Le calcul double

smo
osm
Olympiades Suisses de Mathématiques
n
k=n kn−1 k−1
(a+b)n=Pn
i=0 n
iaibn−i
Pn
i=0 n
i2=2n n
n
0+n+1 1+n+2 2+. . . +n+r r=n+r+1 r
n
X
k=1 n
kk2.
{1,2,...,33}
a1, a2, . . . , an
n
6×6
2n1,2, . . . , n
k k n ≡0
n≡1 (mod 4)
S={1,2, . . . , n}pn(k)S
k
n
X
k=0
k·pn(k) = n!

0< a1< a2< . . . < a101 <5050
ak, al, am, an
5050 |(ak+al−am−an).
n > 1c1, c2, . . . , cn
a{1,2, . . . , n}
S(a) =
n
X
k=1
ck·a(k).
a b S(a)−S(b)
n!
k n S n
S
P S k S
P
k < 1
2+√2n.
S n (X1, . . . , Xn)X1, . . . , Xn
{1,2,...,1000}a= (X1, . . . , Xn)∈
S
E(a) = X1∪. . . ∪Xn.
X
a∈S
E(a).
a b b ≥3
k
k
k
a≥b−1 2b.