Page 125 - MATINF Nr. 13-14

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Rezolvarea problemelor date la concursul de admitere la Ecole Polytechnique, filiera MPI, sesiunea 2024 125

cu punctul 21b, avem

X x X x
= + O (xln 2 (x)) .
p 1 p 2 x→∞ p 1 p 2
p 1 6=p 2 prime p 1 6=p 2 prime
p 1 p 2 ≤x p 1 ,p 2 ≤x

ˆ
In fine
2
ã
Å
X x X 1 X 1
= x − x ,
p 1 p 2 p p 2
p 1 6=p 2 prime p prim p prim
p 1 ,p 2 ≤x p≤x p≤x
ˆın care
1 1
X X
≤
p 2 p 2
p prim p ≥1
p≤x
convergent˘a s , i (cu punctul 20.c)
Å ã 2 Å Å ãã
X 1 1 2
x = x ln 2 (x) + c 1 + O = x(ln 2 (x)) + O (xln 2 (x)) .
p x→∞ ln (x) x→∞
p prim
p≤x
ˆ
In concluzie, avem
X 2
∗
card {n ∈ N : n ≤ x, p 1 |n s , i p 2 |n} = x (ln 2 (x)) + O (xln 2 (x)) .
x→∞
p 1 ,p 2 ≤x
p 1 6=p 2 prime

22. d. Din punctele 22.b s , i 22.c avem

1 X 2 2
(ω(n)) = (ln 2 (x)) + O (ln 2 (x)) .
x x→∞
n≤x

Cu punctul 22.a obt , inem

1 X 2
(ω (n) − ln 2 (x)) = O (ln 2 (x)) .
x x→∞
n≤x

23. Avem
√
√
card {n ≤ x : n ∈ S} = card S ∩ 1, x + card S ∩ x, x .
Deoarece
√ √
card S ∩ 1, x ≤ x,
avem
1 √
lim card S ∩ 1, x = 0.
x→∞ x
a
Astfel mai r˘amˆane s˘ ar˘at˘am c˘a s , i
1 √
lim card S ∩ x, x = 0.
x→∞ x