Sign up ×
MathOverflow is a question and answer site for professional mathematicians. It's 100% free, no registration required.

As I was flipping through the scanned version of Ramanujan's "Lost Notebook" in our library, I came across a result which caught my attention. And as any excited teenager would do, I immediately photographed that page and put it up here.

The result in the photo is as follows.

If $S(N)$ be the number of ways in which $N$ can be expressed as the sum of 2 squares, then the max. order of $S(N)$ $$= \sqrt{\text{max order of }\\ d(N^2 +aN+b)} \cdot e^{O(\log N)^{1/2 + \epsilon}}$$

Has anyone come across any such result before? Is this result true?

share|cite|improve this question
and $d$ is ... –  Ricky Demer Aug 2 '11 at 17:57
$d$ must be the number of divisors of a positive integer, $d(n) = \sigma_0(n).$ Reading from Hardy and Wright, fifth edition paperback, section 16.7. Then the "average order" of $d(n)$ is section 18.2, $r(n)$ defined 16.9, then average order and max order in section 18.7. Oh, H+W $r(n)$ is $S(N)$ above. Estimates on $r(n)$ are called the Gauss Circle Problem, upper bounds on $d(n)$ are… –  Will Jagy Aug 2 '11 at 19:22

1 Answer 1

From Hardy and Wright, fifth edition paperback, I am going to call your $S(N)$ their $r(n).$ Chapter 18 is called The Order of Magnitude of Arithmetical Functions. In section 18.1, Theorem 317 $$ \limsup \frac{d(n) \log \log n}{\log n} \; = \; \log 2$$ with the comment

Thus the true `maximum order' of $d(n)$ is about $$ 2^{\log n / \log > \log n} $$

Then in section 18.7,

There is also a theorem corresponding to Theorem 317; the maximum order of $r(n)$ is $$ 2^{ \frac{\log n}{2 > \log \log n} } $$

So, if one takes the square root of the first, the result is the second. It is a bit unclear what is meant by the numbers $a,b.$ Anyway, there has been considerable work on both items, as separate estimation problems, see What is the lower bound for highly composite numbers?

and, for example, What is the status of the Gauss Circle Problem? where the interior of the circle is included.

share|cite|improve this answer
Thank you for the answer sir. –  Koundinya Vajjha Aug 3 '11 at 9:46

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.