Two unexpected integralities

2.9.1. The first integrality

We shall illustrate strong induction on two further examples, which both have the form of an “unexpected integrality”: A sequence of rational numbers is defined recursively by an equation that involves fractions, but it turns out that all the en-tries of the sequence are nevertheless integers. There is by now a whole genre of such results (see [Gale98, Chapter 1] for an introduction⁶⁷), and many of them are connected with recent research in the theory of cluster algebras (see [Lampe13] for an introduction).

The first of these examples is the following result:

Proposition 2.64. Define a sequence (t₀,t₁,t₂, . . .) of positive rational numbers recursively by setting

t0=1, t1 =1, t2=1, and tn = ¹+tn−1tn−2

tn−3 for eachn ≥3.

67See also [FomZel02] for a seminal research paper at a more advanced level.

(Thus,

t₃ = ¹+t₂t₁ t0

= ¹+1·1 1 =2;

t₄ = ¹+t₃t₂ t1

= ¹+2·1 1 =_3;

t5 = ¹+t4t3

t2

= ¹+3·2 1 =7;

t6 = ¹+t5t4

t3

= ¹+7·3 2 =11, and so on.) Then:

(a)We have t_n+2 =4tn−t_n−2for each n∈ _Z≥2. (b)We have tn ∈_Nfor eachn ∈_N.

Note that the sequence (t0,t1,t2, . . .) in Proposition 2.64 is clearly well-defined, because the expression 1+tn−1tn−2

tn−3

always yields a well-defined positive ratio-nal number when tn−1,tn−2,tn−3 are positive rational numbers. (In particular, the denominator t_n−3 of this fraction is 6= 0 because it is positive.) In contrast, if we had set tn = ¹−_tn−1tn−2

tn−3 instead of tn = ¹+_tn−1tn−2

tn−3 , then the sequence would not be well-defined (because then, we would get t₃ = ¹−1·1

1 = 0 and

t6 = ¹−t5t4

t3

= ¹−t5t4

0 , which is undefined).

Remark 2.65. The sequence (t0,t1,t2, . . .) defined in Proposition 2.64 is the se-quence A005246 in the OEIS (Online Encyclopedia of Integer Sese-quences). Its first entries are

t0 =1, t1 =1, t2 =1, t3=2, t4=3, t5 =7, t6 =11, t7 =26, t8=41, t9=97.

Proposition 2.64 (b) is an instance of the Laurent phenomenon (see, e.g., [FomZel02, Example 3.2]).

Part(a)of Proposition 2.64 is proven by a (regular) induction; it is part(b)where strong induction comes handy:

Proof of Proposition 2.64. First, we notice that the recursive definition of the sequence (t0,t1,t2, . . .)yields

t3 = ¹+t3−1t3−2

t₃−3

= ¹+t2t1

t₀ = ¹+1·1

1 (sincet0=1 andt1 =1 andt2=1)

=_2.

Furthermore, the recursive definition of the sequence(t₀,t₁,t₂, . . .)yields

(a)We shall prove Proposition 2.64(a)by induction onn starting at 2:

Induction base: We have already shown that t₂+2 = 4t₂−t₂−2. In other words, Proposition 2.64(a)holds for n=2. This completes the induction base.

Induction step: Let m ∈ _Z≥2. Assume that Proposition 2.64 (a)holds for n = m.

We must prove that Proposition 2.64(a)holds forn =m+1.

We have assumed that Proposition 2.64 (a)holds for n =m. In other words, we havetm+2 =4tm−tm−2. completes the induction step. Hence, Proposition 2.64(a)is proven by induction.

(b)We shall prove Proposition 2.64(b)by strong induction on nstarting at 0:

Induction step: Let m ∈ _N. ⁶⁸ Assume that Proposition 2.64 (b)holds for every n ∈ _{N satisfying n} < m. We must now show that Proposition 2.64 (b) holds for n=m.

We have assumed that Proposition 2.64 (b) holds for every n ∈ _N satisfying n<m. In other words, we have

tn ∈_Nfor everyn ∈_Nsatisfyingn <m. (99) We must now show that Proposition 2.64 (b) holds for n = m. In other words, we must show thattm ∈ _N.

Recall that(t0,t₁,t2, . . .) is a sequence of positive rational numbers. Thus,tm is a positive rational number.

We are in one of the following five cases:

Case 1: We havem=0.

Case 2: We havem=1.

Case 3: We havem=_2.

Case 4: We havem=3.

Case 5: We havem>3.

Let us first consider Case 1. In this case, we havem =_{0. Thus,} tm =t₀ =₁∈ _N.

Hence,tm ∈ _Nis proven in Case 1.

Similarly, we can prove tm ∈ _N in Case 2 (using t1 = 1) and in Case 3 (using t₂ =1) and in Case 4 (usingt₃ =2). It thus remains to provetm ∈_{Nin Case 5.}

So let us consider Case 5. In this case, we havem >3. Thus,m ≥4 (sincemis an integer), so that m−2 ≥ 4−2 =2. Thus, m−2 is an integer that is ≥ 2. In other words, m−₂ ∈ _Z≥2. Hence, Proposition 2.64 (a) (applied to n = m−_{2) yields} t_(m−2)+2 =4tm−2−t_(m−2)−2. In view of(m−2) +2=m and (m−2)−2=m−4, this rewrites astm =4tm−2−tm−4.

Butm ≥_{4, so that}m−₄∈ _{N, and}m−₄<m. Hence, (99) (applied to n=m−₄₎ yieldst_m−4 ∈_N⊆Z. Similarly,tm−2 ∈_Z.

So we know thattm−2and tm−4are both integers (sincetm−2∈ _Zandtm−4 ∈ _Z).

Hence, 4t_m−2−t_m−4 is an integer as well. In other words,tm is an integer (because tm = 4tm−2−t_m−4). Since tm is positive, we thus conclude that tm is a positive integer. Hence, tm ∈N. This shows thattm ∈_Nin Case 5.

We now have proventm ∈ _Nin each of the five Cases 1, 2, 3, 4 and 5. Since these five Cases cover all possibilities, we thus conclude that tm ∈ _N always holds. In other words, Proposition 2.64 (b) holds for n = m. This completes the induction step. Thus, Proposition 2.64(b)is proven by strong induction.

2.9.2. The second integrality

Our next example of an “unexpected integrality” is the following fact:

68In order to match the notations used in Theorem 2.60, we should be saying “Letm∈_Z≥0” here, rather than “Letm∈N”. But of course, this amounts to the same thing, sinceN=_Z≥0.

Proposition 2.66. Fix a positive integer r. Define a sequence (b0,b₁,b2, . . .) of positive rational numbers recursively by setting

b₀=1, b₁ =1, and bn = ^b

rn−1+1 bn−2

for eachn ≥2.

(Thus,

b₂= ^b

1r +1 b0

= ¹

r+1 1 =2;

b3= ^b

2r +1 b1

= ²

r+₁

1 =2^r+1;

b₄= ^b

r 3+1

b2

= (2^r+1)^r+1

2 ,

and so on.) Then:

(a)We have bn ∈_Nfor eachn ∈ _N.

(b)If r≥_{2, thenbn} | _bn−2+bn+2 for eachn ∈_Z≥2.

Remark 2.67. If r = 1, then the sequence (b0,b1,b2, . . .) defined in Proposition 2.66 is

(1, 1, 2, 3, 2, 1, 1, 2, 3, 2, 1, 1, 2, 3, 2, . . .)

(this is a periodic sequence, which consists of the five terms 1, 1, 2, 3, 2 repeated over and over); this can easily be proven by strong induction. Despite its sim-plicity, this sequence is the sequence A076839 in the OEIS.

Ifr =2, then the sequence(b₀,b₁,b₂, . . .) defined in Proposition 2.66 is (1, f₁, f₃, f₅, f₇, . . .) = (1, 1, 2, 5, 13, 34, 89, 233, 610, 1597, . . .)

consisting of all Fibonacci numbers at odd positions (i.e., Fibonacci numbers of the form f_2n−1 for n ∈ N) with an extra 1 at the front. This, again, can be proven by induction. Also, this sequence satisfies the recurrence relation bn =3b_n−1−b_n−2for all n ≥2. This is the sequence A001519 in the OEIS.

Ifr =3, then the sequence(b0,b₁,b2, . . .) defined in Proposition 2.66 is (1, 1, 2, 9, 365, 5403014, 432130991537958813, . . .);

its entries grow so fast that the next entry would need a separate line. This is the sequence A003818 in the OEIS. Unlike the cases of r = 1 and r = 2, not much can be said about this sequence, other than what has been said in Proposition 2.66.

Proposition 2.66(a)is an instance of the Laurent phenomenon for cluster algebras (see, e.g., [FomZel01, Example 2.5]; also, see [Marsh13] and [FoWiZe16] for ex-positions). See also [MusPro07] for a study of the specific recurrence equation from Proposition 2.66 (actually, a slightly more general equation).

Before we prove Proposition 2.66, let us state an auxiliary fact:

Proof of Proposition 2.66. First, we notice that the recursive definition of the sequence (b0,b1,b2, . . .) yields

Let us first prove the following observation:

Observation 1: Each integern ≥2 satisfiesbn+2 =bn−2b^r_n+1−b^r_n⁻¹H(b_n^r).

[Proof of Observation 1: Let n ≥ 2 be an integer. Thus, n ≥ 2 ≥ 1. Thus, (100) rational numbers). Thus, b^r_n is also a positive rational number. Hence, b^r_n ∈ _Q is nonzero. The definition of H(b_n^r) yieldsH(b_n^r) = (b^r_n+₁)^r−₁

(a)We shall prove Proposition 2.66(a)by strong induction on n:

Induction step: Let m ∈ N. Assume that Proposition 2.66 (a) holds for every n ∈ _{N satisfying n} < m. We must now prove that Proposition 2.66 (a) holds for n=m.

We have assumed that Proposition 2.66 (a) holds for every n ∈ _N satisfying n<m. In other words, we have

bn ∈_Nfor everyn ∈_Nsatisfyingn <m. (104) We must now show that Proposition 2.66 (a) holds for n = m. In other words, we must show thatbm ∈ _N.

Recall that(b0,b₁,b2, . . .) is a sequence of positive rational numbers. Thus, bm is a positive rational number.

We are in one of the following five cases:

Case 1: We havem=0.

Case 2: We havem=1.

Case 3: We havem=_2.

Case 4: We havem=3.

Case 5: We havem>3.

Let us first consider Case 1. In this case, we havem=_{0. Thus,}bm =b₀=₁∈ _N.

Hence,bm ∈ _Nis proven in Case 1.

Similarly, we can prove bm ∈ _N in Case 2 (using b1 = 1) and in Case 3 (using b₂ =2) and in Case 4 (usingb₃ =₂^r+1). It thus remains to provebm ∈ _{Nin Case} 5.

So let us consider Case 5. In this case, we havem >3. Thus,m ≥4 (sincemis an integer), so that m−₂ ≥ ₄−₂ = 2. Hence, Observation 1 (applied to n = m−₂₎ yields b(m−2)+2 = b(m−2)−2b^r_(m−2)+1−b^r_m⁻₋¹₂H b^r_m−2

. In view of (m−2) +2 = m and (m−2)−2=m−4 and(m−2) +1=m−1, this rewrites as

bm =b_m−4b^r_m−1−b^r_m⁻₋¹₂H b^r_m−2

. (105)

But m−2 ∈ _N (since m ≥ 4 ≥ 2) and m−2 < m. Hence, (104) (applied to n = m−_{2) yields} b_m−2 ∈ _N⊆ _{Z. Also,} b_m−2 is a positive rational number (since (b0,b₁,b2, . . .)is a sequence of positive rational numbers) and thus a positive integer (sincebm−2 ∈N), hence a nonzero integer. Thus,b^r_m−2is a nonzero integer as well.

Therefore, Lemma 2.68 (applied to x = b^r_m−2) shows that H b_m^r₋₂

∈ Z. In other words, H b^r_m−2

is an integer. Also, r−1 ≥ 0 (since r ≥ 1), and thus r−1 ∈ _N.

Hence,b^r_m⁻₋¹₂is an integer (since bm−2 is an integer).

Also, m−4 ∈ _N (since m ≥ 4) and m−4 < m. Hence, (104) (applied to n = m−4) yieldsb_m−4∈ _N⊆Z. In other words,b_m−4is an integer.

Similarly,bm−1 is an integer. Thus, b^r_m−1 is an integer.

We now know that the four numbersb_m−4,b_m^r₋₁,b_m^r−₋¹₂and H b^r_m−2

are integers.

Thus, the numberb_m−4b^r_m−1−b^r_m⁻₋¹₂H b^r_m−2

also is an integer (since it is obtained from these four numbers by multiplication and subtraction). In view of (105), this

rewrites as follows: The number bm is an integer. Since bm is positive, we thus conclude thatbm is a positive integer. Hence, bm ∈ N. This shows that bm ∈ _Nin Case 5.

We now have proven bm ∈ _N in each of the five Cases 1, 2, 3, 4 and 5. Thus, bm ∈ _N always holds. In other words, Proposition 2.66 (a) holds for n = m.

This completes the induction step. Thus, Proposition 2.66 (a) is proven by strong induction.

(b)Assume thatr≥2. We must prove thatbn | bn−2+bn+2 for eachn ∈_Z≥2. So let n∈ Z≥2. We must show that bn |bn−2+bn+2.

Fromn∈ _Z≥2, we obtain n≥2, so that n−2∈ _N.

Proposition 2.66 (a) (applied to n−2 instead of n) yields bn−2 ∈ N. Similarly, bn ∈ N and bn+1 ∈ N and bn+2 ∈ N. Thus, all of bn−2, bn+1, bn are bn+2 are integers.

Butbn is a positive rational number (since(b0,b1,b2, . . .)is a sequence of positive rational numbers), and therefore a positive integer (sincebn is an integer). Hence, b^r_nis a positive integer, and thus a nonzero integer. Therefore, Lemma 2.68 (applied tox =b^r_n) shows that H(b^r_n)∈ Z. In other words, H(b^r_n) is an integer.

We haven+2≥2. Hence, the recursive definition of the sequence (b0,b1,b2, . . .) yieldsbn+2 = ^b

r

(n+2)−1+1 b_(n+₂)−2

= ^b

r

n+1+1

bn . Multiplying this equality by bn, we obtain bnbn+₂ =b^r_n+1+1. (106) We haver−₂∈ _N(sincer≥_{2) and}bn ∈ _{N. Hence,} b^r_n⁻² is an integer.

Observation 1 yields bn+2 =bn−2b^r_n+1−b^r_n⁻¹H(b^r_n). Thus, bn+2

| {z }

=bn−2b^r_n+1−b^r−1_n H(b^r_n)

+bn−2

=bn−2b^r_n+1−b^r_n⁻¹H(b^r_n) +bn−2 =bn−2b^r_n+1+bn−2

| {z }

=bn−2(^br_n+1⁺¹)

−b^r_n⁻¹H(b^r_n)

=bn−2 b^r_n+1+1

| {z }

=bnb_n+2 (by (106))

− b^r_n⁻¹

|{z}

=bnb^r−2_n

H(b^r_n) = bn−2bnbn+2−bnb^r_n⁻²H(b^r_n)

=bn

bn−2bn+2−b^r_n⁻²H(b^r_n). (107) But bn−2bn+2−_bn^r−2_H(b^r_n) is an integer (because bn−2, bn+2, b^r_n⁻² and H(b^r_n) are integers). Denote this integer byz. Thus, z = bn−2bn+₂−b^r_n⁻²H(b_n^r). Since bn and

zare integers, we have bn | bn z

|{z}

=bn−2bn+2−b^r−2n H(b^rn)

=bn

b_n−2b_n+2−b^r_n⁻²H(b^r_n)=b_n+2+b_n−2 (by (107))

=bn−2+bn+₂. This proves Proposition 2.66(b).

For a (slightly) different proof of Proposition 2.66, seehttp://artofproblemsolving.

com/community/c6h428645p3705719.

Remark 2.69. You might wonder what happens if we replace “b^r_n−1 +_{1” by}

“b^r_n−1+q” in Proposition 2.66, where q is some fixed nonnegative integer. The answer turns out to be somewhat disappointing in general: For example, if we set r=_{3 and}q =2, then our sequence (b₀,b₁,b₂, . . .) begins with

b0=1, b1 =1, b2= ¹

3+2 1 =3, b3= ³

3+2

1 =29, b4 = ²⁹

3+2

3 = ^{24 391} 3 ,

at which point it becomes clear that bn ∈ _Nno longer holds for all n ∈ _{N. The} same happens for r = 4, 5, . . . , 11 (though the first n that violates bn ∈ _Nis not always 4); this makes me suspect that it also happens for allr >2.

However, bn ∈ _N still holds for all n ∈ _N when r = 2. This follows from Exercise 2.1 below.

Exercise 2.1. Fix a nonnegative integer q. Define a sequence (b₀,b₁,b₂, . . .) of positive rational numbers recursively by setting

b0 =1, b1 =1, and bn = ^b

2n−1+q bn−2

for eachn ≥2.

(Thus,

b2= ^b

12+q b₀ = ¹

2+q

1 =q+1;

b3= ^b

22+q b1

= (q+₁)²+q

1 =q²+3q+1;

b₄= ^b

32+q b₂ = ^q

2+3q+12

+q

q+1 =q³+5q²+6q+1;

and so on.) Prove that:

(a)We have bn = (q+₂)b_n−1−b_n−2 for eachn∈ _Z≥2. (b)We have bn ∈_Nfor eachn ∈ _N.

Im Dokument Notes on the combinatorial fundamentals of algebra (Seite 122-132)