn=

n

X

k=1

n

gcd(n, k)=X

d|n

nϕ(n/d)

d =X

d|n

dϕ(d)

is multiplicative . Since for any prime p and for any positive integer r,

bpr = 1 +

r

X

k=1

pkϕ(pk) = 1 +

r

X

k=1

(p2k− p2k−1) = p2r+1+ 1 p + 1 ,

it follows that for s > 3, the Dirichlet series of the multiplicative function bn reduces to

X

n=1

bn

ns =Y

p

1 +

X

r=1

bpr

prs

!

=Y

p

1 + p p + 1

X

r=1

1

p(s−2)r + 1 p + 1

X

r=1

1 psr

!

=Y

p

 1 + p

p + 1· 1

ps−2− 1 + 1

p + 1 · 1 ps− 1



=Y

p

1 −ps−11



1 − p1s 

1 −ps−21

 =

ζ(s)ζ(s − 2) ζ(s − 1) . Therefore, we may conclude that

X

n=1

an

ns =1 2

X

n=1

1 ns +

X

n=1

bn

ns

!

=ζ(s) 2



1 +ζ(s − 2) ζ(s − 1)

 .



Riferimenti

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