Problem 11117
(American Mathematical Monthly, Vol.111, December 2004) Proposed by M. Nyblom (Australia).
An integer n is a positive power if there exist integers a and k such that a≥ 1, m ≥ 2, and n = am. Let N(x) denote the number of positive powers n such that 1 ≤ n ≤ x. For real x ≥ 4 and with L= ⌊log2x⌋, show that
N(x) =
L−1
X
k=1
(−1)k+1 X
2≤i1<···<ik≤L
⌊x1/lcm(i1,...,ik)⌋.
Solution proposed by Giulio Francot and Roberto Tauraso,
Dipartimento di Matematica, Universit`a di Roma “Tor Vergata”, via della Ricerca Scientifica, 00133 Roma, Italy.
For m ≥ 2 and for x ≥ 4, let Am(x) be the set of m-powers contained in the interval [1, x]. Since
|Am(x)| = ⌊x1/m⌋ then Am(x) = {1} for m > L = ⌊log2x⌋ ≥ 2 and by the inclusion-exclusion principle
N(x) =
L
[
m=2
Am(x)
=
L−1
X
k=1
(−1)k+1 X
2≤i1<···<ik≤L
|Ai1(x) ∩ · · · ∩ Aik(x)|
Let pα11pα22· · · pαrr be the prime factorization of n then n ∈ Am(x) if and only if 1 ≤ n ≤ x and m| αsfor all s = 1, . . . , r. Hence
Ai1(x) ∩ · · · ∩ Aik(x) = Alcm(i1,...,ik)(x), and therefore
N(x) =
L−1
X
k=1
(−1)k−1 X
2≤i1<···<ik≤L
Alcm(i1,...,ik)(x) =
L−1
X
k=1
(−1)k+1 X
2≤i1<···<ik≤L
⌊x1/lcm(i1,··· ,ik)⌋.