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.

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5 responses

    • You can have a look at ch 5 (Ramanujan tau function) of Ram Murty’s “Problems in Modular Forms” for details. This book should be easily available

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  1. This is certainly “easy to understand” as you claim for mere mortals like me…I was really under the impression that all of Ramanujan’s work is inaccessible to me…thanks for cheering me up…:-) 🙂 🙂

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    • I wish I could discuss the proofs given by Ramanujam for these statements, but they rely on the properties of elliptic functions. Anybody who reads his work thinks “How can someone learn so much on his own without help from any established mathematician?” Most of his work involves advanced complex analysis techniques.

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