The stream begins at its due time, maintaining all the characteristic protocol. Pura checks the desk; assured of all its belongings she turns around to Senn, who has made himself comfortable on the couch behind. His being there is itself an issue, nevertheless he is very impressive when invested. She would definitely see to it.

“From today onward, you shall see me and Senn stream together. Well, it would be 90% me and 10% my notebook. The viewership of this programme is increasing by leaps and bounds, and all is possible due to the consistent support you all have provided. As such, I shall present you today with some problems that shows you that the most fundamental knowledge of mathematics that we hold is key to solving many complicated or non-intuitive ideas. I can assure you that these would be very easy to explain and solve, as long as you can get hold of the reasoning behind the problems. Over all, I hope it is a fun ride like always! Let’s begin.

“Let me be clear in my statement, today’s topic is a bit different from those before in the sense that we won’t be following a particular theme as such. Instead, you will learn that often visualising a problem from a different perspective will make the picture much clearer than before. Let us begin with a nudge to our previous session.

Suppose we have a sequence of numbers defined as

\[\begin{aligned} f_0 &= 0\\ f_1 &= 1\\ f_n &= \mu f_{n-1}+\lambda f_{n-2} \text{ }\forall n\geq 2 \end{aligned}\]

These are a generalisation of fibonacci numbers, and the general term is derived as\

\[f_n = \frac{r_+^n-r_-^n}{r_+-r_-}\hspace{0.1 in}\text{where }r_{\pm}=\frac{1}{2}\left(\mu\pm\sqrt{\mu^2+4\lambda}\right)\]

where $n$ is any natural number, and $\mu,\lambda$ are non-zero real numbers. Now say for example, we set $\mu = 7$ and $\lambda = -13$. Then we get $\mu^2+4\lambda = -3<0$, and so $r_+$ and $r_-$ are both complex! This poses a question,

(i) In the sequence ${f_1,f_2,f_3,\ldots}$, how many complex numbers are there? Write $\infty$ if infinitely many. Can you prove your result?

“Did you get the answer? It’s easier than it looks!”

Senn sat up, putting the magazine aside on the table. “Since this session is all about these, you can put the problem that Albert asked me a couple years ago Pura. It is a popular one, but some might not get the idea at first sight.”

“That seems reasonable. Well then, let’s look at another one of these. Suppose on the integer lattice of the Cartesian plane, we have a unit square with vertices at the points (0,0), (0,1), (1,1), (1,0) in clockwise order respectively. At each vertex there is a cricket, which can leap over any of the three others in a symmetric fashion. Say the cricket at (0,0) leaps over the one at (0,1) - it then lands on (0,2). Leaping over (1,0) lands it on (2,0) and leaping over (1,1) lands it at (2,2).

“Note that whatever be the direction of the jump or the position of the cricket, it will always be an integer point (where both coordinates are integers) due to the very nature of the jump and starting position. Initially they were sitting of the vertices of a square of unit side and then at every step, one cricket jumps over any one of the others.

(iii) Can you give at least one sequence of steps such that at the end of it, the crickets sit on the vertices of a square of side 2?

(iv) Is there a sequence of steps such that the crickets now sit on the vertices of a square of integer side n, where n is any arbitrary integer greater than 1? Could you prove your answer, whether yes or no?

“You should note that the solution to this problem isn’t apparently rooted in the permutations of the various possible steps, but somewhere else. Can you find it?”

Pura began going through her scribble-book, trying to check back on the things she had planned. There could be a large list of all such problems, simple yet cleverly disguised. She started on her next problem, but it was Senn’s turn.

“Pura’s third problem is pretty standard, so I’ll tweak it a bit. There is a finite series of lightbulbs arranged along a straight line. Each bulb has 3 possible states: Off (O), Dim (D) or Bright (B). Initially all bulbs shine bright. Now an user has the following three valid operations in hand.

• Change $B\to O$.
• Change $D,O\to O,B$ in left to right order.
• Change $B,O\to D,B$ in left to right order.

“The user is free to go on for as long as he/she likes carrying out one operation after the other. But can he/she go on forever?

(v) Show that the user will run out of valid operations after a finite number of steps.

“All of these problems can be represented visually, which might prompt you to work with examples just in case.” Senn went back to the couch and plopped on it soundly. “But there can be problems very complicated, in fact, really difficult - which become almost trivial based on the questions we ask. For example, we can often easily compute the limit of a function, but having to prove the same can be tricky. In some other cases, computation is the difficult one while the proof is easier than you think. Didn’t you put up something similar, Pura?”

“Yeah, I remember it. The first time I saw it, I was stumped as well due to its wording, so I’ll try to explain it properly.

“Let us define something called an Random Sum. It is a sum where the terms depend upon a random choice by the user. The expectation value of such a sum is calculated by something known as Wald’s Lemma given we know the probability distributions from which the terms are drawn, but we will ignore that for today. We will consider uniform distributions where the user can choose any value with equal probability. An easy example is:

\[\sigma = \sum_{i=1}^{N}\chi(i)\]

where $\chi(i)$ can take any integer value in the set ${1,2,3,\ldots,i-1}$ uniformly. Clearly, $\sigma$ may be different each time you calculate the sum - you can check by trying to programme the sum in any programming language. Now consider two functions that go as follows:

• $\Gamma(n)=1+\frac{200}{\log_2{n}}$
• $\chi(n) = |p_1-p_2-1|$ for two random primes $10^n>p_1>p_2>1$ such that $p_1-p_2<n$. If there are no such primes then $\chi(n)=1$.

“This should be pretty clear to all. $\Gamma$ is fully deterministic, while $\chi$ is a random variable which draws two distinct primes at random such that their difference is less than the argument and they are bounded by $10^n$, and gives the value $|p_1-p_2-1|$. Now consider the infinite sum

\[\tau = 1+\sum_{n=2}^{\infty}\Gamma(n)\chi(\Gamma(n))\]

(vi) Given any choice of the random primes for each and every term in the sum, will it converge to a finite value? Prove if yes. If not, show a case where it shall not converge to a finite value.

“Do you want a hint? All I would say is that the answer lies in the underlying behaviour of the given function/random variable. You would probably be surprised when you witness how easy it is to get fooled by such a morbid looking sum.

Since we’re almost at the end,” Pura sat up, de-stressing her shoulders, “let us put one last problem for those who have cleverly found their way across these simply nasty ones. The last problem will also invoke a similar idea, and will be a fun exercise for those who got it. If you didn’t, nothing to be sad about, for good times shall come.”

(vii) A k-length partition of a positive integer $N$ is defined as a decreasing sequence ${a_1,a_2,a_3,\ldots,a_k}$ (not strict), such that $a_1+a_2+\ldots+a_k=N$. Show that the number of $r$-length partitions is the same as the number of partitions with $a_1=r$.

“Good times are here!”

Pura turned back to see Senn (who had sneaked out in the middle) back with two packets of dessert; a wide grin on his face.

Submitting

Send in your solutions at maths.club@iiserkol.ac.in.