denombrement - Institut de Mathématiques de Toulouse

C
n∈N?Tn2n
A
n≥3n A, C, F, H n
D, B, E, G n
Sn2n A 2n
S0= 1
n∈NSn+1 =Sn+ 2 ·32n
n∈NSn= 1 + 32n−1
4
A2n
A2n
n∈N?pnA
2n
lim
n→+∞(p1+p2+··· +pn)
P∩1≤k≤nVk>0P∩1≤k≤nVk=P(V1)·
P(V2/V1)...P(Vn/V1∩ · · · ∩ Vn−1)

p1= 1/3pn=12·7n−2
9nn≥2
E1p2 + √3≤ |1 + z|
Zz2010 −1=0
u∈Zp2 + √3≤ |1 + u|

Tn2n
Tn= 32n
{B, D, E} {A, C, F, H}
{B, D, E, G} {A, C, F, H}
A2n+ 2
A2n
A2n Sn
A2n+ 2 ABA, ADA, AEA 3Sn
2n A 32n−Sn
2n C F H
HDA, HEA, FBA, F EA, CBA, CDA
A2n+ 2 2(32n−Sn)
Sn+1 = 3Sn+ 2(32n−Sn) = Sn+ 2 ·32n
0≤k≤n−1
Sn=S0+ 2(30+ 32+··· + 32(n−1)) = 1 + 2 32n−1
9−1= 1 + 32n−1 4.
pn=1 + 32n−1
4
32n
VnA2n
∩1≤k≤nVk
P∩1≤k≤nVk=P(V1)·P(V2/V1). . . P (Vn/V1∩ ··· ∩ Vn−1)
=P(V1)·P(V2/V1). . . P (Vk/Vk−1). . . P (Vn/Vn−1)
=P(V1)·P(V2/V1)n−1
k
k−1
P(V1)=6/9
P(V2/V1)V1
H, C F H
HGF, HGC, HGH, HDA, HDH, HDC, HEF, HEA, HEH
A C F 21
27 P(V2/V1) = 21/27 = 7/9
P∩1≤k≤nVk=P(V1)·P(V2/V1)n−1=6
9·7 9n−1
=6·7n−1
32n.
32nP∩1≤k≤nVk= 6 ·7n−1
p1=P(Vc
1)=1/3n≥2
pn=PV1∩V2∩ ··· ∩ Vn−1∩Vc
n
=P(V1)·P(V2/V1). . . P (Vk/Vk−1). . . P (Vn−1/Vn−2)P(Vc
n/Vn−1)
=6
9·7 9n−22 9=12 ·7n−2 9n

p1+p2+··· +pn=1
3+
n
X
k=2
12 ·7n−2
9n=1 3+12 92
n−2
X
k=0 7
9k
=1
3+12 81 ·1−(7/9)n−1 1−7/9=1 3+2 31−(7/9)n−1−→
n→∞
1
3+2 3= 1.
z1z= cos(θ) + isin(θ)p2 + √3≤ |1 + z|2 + √3≤
|1 + u|2= (1 + cos(θ))2+ sin2(θ) cos(θ)≥√3
2E{eiθ, θ ∈[−π/6, π/6]}
2011 e2ikπ/2011,(0 ≤
k≤2010) E0≤2kπ
2011 ≤π 60≤k≤2010 12 = 167.5
Ox167 + 167 + 1 = 335
E1335
2010 ∼0.1666...
Y Y