# Happy Birthday Ramanujam

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Today is the 130th birthday of Srinivasa Ramanujam Iyengar.

I will discuss the easiest-to-follow work of Ramanujam, from G. H. Hardy’s Ramanujan: Twelve lectures on subjects suggested by his life and work.

A partition of $n$ is a division of $n$ into any number of positive integral parts. Thus, the sum of digits of 130 = 1+3+0=4 has 5 partitions:

$\displaystyle{4 = 3+1 = 2+2 = 2+1+1 = 1+1+1+1}$

The order in which the partitions are arranged is irrelevant, so we may think of them, as arranged in descending order. We denote the number of partitons of $n$ by $p(n)$; thus $p(4) = 5$. Also, by convention, we define $p(0) = 1$.

Very little was known about the arithmetical properties of $p(n)$; when Ramanujam started his investigations. Though we still don’t know when $p(n)$ is even or odd, there has been a lot of progress in this domain of research. For an overview, see the first section of Ken Ono’s “Distribution of the partition function modulo m” (it’s 17-year-old paper…)

Ramanujam was the first, and up to his death, the only, mathematician to discover the arithmetical properties of $p(n)$. His theorems were discovered by observing Percy MacMahon‘s table of $p(n)$ for the first 200 values of $n$. Ramanujan observed that the table indicated certain simple congruence properties of $p(n)$. In particular, the numbers of the partitions of numbers $5m+4, 7m+5$ and $11m+6$ are divisible by 5, 7 and 11 respectively, i.e.

• $p(5m+4) \equiv 0 \pmod{5}$
• $p(7m+5) \equiv 0 \pmod{7}$
• $p(11m+6) \equiv 0 \pmod{11}$

Hence, for example, for $n=130+1$ (Chinese way of calculating age) $p(131) \equiv 0 \pmod{7}$. And we can verify this using SageMath:

Now, to check its divisibility by 7, take the last digit of the number you’re testing and double it. Then, subtract this number from the rest of the remaining digits. If this new number is either 0 or if it’s a number that’s divisible by 7, then the original number is divisible by seven. [Derive it yourself!]

This process is lengthy but it converts the process of division by a simpler operation of subtraction.

Here, we have:

$5964539504 \rightarrow 596453942\rightarrow 59645390\rightarrow 5964539\rightarrow 596435\rightarrow 59633\rightarrow 5957\rightarrow 581\rightarrow 56 = 7\times 8$.

If you know how SageMath calculates the number of partitions, please let me know in the comments below.

# Happy Birthday Ramanujam

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Today is 129th birthday of Srinivasa Ramanujam Iyengar. Let’s see some properties of the number 129:

♦  It is sum of first ten prime numbers. [Tanya Khovanova’s “Number Gossip”]

$\displaystyle{2+3+5+7+11+13+17+19+23+29 = 129}$.

♦  It is the smallest number that can be written as the sum of 3 squares in 4 ways. [Erich Friedman’s “What’s Special About This Number?”]

$\displaystyle{11^2+2^2+2^2 = 10^2+5^2+2^2 = 8^2+8^2+1^2 = 8^2+7^2+4^2 = 129}$.

Though the first property appears like a coincidence (to me!), but the second fact is connected to a well-known theorem in number theory.

Legendre’s three-square theorem: A non-negative integer $n$ can be represented as sum of three squares of integers if and only if $n$ is NOT of the form $4^a (8b+7)$ for some integers $a$ and $b$.

Therefore, to make sure that a number can be written as sum of three squares or not, we just need to check its divisibility with 4 and 8. Since 129 is a small number (and a multiple of 3) we can easily factorize it as $129 = 3 \times 43$.  Now, since 4 is not a factor, we get $a=0$ and just need to check $8b+7= 129 = 8\times 16 + 1$, but no integer $b$ can satisfy this condition. Completing the verification of our example.

It’s not difficult to prove that no integer $n=4^a(8b+7)$ can be sum of three squares. But, it’s difficult to prove the converse of this statement. Till now I didn’t know the complete proof of this theorem. Interestingly, the two available proofs use some deep results like:

◊  quadratic reciprocity law + Dirichlet’s theorem on arithmetic progressions + equivalence class of the trivial ternary quadratic form.

◊  quadratic reciprocity law + Minkowski’s theorem on lattice points contained within convex symmetric bodies + Fermat’s theorem on sums of two squares

I would prefer proving this result using second approach, since proving Dirichlet’s theorem is very difficult (as compared to the main theorem we wish to prove). The proof using second approach was given by Nesmith . C. Ankeny, in  “Sums of Three Squares,” Proceedings of the American Mathematical Society, vol. 8, no. 2, p. 316, Apr. 1957. Also, note that Ankeny prove the three-square theorem only when $n$ is square-free because it’s easy to prove that

Lemma: If an integer is a sum of squares of three positive integers, so is its square.

For proof see this article by Alexander Bogomolny, “Sum of Three Squares” from Interactive Mathematics Miscellany and Puzzles

Now, what remains to determine is the number of ways we can write a non-negative integer which satisfies Legendre’s three-square theorem as sum of three squares. Interestingly, this is a research level problem if we are talking about a formula to calculate the number of presentations of a given number as sum of three squares. Indeed, there is a (very long) paper by Paul T. Bateman titled “On the Representations of a Number as the Sum of Three Squares”, Transactions of the American Mathematical Society, Vol. 71, No. 1 (Jul., 1951), pp. 70-101. Though I didn’t have patience to read this paper, this MathOverflow discussion gives an overview of the Sum of Squares Function. The output of this function is available as A005875: Number of ways of writing a nonnegative integer n as a sum of 3 squares (zero being allowed) but it is useless since it counts the “tuples”. For example, 129 can be represented by 144 =  (3+6+6+3)8 tuples because we are allowed to replace positive by negative integers before squaring them (hence multiplied by 8).

So, we will have to “manually” check the number of ways we can write the non-negative integers less than 129 as sum of three squares (A000408: Numbers that are the sum of three nonzero squares). As of now, I don’t know how to find the distinct representations. Will try to answer this question in future.

# Happy Birthday Ramanujam

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Today is 128th birthday of Srinivasa Ramanujam Iyengar.

[Please note that Britishers gave wrong English spelling for his name, they called him “Ramanujan”,  but it should be “Ramanujam” which means  “The younger brother of Lord Rama”]

John Littlewood said (as recalled by G. H. Hardy (1921), in “Obituary Notices: Srinivasa Ramanujan”, Proceedings of the London Mathematical Society 19, p. lvii) :

Every positive integer is one of Ramanujan’s personal friends.

So, let’s talk about numbers on his birthday.

128 is 7th power of 2 and leads to 4th Mersenne Prime ($2^7 - 1=127$ is a prime)

128 is the largest number which is not the sum of distinct squares. [https://oeis.org/A001422]

and here is my present for him:

RAMANUJAM’S MAGIC SQUARE

When I was in High School, I learned making such “Special Date Magic Squares” form P.K. Srinivasan’s (1924-2005) book regarding Ramanujam’s work. [see: https://nrich.maths.org/1380 ]

Magic sum of this square is 69. Thus all rows, columns and diagonals add up to the same total of 69 and today’s date has been placed in first row.

69 is a value of $n$ where $n^2$ and $n^3$ together contain each digit once. Since, $69^2=4761$ and $69^3 = 328509$

Ramanujam also studied “Partition Function”, this function gives the number of ways of writing the integer n as a sum of positive integers, where the order of addends is not considered significant. Observe that, in the magic square above we created 10 distinct 4-partitions of 69. (though 2376 distinct 4-partitions of 69 are possible!!)

Also, the total sum of numbers in our magic square is $69\cdot 4 = 276 = 1^5 + 2^5 + 3^5$

Once more, Happy Birthday Ramanujam!

# Happy Birthday Kaprekar

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Today is 110th birthday of great Indian mathematician D. R. Kaprekar

Dattaraya Ramchandra Kaprekar (17 January 1905 – 1986) was an Indian recreational mathematician who described several classes of natural numbers. For his entire career (1930–1962) he was a schoolteacher at Nasik in Maharashtra. He published extensively, writing about such topics as recurring decimals, magic squares, and integers with special properties. Kaprekar was once laughed at by most contemporary Indian mathmaticians for his so-called ‘trivial’ play with numbers. It required G. H. Hardy to recognize Ramanujan while Kaprekar’s recognition came through Martin Gardner (he wrote about Kaprekar in his “Mathematical Games” column in March 1975 issue of “Scientific American”)

Here I would discuss a mathemagical trick re-disvovered by Kaprekar called “Gap Filling Process” (though claimed to be present in vedic mathematics)

Gap filling process

This process is magical one and will make you Mathemagician

Let, $(a)_n$ stand for $a$ repeated $n$ times (called Repunit $a$).

Then, we shall denote $(a)_n ^m$ for $m$-th power of $(a)_n$

$(9)_n ^m$ can be obtained by remembering the expansion for $(9)^m$ and inserting in the gaps between digits of expansion of $(9)^m$ with the numbers $(9)_{n-1}$ and $(0)_{n-1}$ alternately, beginning from left to right. No gap is counted after the unit digit.

Let’s see an example:
Find the value of $(99999)\times (99999)\times (99999) = (99999)^3 = (9)_5 ^3$.

Even my scientific calculator fails to calculate this exact value !

We know $9^3 = 729$

Then the gaps are: $----7----2----9$
Now fill the blanks alternately with $(9)_{5-1} = (9)_4$ and $(0)_{5-1} = (0)_4$.
We get: $9999\textbf{7}0000\textbf{2}9999\textbf{9}$
Hence, $(99999)\times (99999)\times (99999) = 999970000299999$

# A Mathematical December at NISER

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This December I unofficially [just knowledge, no certificate] attended Annual Foundation School – 1,  held at National Institute of Science Education and Research. This was possible due to Sagar Shrivastava.

This winter I just wanted to explore maximum number of mathematics topics unknown to me. Also this gave me an excellent opportunity to meet aspiring as well as established mathematicians from all over India. It was wonderful experience to have mathematics being talked all around me.

In this post I will tell about “what new things I learned this December

Week 1

Real Analysis 1 by Pradipta Bandyopadhyay: First topic to be covered was Measure theory and integration. I could understand just some basic definitions and 2.5 out of 4 classes were fruitful for me.

Topology by Parameswaran Sankaran: He spent one class on discussing basics and then jumped to problem solving, so that all students who have done the course can also benefit from his classes. From the time I saw proof of Euler’s polyhedron formula, I was very much attracted to Topology and thus enjoyed these classes. I can illustrate this with following incident:
On 7th December 2014 (first Sunday of AFS 1) I was sitting outside barber shop (as inside it was already very much crowded), I started thinking about Klein Bottle and how to transform it into a sphere with 2 discs removed and replaced by Möbius strips. Then the barber came out (after half an hour) and called me in, but I didn’t want to be disturbed and rather sent a man who had just come, inside. Then after I was done with my thinking I realized my mistake and as a result of which I had to wait for another 15 minutes.

Group theory by Brundaban Sahu : All basics were discussed, thus this was most fruitful course for me (also I had already done some homework.)

Week 2

Group theory by Brundaban Sahu  : Learnt awesome awesome new proof for “there exist only 5 platonic solids” (earlier I knew proof by topology only)

Real Analysis 1 by Saugata Bandyopadhyay :Saw the most dreadful proof of real analysis [Resiz  Representation Theorem] (around 13 different steps, took 3 classes to complete the proof)

Real Analysis 2 by Varadharajan Muruganandam: Discussed, the topic “Functions of several real variables” – This was most weird course for me. I just made myself familiar with some elementary theorem (could not understand their proofs) but this course forced me to kick start a systematic study of Real Analysis and Linear Algebra.

Week 3

Group theory by Binod Sahoo : Learnt awesome awesome new proof for ” Fundamental Theorem of Arithmetic” using Jordan-Hölder Theorem. Also a bit of  Linear Algebra was discussed in last class.

Real Analysis 1 by Sanjay Parui: Application of vitalli cover and Lebesgue integration were discussed.

Real Analysis 2 Varadharajan Muruganandam:  Got a flavor of differential geometry.

Week 4

This week started with celebration of Ramanujan’s birthday. This week was rather Mathemusical., as every evening I attended Spic Macay, Rural School Intensive programe..

Group theory by Varadharajan Muruganandam: He, taught topology groups, so as to prove that SO(n) groups are simple.

Real analysis 1 by S. Thangavelu : Measure theory was brought to it’s end by discussing convolutions etc.

Topology by Samik Basu [Dept. of Mathematics, Ramakrishna Mission Vivekananda University, Belur Math, W. B.] : Main focus was on quotient topology and its applications.

Mathematical December at NISER

# Happy Birthday Ramanujam

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Today is 127th birthday of our beloved mathematician Srinivasa Ramanujam Iyenger.
To celebrate his birthday I would like to pose a puzzle to all the readers and urge you  to solve it in your head, as Ramaujam did!!

In a certain street, there are more than fifty but less than five hundred houses in a row, numbered from 1, 2, 3 etc. consecutively. There is a house in the street, the sum of all the house numbers on the left side of which is equal to the sum of all house numbers on its right side. Find the number of this house.

[Hint: Ramanujam used concept of continued fractions]