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Old 2019-04-08, 06:18   #42
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"Tilman Neumann"
Jan 2016

6648 Posts

Hi nesio,
sorry if I have been unpatient. I do not understand Russian, so I cannot get all information from your paper. And it took some time to get some interesting information from you.

I will try to implement your particular proposition of sorting m by

score(m) = S(m) / n^(1/3) with S(m) = sum of divisors(m) without 1 and m
this evening.

Maybe we made an error so far in our evaluations of the smoothness of m's. We will see.

You said that the computation of score(m) above is a very rough model. What would a more elaborate model look like?
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