MATH IN MOSCOW
5. FIBONACCI NUMBERS.
COMBINATORICS, SPRINT 2016
1. Synopsis
Fibonacci numbers is a sequence of integers x0 ; x1 ; : : : such that x0 = 0, x1 = 1 and xn = xn−1 + xn−2 for all
n ≥ 2.
P
t
n
Theorem 1. Let X = ∞
n=0 xn t . Then X = 1−t−t2 .
P∞
n+2 + tn+1 − tn ) = −x +(x − x )t + P∞ (−x + x
n
0
0 1
n
n−1 + xn−2 )t = −t. ¤
n=2
n=0 xn (t
³¡ √ ¢
¡ √ ¢ ´
Theorem 2. xn = √15 1+2 5 n − 1−2 5 n .
√
√
2
Proof. The numbers '1 = (−1 + 5)=2 and '2 = −(1 + 5)=2 ³are roots of the
t2 +
´ polynomial
³ t + t − 1, so ´
'
'
1
1
1
1
t
1
2
t − 1 = (t − '1 )(t − '2 ). Therefore one has 1−t−t2 = '2 −'1 t−'1 − t−'2 = '1 −'2 1−t='1 − 1−t='2 =
´
³
1 P∞
1 − 1 tn . Once 1=' = (1 + √5)=2 and 1=' = (1 − √5)=2 and ' − ' = √5, the result
1
2
1
2
n
n
'1 −'2 n=0 '1
'2
follows.
¤
√
1+
5
Corollary 1. limn→∞ xn =xn−1 = 2 (the \golden ratio").
´n ³
³√
¯ √ ¯
¡ 1−√5 ¢n ´
¡ √ ¢
√
1
−
Proof. xn = √15 5+1
. Since ¯ 11+−√55 ¯ < 1, one has yn def
= 11+−√55 n → 0 as n → ∞. Therefore
2 √
1+ 5
√
limn→∞ xn =xn−1 = 1+2 5 limn→∞ 1+1+ynyn−1 = 1+2 5 .
¤
Proof. One has (t2 +t−1)X =
A sequence xn is called a generalized Fibonacci sequence (abbreviate it as GFS) if xn = xn−1 + xn−2 for all
n ≥ 2.
Theorem 3. xn is a GFS if and only if xn = A='n1 + B='n2 for some A; B ∈ C.
Proof of the \if" part of the theorem. Note rst that if xn is a GFS, and A ∈ C then Axn is a GFS, too. Also, if
xn and yn are GFSs then their sum xn + yn also is. Thus to prove the \if" part of the theorem it suces to check
that the sequences xn = 1='n1 and xn = 1='n2 are GFS.
Let i = 1; 2. Indeed, xn −xn−1 −xn−2 = '1ni (1 −'i −'2i ) = 0 because 'i are roots of the equation 1−t−t2 = 0. ¤
Give two proofs of the \only if" part of the theorem.
Proof 1 of the \only if" part. Let xn be a GFS. Find A and B such that the sequence yn = A='n1 + B='n2 satises
the equations y0 = x0 and y1 = x1 . Indeed, it means that A + B = x0 and A='1 + B='2 = x1 . This is
a system of linear equations with respect to the variables A and B ; for any x0 ; x1 it has the unique solution
A = '2'−1'1 (x1 '2 − x0 ); B = '1'−2'2 (x1 '1 − x0 ). So, both xn and yn are GFS, and have two common initial terms.
An evident induction shows that xn = yn for all n.
¤
P
n
Proof 2 of the \only if" part. Let xn be a GFS; denote X = ∞
n=0 xn t . Then, similar to the proof of Theorem
(
x
−x0 )t−x0
(
x
−x0 )t−x0
1
1
1. obtain the equality X = t2 +t−1 = (t−'1 )(t−'2 ) . Prove that there exist constants P and Q such that
X = P=(t − '1 ) + Q=(t − '2 ). Indeed, this is equivalent to the equality P (t − '2 ) + Q(t − '1 ) = (x1 − x0 )t − x0
between polynomials of degree 1. Taking t = '1 and t = '2 , obtain P = ((x1 − x³0 )'2 − x0 )=('´2 − '1 ) and
P
P∞
P='1
Q='2
Q
P
n
Q = ((x1 − x0 )'1 − x0 )=('1 − '2 ). So, ∞
n=0 xn t = − 1−t='1 − 1−t='2 = − n=0 'n1 +1 + 'n2 +1 . So, xn =
A='n1 + B='n2 where A = −P='1 and B = −Q='2 .
¤
2. Exercises
Exercise 1. Prove that the number of subsets A ⊂ {1; : : : ; n} containing no two adjacent elements (that is, if
a ∈ A then a ± 1 ∈= A), is the Fibonacci number.
Exercise 2. Prove that the number of ways to tile the rectangle 2 × n by rectangles 1 × 2 is the Fibonacci number.
Exercise 3. Prove that the number of ways to represent n − 1 as a sum of 1s and 2s (the order of summands
matters) is the n-th Fibonacci number. For example, 4 = 2 + 2 = 2 + 1 + 1 = 1 + 2 + 1 = 1 + 1 + 2 = 1 + 1 + 1 + 1
| totally 5 = x3 representations.
1
Exercise 4. Prove that the iterated fraction 1 + 1+
1
1+
1
1
1+
numbers.
(n levels) is the ratio of two subsequent Fibonacci
...
Exercise 5. (a) Prove that xn is even if and only if n is divisible by 3. (b) For what n is xn divisible by 3?
(c) Prove that for any m the sequence of residues xn mod m is periodic. (d) Let yn = xn mod 5. Prove that
P
t
n
n−1 mod 5.
Y def
= ∞
n=0 yn t = (t−2)2 ∈ (Z=5Z)[[t]], and yn = n · 3
3. Homework
The problems marked with an asterisk do not inuence the homework score. Such problem require more thinking
than usual: they may be more dicult, deeper, or just less accurately formulated.
Problem 1. Prove the following identities.
P Try ton give at least two proofs: by induction (or using the explicit
formula) and using generating series X = ∞
n=0 xn t for Fibonacci numbers.
2
(a) x0 + · · · + xn = xn+2 − 1, (b) xn − xn+1 xn−1 = (−1)n+1 , (c) x2n+1 − x2n−1 = x2n .
P∞ ¡n−k−1¢
k=0
k
¡ ¢
(recall that if m > n then mn = 0, so the sum is actually nite).
Try to give at least two proofs: using induction and using Exercise 1.
Problem 3*. Prove that for any n and k the number xnk is divisible by xn .
Problem 2. Prove that xn =