# Power Series Summation

Here is a problem I came up with: Let $P(x)$ be the power series $a_0x^0+a_1x^1+a_2x^2+a_3x^3+\cdots$ defined as $\displaystyle\sum^{\infty}_{i=1}\displaystyle\sum^{\infty}_{j=1}x^{ij}$. There are certain $n$ such that $a_n$ is odd. Find the number of such $n$ satisfying $n<1000$.

Hint: this is a classic problem in disguise.

EXTENSION: What about the power series $\displaystyle\sum^{\infty}_{i=1}\displaystyle\sum^{\infty}_{j=1}x^{i^2j}$? Note by Daniel Liu
6 years, 2 months ago

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Oops, how did that tag get on there...

- 6 years, 2 months ago

This is purely cosmetic remark, but either at least one of indices should start from 0 or $a_{0} = 0$. Or am I missing something? But this is actually very interesting problem.

- 6 years, 2 months ago

If either index begins with $0$, then $a_0 = \infty$.

- 6 years, 2 months ago

This is an interesting way to look at this problem in the context of generating functions. Very enjoyable. Thanks for sharing. :)

- 6 years, 2 months ago

Basically we count the number of times the exponent $n$ can be expressed as $ij$ for some positive integers $i,j$. But this also means the number of divisors of $n$; for any $i$ divisor of $n$, there is exactly one $j$ satisfying the condition. So now we count the number of integers $n \in [1,999]$ such that it has an odd number of divisors. Now it's the classic problem :)

The second one is interesting. It basically counts the number of squares that divides $n$, but is there a particular property (that is relatively easy to check instead of this bruteforcing) that is shared by all $n$ having odd number of square divisors (and none by those having even number of square divisors; or the other way around)?

- 6 years, 2 months ago

I don't know either. I was hoping one of you smarter people could figure one out.

- 6 years, 2 months ago