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# 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
4 years ago

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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)?

- 4 years ago

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

- 4 years ago

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

- 4 years 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.

- 4 years ago

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

- 4 years ago

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

- 4 years ago