Power Series Summation

Here is a problem I came up with: Let P(x)P(x) be the power series a0x0+a1x1+a2x2+a3x3+a_0x^0+a_1x^1+a_2x^2+a_3x^3+\cdots defined as i=1j=1xij\displaystyle\sum^{\infty}_{i=1}\displaystyle\sum^{\infty}_{j=1}x^{ij}. There are certain nn such that ana_n is odd. Find the number of such nn satisfying n<1000n<1000.

Hint: this is a classic problem in disguise.

EXTENSION: What about the power series i=1j=1xi2j\displaystyle\sum^{\infty}_{i=1}\displaystyle\sum^{\infty}_{j=1}x^{i^2j}?

Note by Daniel Liu
5 years, 11 months ago

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

Daniel Liu - 5 years, 11 months ago

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This is purely cosmetic remark, but either at least one of indices should start from 0 or a0=0a_{0} = 0. Or am I missing something? But this is actually very interesting problem.

hubert światkiewicz - 5 years, 11 months ago

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If either index begins with 00, then a0=a_0 = \infty.

Ivan Koswara - 5 years, 11 months 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 - 5 years, 11 months ago

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Basically we count the number of times the exponent nn can be expressed as ijij for some positive integers i,ji,j. But this also means the number of divisors of nn; for any ii divisor of nn, there is exactly one jj satisfying the condition. So now we count the number of integers n[1,999]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 nn, but is there a particular property (that is relatively easy to check instead of this bruteforcing) that is shared by all nn having odd number of square divisors (and none by those having even number of square divisors; or the other way around)?

Ivan Koswara - 5 years, 11 months ago

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

Daniel Liu - 5 years, 11 months ago

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