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Recurrence relations + Christmas Blog

 Hey everyone! Welcome back to my blog! Today's topic is Recurrence relations! 

The handouts/ books I have referred to are GRAMOLY's recurrence in combinatorics, Recursion in the AIME , Combinatorics: A problem-oriented approach and principles and techniques in Combinatorics.

Recursion is as cute as a rabbit ( not really)


Here are my notes!

Recurrence Relation: A recurrence relation is a formula or rule by which each term of a sequence can be determined using one or more of the earlier terms.


Problem 1: For each of the following sequences find a recurrence pattern.
a. 1,10,100,\dots

b. 1,3,6,10,\dots

c. 1,2,6,24,120,\dots

d. 1,1,2,3,5,8,13,\dots

e. 0,1,9,44, 265, \dots

Solution: 

a. a_n=10\cdot a_{n-1}

b. a_n=a_{n-1}+n

c. a_n=n\cdot a_{n-1}

d. This is the Fibonacci sequence. We have F_0=1,F_1=1, F_2=2,\dots F_n=F_{n-1}+F_{n-2}

e. This is the derangement number. We have D_n= n \cdot D_{n-1}+(-1)^n. We also have D_n=(n-1)(D_{n-1}+D_{n-2})

The Stamp problem: 

Suppose we have 1,2,5 stamps. The problem is to find the number of ways these can be arranged in a row so that they can add up to a give value n.

Solution: Let a_n be the number of ways the stamp can add up to n.

Then we have three cases considering the value of the last stamp. 

Case 1: If the last stamp is 1. Then the total value of the remaining stamps must be n-1. Therefore the number of ways in which these remaining stamps can be selected is a_{n-1}.

Case 2: If the last stamp is 2. Then the total value of the remaining stamps must be n-2. Therefore the number of ways in which these remaining stamps can be selected is a_{n-2}.

Case 3: If the last stamp is 5. Then the total value of the remaining stamps must be n-5. Therefore the number of ways in which these remaining stamps can be selected is a_{n-5}.

So \boxed{a_n=a_{n-1}+a_{n-2}+a_{n-5}}

Word with no two consecutive As :

Find the number of n letter of letter words using letters from the set \{A,B\} in which no two consecutive A can appear.

Solution: Let w_n= the number of n letter word satisfying the given condition 

a_n= the number of n letter word ending with A

b_n= the number of n letter word ending with B

So w_n=a_n+b_n.

But note that if the last letter of a word is A then the second last letter of the word is B. So a_n=b_{n-1}.

And b_n= b_{n-1}+a_{n-1}\implies b_n=w_{n-1}\implies a_n=w_{n-2}

So \boxed{ w_n= w_{n-1}+w_{n-2}}.

Subsets with no two consecutive digits: 

Find the number of subsets of the set of digits \{0,1,2,\dots, n\} such that the subset contains now two consecutive digits.

Solution:  Let X_n denote the subset with highest element n.

Here are some  examples

X_0=1;\{0\}
X_1=1; \{1\}
X_2=2; \{2\}, \{0,2\}
X_3=3;\{3\},  \{0,3\}, \{1,3\}
X_4=5; \{4\}, \{0,4\}, \{1,4\},\{2,4\}, \{0,2,4\}

Note that \boxed{X_n= 1+ X_1+\dots+X_{n-2}}

where 1 is for the single set \{n\} and rest are because n is added to the subset.

Classical Stairs:

There is a n stair staircase, one can climb 1 or 2 stairs (1 or 2 steps) at a time, in how many ways he can climb the entire staircase?

Solution: Let a_n denote the number of ways he can climb the n stairs. Then we get two cases considering the last step.

Case 1: If the last step was 1 stair then he was at n-1 staircase. There are a_{n-1} ways to reach the n-1th staircase

Case 2: If the last step was 2 stair, then we previously at n-2 th staircase. There are a_{n-2} ways to reach the n-2th staircase.

So \boxed{a_n=a_{n-1}+a_{n-2}}

Tower of Hanoi: 

Given a stack of n disks arranged from largest on the bottom to smallest on top placed on a rod, together with two empty rods, the tower of Hanoi puzzle asks for the minimum number of moves required to move the stack from one rod to another, where moves are allowed only if they place smaller disks on top of larger disks. 

Solution: The minimum number of moves is \boxed{2^n-1}.

Proof: Let the discs be [1,2,3,\dots, n] with [1]<[2]<\dots<[n]. Denote A_n as the number of moves requires to move discs [1,2,3,\dots, n].

Also let's name the rods as A, B, C.

Claim: we need atleast require  2\cdot A_{n-1}+1 moves.

Proof: Now, consider the step when we keep disk [n] from some rod X to rod C. This will require us 1 move. Also during this move the [n-1] discs must be in rod other than X and C. ( arranging the [n-1] discs would have took us at least A_{n-1}) And then to bring back the n-1 disks back to rod C would have took us at least another A_{n-1} moves.

So we know that the required moves is \ge 2\cdot A_{n-1}+1.

Claim: 2\cdot A_{n-1}+1 works

Proof: We shift the [1,2,\dots , n-1] disc from A to B which will be done in A_{n-1} moves, shift [n] from A to C in one move and then shift [1,2,\dots , n-1] disc from B to C   in A_{n-1} moves.

So A_n= 2\cdot A_{n-1}+1.

Another way of proving the bound was just strong induction.

We have A_1= 1, A_2=3, A_3=7, \dots and by induction we get \boxed{A_n=2^n-1}.

Okay, now we should know how to "solve" linear recurrence relations..

Open and closed Brackets: 

How many ways are there to arrange n open brackets ”[” and n closed brackets ”]” such that when read left to right, the number of closed brackets is less than or equal to the number of open brackets?

Solution: The answer is the Catalan number.

Let C_n denote the number of ways.  

Let us consider  the sequence of closed and open brackets as a_1,a_2,\dots a_{2n} where a_i is either [, ] 

Let f(j) denote the number of [ which are there in the sequence from a_1,\dots ,a_i and g(j) denote the number of ] which are there in the sequence from a_1,\dots ,a_i.

Now we consider cases. At first, we have f(1)>g(1) but at the end of the sequence, we have f(2n)=g(2n). So there would be an index k where f(k)=g(k). Note that in this index our first open bracket will complete.

So we have the sequence of the form [A]B

Now, both A and B also have to balance and can max contain n-1 pairs inside.

So C_n=\boxed{C_0\cdot C_{n-1}+C_1\cdot C_{n-2}+C_3\cdot C_{n-4}+\dots + C_{n-1}\cdot C_0}

Also, we have \boxed{C_n=  \frac{1}{n+1}\binom{2n}{n}}

Dominoe tiling:

How many ways are there to fill in a 2 \times n  grid with 1 \times 2 or 2\times 1 dominos?

Solution:  Now we consider the last 2 x 1 cell. We consider in what ways we can tile. Denote T_n as the number of ways to tile. We consider cases.









So we get \boxed{T_n=T_{n-1}+T_{n-2}}



Linear Recurrence:  A linear recurrence


is a recurrence relation where a_n is expressed as a linear combination of previous terms of the sequence.

Wait what about other non-linear recurrence relations? It's hopeless to solve them :) 

Also, a recurrence relation is homogeneous if it doesn’t contain any constant terms. Else, not homogeneous.


So a linear homogeneous recurrence is c_0a_n+c_1a_{n-1}+\dots +c_ra_{n-r}=0

The  characteristic equation  of the recurrence is c_0x^r+c_1x^{r-1}+\dots +c_r=0

If \alpha_1,\dots \alpha_r are the distinct roots then a_n= A_1(\alpha_1)^n+A_2(\alpha_2)^n+\dots +A_r(\alpha_r)^n

What about repetitions?



So for two-degree equation whose root repeat we get,

a_n= (A+Bn)r^n


Solving Fibonacci: 

We have to solve the recurrence a_n-a_{n-1}-a_{n-2}=0

So the characteristic equation is x^2-x-1=0

and it's roots are \alpha_1=\frac{1+\sqrt{5}}{2} and \alpha_2=\frac{1-\sqrt{5}}{2} 

Hence a_n= A\Big(\frac{1+\sqrt{5}}{2}\Big)^n+B\Big(\frac{1-\sqrt{5}}{2}\Big)^n

Since a_0=a_1=1\implies A+B=1, A\Big(\frac{1+\sqrt{5}}{2}\Big)+B\Big(\frac{1-\sqrt{5}}{2}\Big)=1

Solving gives us A= \frac{1+\sqrt{5}}{2\sqrt{5}}, B=\frac{-1+\sqrt{5}}{2\sqrt{5}}

So a_n= \frac{1}{\sqrt{5}}\Big[ \Big(\frac{1+\sqrt{5}}{2}\Big)^{n+1}-\Big(\frac{1-\sqrt{5}}{2}\Big)^{n+1}\Big]


Another example: 

Find solution to the recurrence relation a_n=3a_{n-1}+4a_{n-2} with a_0=2, a_1=3.

Solution: We have x^2-3x-4=0\implies (x-4)(x+1)=0

So a_n=A\cdot 4^n+B\cdot (-1)^n
So a_0=2\implies A+B=2

a_1=3\implies 4a-B=3

\implies A=1,B=1

Hence a_n= 4^n+(-1)^n

Olympiad Problems:

Problem [India postal 2006]: Consider the set A = \{1, 2, 3, ..., 2008\}. We say that a set is of type r, r \in \{0, 1, 2\}, if that set is a nonempty subset of A and the sum of its elements gives the remainder r when divided by 3. Denote by X_r, r \in \{0, 1, 2\} the class of sets of type r. Determine which of the classes X_r, r \in \{0, 1, 2\}, is the largest.

Proof: Consider set B=\{1,2,\dots , 3k\}, k>1.

Claim 1:X_0 is the largest for set B.

We can say by symmetry, |X_1|=|X_2|. And we can show this by induction on k. 

For base case take k=2, we get |X_1|=|X_2|=2 and |X_3|=3. So X_3 is largest. Now if we add \{4,5,6\}, we will get previous set using only 

E=\{1,2,3\}  , sets  using only F=\{4,1,2,3\} , sets  using only G=\{5,1,2,3\} ,sets  using only H=\{6,1,2,3\}, sets  using only I=\{4,5,1,2,3\},  sets  using only J=\{4,6,1,2,3\},  sets  using only K=\{5,6,1,2,3\},  sets  using only L=\{4,5,6,1,2,3\}.

For now sets F,G,H,I,J,K,L will use at least two elements from there sets with F must having 4, G having 5, and so on.

Now initially we has X_1=2,X_2=2, X_3=3 . Now if we see sets, \{4\},\{5\},\{6\},\{4,5\},\{4,6\}, \{5,6\}. Now X_1=3,X_2=3,X_3=5.

Now set E's all non empty subset ahs been counted. Now consider set F. Since F's subsets will must have 4, it's same as having E's subsets, but everything shifting to 1\mod 3, hence it will have X_1=3,X_2=2,X_3=2.

Similarly Since G's subsets will must have 5, it's same as having E's subsets, but everything shifting to 2\mod 3, hence it will have X_1=2,X_2=3,X_3=2.

And so on.. adding we get X_1=20,X_2=20,X_3=23. For set \{1,2,3,4,5,6\}.

Now for \{1,2,3,4,5,6,7,8,9\}, we will have X_3 and infact by this method, we get |X_3| - |X_1|=2\cdot 3+1=7.

For \{1,2,3,4,5,6,7,8,9,10,11,12\}, we will have X_3 and  we get |X_3| - |X_1|=2\cdot 7+1.

So this forming recurrence stuff. Clearly K grows more, the difference between X_3 and X_1 grows very much too.

Now consider C=\{1,2,\cdots,3k,3k+1\}, It's just the set B but we added {3k+1}. So there are 3 types of subsets, 

Subset's of B, \{3k+1\}, D=Subsets must containing 3k+1.

Claim2: X_1  is the largest for C.

Now if for B, we have by our claim, X_1=X_2=a and X_3=b and b>>a. 

Again, note that D's subets are just B's subsets, but everything shifted with 1\mod 3. 

So D's contribution to X_1=b,X_2=a,X_3=a and the single set \{3k+1\} contributes 1 to X_{1}.

So for set C, we get X_1=a+b+1,X_2=a+a,X_3=a+b. Hence we get X_1 to be the largest. 

Note that 2008=2007+1= 3l+1, so by our Claim2, we get X_1 to be the largest. And we are done!

Problem[ Modified INMO from Gramoly handout]: For any natural number n, consider a 1 \times n rectangular board made up of n unit squares. This is covered by 3 types of tiles : 1 \times 1 red tile, 1 \times 1 green tile and 1 ×\times 2 domino. Let T_n denote the number of ways of covering 1 \times n rectangular board by these 3 types of tiles. Find a recursion for T_n

Proof: Clearly, T_n=2\cdot T_{n-1}+ T_{n-2}.
Solving, we x^2-2x+1=0 \implies a_n= A(1+\sqrt{2})^n+B(1-\sqrt{2})^n

Then a_1=2, a_2 =3\implies A+B=2, A-B=\frac{3}{\sqrt{2}}

Problem [2018 USAJMO]: For each positive integer n, find the number of n-digit positive integers that satisfy both of the following conditions:

no two consecutive digits are equal, and the last digit is a prime.

Proof: Note that the unit digit can be filled up in 4 ways and then we can fill up the rest n-1 digits in 9^{n-1} ways ( allowing 0 to be in the first digit). Now, note that the no. of numbers has 0 in the first place is a_{n-1}.

Hence 4\cdot 9^{n-1}=a_n+a_{n-1}\implies a_n+a_{n-1}= \frac{ a_n+a_{n-1}}{9}

\implies 8a_{n-1}+9a_{n-2}=a_n And a_1=4, a_2=32

Solving we get x^2-8x-9=0\implies \alpha_1= -1, \alpha_2=9 \implies a_n= A(-1)^n+B(9)^n

\implies A=\frac{-2}{5}, B=\frac{2}{5}

So \boxed{a_n=\frac{2}{5}\left(9^n-(-1)^{n}\right)}

Problem[ 2020 C1]: Let n be a positive integer. Find the number of permutations a_1, a_2, \dots a_n of the sequence 1, 2, \dots , n satisfying

a_1 \le 2a_2\le 3a_3 \le \dots \le na_n

Proof: Let F_0=1,F_1=1,F_2=2, F_n=F_{n-1}+F_{n-2}. We claim that the number of permutation is F_n. 
We use strong induction. For n=1 it is true. For n=2 we get (1,2) and (2,1).

Claim:  k+1 is a_k or a_{k+1} in a valid permutation

Proof: If not then let k+1 be a_y. Then let y be a_i. Also i\cdot y\le (k+1) y for any i.

So i \le y. Hence by PHP, there is a number x<y which is a_j and j>y. But then we must have j\cdot x\ge (k+1) y which is false. A contradiction.

So k+1 is a_k or a_{k+1}.

Case 1: If k+1 is a_{k+1} then by induction, we have F_k valid permutaion.

Case 2: If k+1 is a_{k} then a_{k+1} must be k, then by induction, we have F_{k-1} valid permuations.

So \boxed{F_{k+1}=F_k+F_{k-1}}.

Yayy!! That's it for today's recurrence blog post!! There will be a sequel of harder problems ( I asked Rohan bhaiya to send some handouts, which he did and the problems look hard)! 

Also a short Christmas blog :P

So I had an appointment with the dentist ( cause I have braces). So, we went for a regular checkup. After the check-up, we went to Rudrakh mall which has my favourite shop in Guwahati! MUMUSO!! It has so cute stationery items :pleading: and as always I bought so cute stuff. Although, my mom always says the items are just useless -_-..

I bought these paper clips.. They are so cute!

After Mumuso, Suhan played a virtual reality cricket game, where he lost terribly :P And then we realised Pantaloons had sale! My mom after hearing the word "sale" turned into Sinchan's mom 😂

The sale was " Buy 5000+ rupees cloth,  get 5000 rupee cloth for free." And boom! We spent 3 hours buying clothes 😇. My mom was very proud of herself.

The mall looked so beautiful


After that, we went to dominoes, where I ate my fav food item, Moroccan chicken chunks ( try it!!) 😋

Then we went to a Turkish ice cream stall, where the ice cream maker guy played tricks like not giving Suhan his ice cream ( you know those viral shorts videos?)
And the icre cream was so nice!!


Suhan was looking weird :P

Yeah! That's what we did yesterday! It was quite fun!! I hope you guys are also having fun during this vacation! 


Stay safe
Sunaina 💜



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