Tag Archives: ramanujan

Listening Maths


Earlier I told about new mediums now available to enjoy maths. You can find a list of YouTube channels, to begin with here. And a list of dedicated math podcasts here.

In this post, I want to point out some episodes from the  podcast “In Our Time” which discuss mathematics:

Apart from them, another podcast worth listening to is “The Wizard of Mathematics, Srinivasa Ramanujan” by Prof. Srinivas Kotyada on FM Gold:


Happy Birthday Ramanujam


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.

Childhood Maths – I


I found two documents that I was very proud of as a child. Both were the result of trying to understand the kind of things Ramanujan did in free time, a result of the little AMTI books I read as a child. I will share one document in this post and another one in the next since both are not related to each other in substance.

Following is a calculation table generated using some spreadsheet software:


As you can see, it was motivated by the famous Taxicab number story.

Happy Birthday Ramanujam


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.


Celebrity Mathematicians


In my opinion, currency notes are one of the biggest motivation to learn arithmetical operations (like addition, multiplication,…). In fact, most of our elementary school problems are about buying a particular quantity of something.

Historically, there had been currencies notes featuring great mathematicians like Carl Friedrich Gauss, Leonard Euler and Rene Descartes. But, today there are no currencies featuring mathematicians. The database of currency notes featuring mathematicians is available here: http://web.olivet.edu/~hathaway/math_money.html

Since honouring people by featuring them on currency notes is politically challenging, government rather issues special postage stamps. The database of stamps featuring mathematicians is available here:  http://jeff560.tripod.com/stamps.html

Apart from illustrating various mathematical concepts (like graphs, metric system, binomial theorem… ) on stamps, India Post has issued stamps to honour mathematicians like Damodar Dharmananda Kosambi , Srinivasa Ramanujan Iyengar and Bertrand Russell.