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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)?

Ivan Koswara - 4 years ago

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

Daniel Liu - 4 years ago

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

Bob Krueger - 4 years ago

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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.

Hubert Światkiewicz - 4 years ago

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

Ivan Koswara - 4 years ago

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

Daniel Liu - 4 years ago

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