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Nov 16 at 11:52 history edited Babar CC BY-SA 4.0
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Nov 13 at 22:21 answer added Henri Cohen timeline score: 1
Nov 13 at 20:57 comment added Henri Cohen I managed to find a good guess for $C(4)$: $$C(4)=\dfrac{\zeta(3)}{3}+\zeta(2)+\dfrac{1}{2}$$ Thus probably $C(m)$ is a simple linear combination of 1 and $\zeta(k)$ for $k<m$.
Nov 11 at 10:06 history edited GH from MO
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Nov 8 at 9:04 history edited Babar CC BY-SA 4.0
I added a conjecture drawing a parallel with Dirichlet's divisor problem.
Nov 7 at 13:03 history edited Babar CC BY-SA 4.0
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Nov 7 at 12:36 history edited Babar
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Nov 6 at 21:48 history asked Babar CC BY-SA 4.0