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Lesson 6: Sequences. Adithya B., Brian L., William W., Daniel X. June 2020

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Lesson 6: Sequences

Adithya B., Brian L., William W., Daniel X.

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Problem of the Week

PotW Define f(n) = ∞ X i=0 gcd(n,i)·i xi Iff(n) = Pn(x)

Qn(x) for integer polynomials Pn,Qn which don’t share any

common factors other than ±1, find the minimal positive integern such

that 2019|Pn(1)

This sum at first glance looks pretty intractable. In particular, we want to get rid of gcd(n,i)

We can try splitting the sum into sums based on gcd. For example, we want to write it as f(n) =X d|n ∞ X i=0 d·(di) xdi

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Problem of the Week

This is incorrect! The coefficient of (i)xi is now

P

d|gcd(n,i)

d instead of

gcd(n,i).

However, recall that P

d|gcd(n,i) φ(d) = gcd(n,i), so if we write f(n) = ∞ P i=0 i xi P d|gcd(n,i)

φ(d) we can now sumswap as

X d|n φ(d) ∞ X i=0 (di) xdi Recall ∞ P i=0 i xi = x (x−1)2, so this inside is dxd (xd1)2 So,f(n) =P d|n dφ(d)xd

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Problem of the Week

All denominators go into (xn1)2, so we should have

Qn(x) = (xn−1)2

This means that

Pn(x) = X d|n dφ(d)xd(xn−1)2 (xd1)2 = X d|n dφ(d)xd(1 +xd+. . .+xn−d)2

When we plug in 1, we get

Pn(x) = X d|n dφ(d)n d 2 =nX d|n φ(d)n d

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Problem of the Week

P

d|pk

φ(d)dn =pk+pk−1(p1) +pk−2·p(p1) +. . ., which is

pk−1(p+k(p−1))

So, if q|Pn(1), we either have q|n or q|(p+k(p−1)) for somep|n.

Now, we are ready to finish. If 2019|Pn(1), we need 3,673|Pn(1).

How can we get 3?

We either choose 3|n or 3|(p+k(p−1)). (p,k) = (2,1) works and 21 <3

For 673, if 673|(p+k(p−1)), then we should have

(k+ 1)(p−1)>673. This basically shows that ifk >1, then pk will

be at least k+1673k which is much too large. On the other hand,

k = 1 admits (p,k) = (337,1) which is small enough

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Basic Sequences

An arithmetic sequence is a sequence in which the difference between any two consecutive terms is a constant

a,a+d,a+ 2d,a+ 3d, . . .

a is the first term,d is thecommon difference

Each term is the average of the two adjacent terms

Ageometric sequence is a sequence in which the ratio between any consecutive two terms is a (nonzero ) constant

a,ar,ar2,ar3, . . .

a is the first term,r is the common ratio

Each term is the geometric mean of the two adjacent terms

It is very useful to write arithmetic/geometric sequences in terms of two parameters: one term and the common difference/ratio

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Basic Sequences

2003 AIME I #8

In an increasing sequence of four positive integers, the first three terms form an arithmetic progression, the last three terms form a geometric progression, and the first and fourth terms differ by 30. Find the sum of the four terms.

Let’s write the first three terms as a−d,a,a+d instead of

a,a+d,a+ 2d

This makes the fourth term (a+d)a 2:

a−d,a,a+d,(a+d) 2

a

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2003 AIME I #8

We can simplify: (a+d)2−a(a−d) = 30a d2+ 3ad = 30a

Let’s solve for ain terms ofd because this is a linear equation:

a= 30d23d

We know a,d must be positive integers, so 1≤d ≤9

We can narrow down d more: since 30−3d is divisible by 3 we know

that d2 is divisible by 3 So d ∈ {3,6,9}

d = 3 =⇒ a= 37

d = 6 =⇒ a= 3; but a−d <0 so this doesn’t work

d = 9 =⇒ a= 27; gives

18,27,36,48 So answer is 18 + 27 + 36 + 48 = 129

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Basic Sequences

2004 AIME II #9

A sequence of positive integers with a1 = 1 anda9+a10= 646 is formed

so that the first three terms are in geometric progression, the second, third, and fourth terms are in arithmetic progression, and, in general, for alln ≥1, the terms a2n−1,a2n,a2n+1 are in geometric progression, and

the terms a2n,a2n+1, anda2n+2 are in arithmetic progression. Let an be

the greatest term in this sequence that is less than 1000. Find n+an.

Not completely arithmetic or geometric, but we can write

a1 = 1,a2 =r,a3=r2 a2,a3,a4 in arithmetic progression soa4 = 2a3−a2 = 2r2−r a3,a4,a5 in geometric progression soa5 = a2 4 a3 = (2r2r)2 r2 = (2r−1)2 a4,a5,a6 in arithmetic progression so a6 = 2a5−a4 = 2(2r−1)2−r(2r−1) = (3r−2)(2r−1)

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2004 AIME II #9

a5,a6,a7 in geometric progression so a7 = a2 6 a5 = (3r−2)2(2r1)2 (2r−1)2 = (3r−2)2 Can continue: a8= (4r−3)(3r−2) a9= (4r−3)2 a10= (5r−4)(4r−3) We know a9+a10= 646 so (4r−3)2+ (5r−4)(4r−3) = 646

Solve this to get the positive solutionr = 5

Now we can calculate all terms of the sequence: 1,5,25,45,81, . . .

Can get that a16= 957 is largest term less than 1000 so answer

16 + 957 = 973

Recommend trying the calculations on your own; good exercise in keeping computations neat and expressions factored

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Recursions

Broadly speaking, recursions are sequences where thenth term is

defined by previous ones, such asxn =xn−1+. . .+x0 orxn= xn11+ 1

As they are a very broad class of sequences, there is no particular global method. However, keep an eye out for

Periodicity: Check if a recursive equation might actually be hiding periodicity, such asxn=xn−11 for an easy example

Large scale behavior: See how the sequence behaves as a whole Generating functions: Many sequences are susceptible to be solved using generating functions, in a method similar to Snake Oil

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Recursions

2012 AIME I #11

Let f1(x) = 233x+13 , and forn≥2, define fn(x) =f1(fn−1(x)). The

value ofx that satisfies f1001(x) =x−3 can be expressed in the form mn,

where mand n are relatively prime positive integers. Find m+n.

To get a better feel for the sequence, let’s write out a few of the terms and see if we notice anything special.

Note that f1(x) = 2 3− 3 3x+ 1, f2(x) = 2 3− 3 3(233x+13 ) + 1 = 2 3 − 3x+ 1 3x−2, f3(x) = 2 3− 3 3(23 −3x+1 3x−2) + 1 =x.

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2012 AIME I #11

Then,f4(x) =f1(x), so we have the period is 3.

f1001(x) =f2(x) = 23 −3x+13x−2 =x−3.

Multiply both sides by 3(3x−2):

2(3x−2)−3(3x+ 1) = 3(3x−2)(x−3)

−3x−7 = 9x2−33x+ 18 =⇒ 9x2−30x+ 25 = 0

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Recursions

2018 PUMaC Algebra #7

Let the sequence {an}∞n=−2 satisfya−1 =a−2= 0, a0= 1, and for all

non-negative integers n, n2 = n X k=0 an−kak−1+ n X k=0 an−kak−2.

Given a2018 is rational, find the maximum integer m such that 2m divide

the denominator of the reduced form of a2018.

The sums look sort of similar to the Catalan number recursion:

Cn+1 =Pnk=0CkCn−k.

Let’s try generating functions. Let A(x) =P∞n=0anxn.

The sum in the question can be written as

n2=Pn

k=0an−k(ak−1+ak−2).

Let’s definebk =ak−1+ak−2 so thatn2 =

Pn

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2018 PUMaC Algebra #7

Define the generating function forbi to be B(x). Let’s find A(x)B(x).

(a0+a1x2+a2x2+· · ·)(b0+b1x+b2x2+· · ·) =a0b0+ (a0b1+a1b0)x+ (a0b2+a1b1+a2b0)x2+· · · = 02+ 12·x+ 22·x2+· · ·= ∞ X n=0 n2xn

Remember the following: 2 (1−x)3 = ∞ X n=0 2 n+ 2 2 xn= ∞ X n=0 (n2+ 3n+ 2)xn. 1 (1−x)2 = ∞ X n=0 n+ 1 1 xn= ∞ X n=0 (n+ 1)xn 1 1−x = ∞ X n=0 xn

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2018 PUMaC Algebra #7

Therefore, ∞ X n=0 n2xn= 2 (1−x)3 − 3 (1−x)2 + 1 1−x = x+x2 (1−x)3 =A(x)B(x)

Now, our goal is to find A(x), so we want to find a way to relateA(x) andB(x). B(x) = ∞ X n=0 (an−1+an−2)xn=x ∞ X n=0 an−1xn−1+x2 ∞ X n=0 an−2xn−2 = (x+x2)A(x) Therefore,A(x)B(x) = (x+x2)A(x)2= (1x+xx)23, so A(x) = 1 (1−x)32 .

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2018 PUMaC Algebra #7

Lemma 1 (1−x)32 = ∞ X n=0 2n+ 1 4n 2n n xn. Proof.

Try to expand with the binomial theorem! Note thata2018 = 403742018 4036 2018 . Note that ν2 4036 2018 =ν2(4036!)−2ν2(2018!) = 4029−2·2011 = 7.

There are 4036 powers of 2 in the denominator and 7 in the

numerator, so the largest power in the denominator after simplifying is 24029.

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Linear Recursions

A linear recurrence is one of the form

an=c1an−1+c2an−2+. . .+ckan−k

Linear recurrences appear a lot, and one of the most important steps towards understanding them is to find a closed form.

Let’s use ideas from last week on generating functions to try to find a closed form!

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Linear Recursions

ConsiderS =P∞i=0aixi. What isc1xS+c2x2S+. . .+ckxkS?

The coefficient of xn for n≥k in this expression is

c1an−1+c2an−2+. . .+ckan−k =an. So, it is aroundS. However,

we aren’t guaranteed thatxn for n<k has the right coefficient, so we actually have

c1xS+. . .+ckxkS =S−P(x)

for P(x) with degree<k

So, solving forS,

S = P(x)

1−c1x−c2x2−. . .−ckxk

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Linear Recursions

The denominator is−ck(x−r1)(x−r2). . .(x−rk). So, we can use

partial fraction decomposition to get

S = m1r1 r1−x + m2r2 r2−x +. . .+ mkrk rk−x

(for now we assume that all ri are distinct.)

Now, if we write each term as its own generating function, remember

r1 r1−x = ∞ P i=0 x r1 i , so we will get S = ∞ X n=0 xn k X i=1 1 ri n mi

Equating coefficients, an=αn1m1+αn2m2+. . .+αnkmk whereαi = r1

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Linear Recursions

α1, . . . , αk are the roots ofC(x) =xn−c1xn−1−. . .−ck. We call

this the characteristic polynomial

The constantsm1,m2, . . . ,mk can be solved for given the initial

conditions

Now, we will address root multiplicity. If a root has multiplicity 2, for example, it will appear in the partial fraction decomposition as (rm+bxx)2 instead of rmx. In the generating function, this will manifest as (p+qn)αn for somep,q. This idea generalizes, so if a root has multiplicityd, then the closed form hasP(n)αn whereP(n) is a

polynomial inn of degree d−1.

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Linear Recursions

Brilliant

A sequence xn is defined by x0=−1, x1 = 0,x2= 1 and the recurrence

relation

xn= 6xn−1−12xn−2+ 8xn−3.

Find the closed form of xn.

This is a linear recurrence! Using our new terminology, what is its characteristic polynomial?

Its characteristic polynomial is x36x2+ 12x8

Notice that this can be factored into (x−2)3

Thus, we see that we can write xn as P(n)2n for some degree 2

polynomialP

Notice that we need P(0) =−1,P(1) = 0, P(2) = 14.

We writeP(n) =an2+bn+c and solve the resulting system.

We get a=−3

8,b = 11

8,c =−1, so the closed form is

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Linear Recursions

1990 AIME #15

Find ax5+by5 if the real numbersa,b,x, andy satisfy the equations

ax +by = 3,

ax2+by2 = 7,

ax3+by3 = 16,

ax4+by4 = 42.

We’ve already done this question before. How can we apply our knowledge of linear recurrence to it?

Notice that these terms look like consecutive terms of a linear

recurrence with characteristic polynomial with roots atx,y, so

λ2−(x+y)λ+xy. Call this sequence tn=axn+byn

Suppose we letu =x+y,v =−xy. Then, we have

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1990 AIME #15

Now, we have 7u+ 3v = 16, 16u+ 7v = 42.

Solving the system givesu =−14,v= 38.

Thus, we find that the next term in the sequence,

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Weird Sequences

Unfortunately, most sequences elude easy characterization

The majority of sequences are ”weird” and require ad hoc methods in order to solve

The ideas to solve such questions are similar to those for general recursions:

Try to compute small terms if possible. Guess a pattern through engineers induction and try to prove it

Look at how the sequence behaves as a whole, and if there are any overarching global patterns

Be on the look out for manipulations, such as factorizations and substitutions which will simplify how the sequence looks

If initial conditions are given, see if they are special by trying the question with your own conditions. This will tell you if they are significant or if you can replace them with variables

Try to prove subresults to get a better intuition with the sequence. Jot down ideas you may have or qualities of the sequence you see, even if they have little to do with what we are actually trying to prove

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Weird Sequences

2017 CMIMC Algebra & Number Theory #9

Define a sequence {an}∞n=1 via a1 = 1 andan+1=an+b

anc for all

n ≥1. What is the smallest N such that aN >2017?

Let’s first try to get a feel for the sequence. What happens when we start at ak =n2?

After 2 moves, we are at ak+2=n2+ 2n = (n+ 1)2−1, and after a

3rd, we get toak+3 = (n+ 1)2+ (n−1)

How about if we start atak =n2+i with 0<i <n?

After two turns we get to ak+2 = (n+ 1)2+ (i−1)

After a squaren2, it takes three turns to increaseb√akc by 1, and we

overshoot the next square byn−1. After a nonsquare, we increase

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2017 CMIMC Algebra & Number Theory #9

So, if we start at a squaren2, it will take 3 moves to get to

(n+ 1)2+ (n−1), and 2∗(n−1) more to get to the next square at

(2n)2.

This tells us that the only squares in the sequence are 1,4,16,64, . . .,

and 4n+1 occurs 2(2n−1) + 3 = 2n+1+ 1 terms after 4n

As a1= 40, 4n occurs at position

1 + (2 + 1) + (4 + 1) +. . .+ (2n+ 1) = 2n+1+ (n−1) So,a26+4 =a68= 1024

By previous logic,a71= (32 + 1)2+ 31, and after each pair of terms,

the argument of the square increases by 1 while the outside ”overshoot” decreases by 1

We want to get close to 452, which is 12 away from 33. So, add 24

more terms

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2017 CMIMC Algebra & Number Theory #9

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Weird Sequences

Romania TST 2003/1

Let (an)n≥1 be a sequence for real numbers given bya1 = 1/2 and for

each positive integer n

an+1=

a2n a2

n−an+ 1 .

Prove that for every positive integern we havea1+a2+· · ·+an<1.

The recurrence doesn’t look very nice. How can we simplify it? Substitutingbn= a1n, we getbn+1 =bn2−bn+ 1

Now, it’s easy to compute terms of {bi}. We haveb1 = 2,b2= 3,

b3= 7, b4 = 43,b5= 1807. Any patterns?

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Romania TST 2003/1

To prove this, rearrange our equality as bn+1−1

bn−1 =bn. Now, if we multiply this telescopes

b1b2· · ·bn= bn+1b111 =bn+1−1

Now, we can substitutean+1 back in. We get that

an+1 = 1 1 +b1b2· · ·bn = a1a2· · ·an 1 +a1a2· · ·an We now get a1· · ·an 1 +a1· · ·an = (a1· · ·an)− (a1· · ·an)2 1 +a1· · ·an = (a1· · ·an)−(a1· · ·anan+1)

a1+a2+. . .+an telescopes! It becomes 1−a1a2· · ·an, so we are

References

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