rng_58's blog

CF #146 Div1 E
By rng_58, 12 years ago, In English

My solution depends on the following hypothesis. Can anyone prove this?

Sorry, the image doesn't work for some reason. \sum_{i=1}^{a} \sum_{j=1}^{b} \sum_{k=1}^{c} d(ijk) = \sum_{gcd(i,j)=gcd(j,k)=gcd(k,i)=1} [\frac{a}{i}][\frac{b}{j}][\frac{c}{k}] Now it works.

Sorry, please type "http://rng.gozaru.jp/cf.png" directly to the address bar.

Finally proved: "http://rng.gozaru.jp/b.pdf"

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12 years ago, # |
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Image is not shown because of 403 error

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12 years ago, # |
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I have thought about it for a long time...still can't understand it...It's like magic...

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12 years ago, # |
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I think it can even be extended to 4 or larger number.

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12 years ago, # |
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Does this couclusion can still be hold in more general form ? .. .

Can we prove it by Möbius_inversion_formula ? .

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12 years ago, # |
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Your proof one could easily reformulate without telescopy. We have sum of d(ijk) -- number of triples x, y, z pairwise coprime divisors of i, j, k respectively. Consider how many times we count any triple of coprime integers x, y, z. Obviously it is a product of their multipliers numbers, it equals

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12 years ago, # |
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Thanks for the neat and instructive proof.

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