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May 11, 2016 at 14:18 vote accept Clark Kimberling
Jan 20, 2016 at 14:49 vote accept Clark Kimberling
Jan 20, 2016 at 14:49
Jan 19, 2016 at 21:40 comment added Asaf Karagila Interesting to whom?
Jan 19, 2016 at 15:09 answer added Eric Naslund timeline score: 11
Jan 19, 2016 at 15:04 comment added Douglas Zare $H(k) = \log k + O(1)$ so for $x=1$, the series will diverge for $y \le 1$, and converge for $y \gt 1$. It would not affect the answer, but shouldn't that be $H(k^y)$?
Jan 19, 2016 at 15:02 comment added Johannes Trost Do I interpret your notation correctly, by assuming that you sum over $k$ from 1 to infinity, and within that summation you sum over $h$ from 1 to an $h_{*}$ defined by $h_{*}^{x}=\left\lfloor k^{y} \right\rfloor$ ?
Jan 19, 2016 at 14:43 history asked Clark Kimberling CC BY-SA 3.0