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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?

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and $d$ is ... – Ricky Demer Aug 2 2011 at 17:57
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$d$ 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 mathoverflow.net/questions/43103/… – Will Jagy Aug 2 2011 at 19:22

1 Answer

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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 http://mathoverflow.net/questions/43103/what-is-the-lower-bound-for-highly-composite-numbers/43105#43105

and, for example, http://mathoverflow.net/questions/19079/what-is-the-status-of-the-gauss-circle-problem where the interior of the circle is included.

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Thank you for the answer sir. – Koundinya Vajjha Aug 3 2011 at 9:46

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