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Dec22

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:

part

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.