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I am trying to find solutions to the well known equation: $$\sum\limits_{n=1}^{n=k} \frac{1}{n^s}=0$$

Now with this program below I have found that for certain values of the integer $k$ one can find one of the solutions to partial Riemann zeta function.

(*Mathematica 8 start*)Clear[nn, cc, s, x, y];
Print["k can be varied to any integer greater than or equal to 2:"]
k = 16;
cc = 4000;
s = 0;
Do[s = (2 I \[Pi]*(k - 1))/Log[k] + 
   N[Round[Log[1/(-Sum[1/n^s, {n, 1, k - 1}])]/Log[k]*10^120]/10^120, 
    120], {i, 1, cc}]
Print["s"]
s
Print["If the result below is zero, then s is a solution to the \
equation in the question."]
sumx = Sum[1/n^(s), {n, 1, k}]
(*end*)

So in the program if you read it I have put $k=16$. What is the pattern in the sequence of different values of $k$? For some values of $k$ it works, for others it does not.

The values:

$$k=4, k=5, k=7, k=8, k=10, k=11, k=16$$

work.

But the values:

$$k=3, k=6, k=9, k=12$$

don't work.

What is the pattern?

Improve this question
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  • $\begingroup$ Do values 23 and 48 work as well? $\endgroup$
    – user85319
    Commented Jan 15, 2016 at 15:53
  • $\begingroup$ k=23 works, but k=48 does not work $\endgroup$
    – Mats Granvik
    Commented Jan 15, 2016 at 15:56
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    $\begingroup$ I think this question is a copy of math.stackexchange.com/questions/1613371/…. $\endgroup$
    – Meysam Ghahramani
    Commented Jan 15, 2016 at 16:26
  • $\begingroup$ Are you asking if solutions exist for $k=3,6$? $\endgroup$
    – joro
    Commented Jan 15, 2016 at 16:28
  • $\begingroup$ Not really. I am asking for which integers k the method works. But I was very excited when I wrote this question and totally forgot to consult the Online-Encyclopedia of Integer Sequences (OEIS), before posting. $\endgroup$
    – Mats Granvik
    Commented Jan 15, 2016 at 16:30

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