• Non ci sono risultati.

# Letµ be the M¨obius function of number theory

N/A
N/A
Protected

Condividi "Letµ be the M¨obius function of number theory"

Copied!
1
0
0

Testo completo

(1)

Problem 11344

(American Mathematical Monthly, Vol.115, February 2008) Proposed by A. Stadler (Switzerland).

Letµ be the M¨obius function of number theory. Show that ifn is a positive integer and n > 1 then

n

X

j=1

µ(j) = −

⌊(n−1)/2⌋

X

j=1

j

⌊n/(2j+1)⌋

X

k=⌈(n+1)/(2j+3)⌉

µ(k).

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

In the RHS, the range of the integer k is n + 1

2j + 3 ≤ k ≤ n 2j + 1, hence

1 2

n k− 1

− 1 + 1 2k = 1

2

 n + 1 k − 3



≤ j ≤1 2

n k − 1

.

It’s easy to see that there is only one integer j that satisfies the above inequalities and therefore, by exchanging the two sums, the RHS can be written as

⌊(n−1)/3⌋

X

k=1

µ(k) 1 2

n k− 1

= −

n

X

k=1

µ(k) 1 2

n k − 1 so it suffices to prove that

n

X

k=1

µ(k)g(n/k) =X

k≥1

µ(k)g(n/k) = f (n) = [n = 1]

where g(x) = 1 +1

2(x − 1) and f (x) = [⌊x⌋ = 1]. Note that X

j≥1

f (x/j) =X

j≥1

 x j



= 1



=X

j≥1

[(x/2) < j ≤ x] = 1 + 1 2(x − 1)



= g(x),

Finally, sinceP

k|mµ(k) = [m = 1] andP

j≥1

P

k≥1|f (n/(kj))| < ∞ X

k≥1

µ(k)g(n/k) = X

k≥1

µ(k)X

j≥1

f (n/(kj)) =X

j≥1

X

k≥1

µ(k)f (n/(kj))

= X

m≥1

f (n/m)X

k≥1

µ(k)[m = kj] = X

m≥1

f (n/m)X

k|m

µ(k)

= X

m≥1

f (n/m)X

k|m

µ(k) = X

m≥1

f (n/m)[m = 1] = f (n).



Remark. This inversion law for the M¨obius function holds for more general f and g:

g(x) =X

k≥1

f (x/k) if and only if f (x) =X

k≥1

µ(k)g(x/k)

as soon asP

j≥1

P

k≥1|f (n/(kj))| < ∞ (see for example Concrete Mathematics by Graham, Knuth, and Patashnik).

Riferimenti

Documenti correlati

We look for small (sometimes “minimal”) set of generators for the fields K(E[m]) generated by the m-torsion points of an elliptic curve E (where K is any number field).. For m = 3

[r]

[r]

(American Mathematical Monthly, Vol.119, November 2012) Proposed by Mircea Merca (Romania). Let p be the Euler partition function, i.e., p(n) is the number of nondecreasing lists

Fon-Der-Flaass (Russia)

Solution proposed by Roberto Tauraso, Dipartimento di Matematica, Universit` a di Roma “Tor Vergata”, via della Ricerca Scientifica, 00133 Roma, Italy. We first show that if n is

in such a way that U 1 receives at least one ball, and while any balls remain, each successive urn receives at least as many balls as in all the previous urns combined.. Let b(n)

[r]