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Threads: 126
Joined: Dec 2010
(02/22/2023, 12:49 AM)tommy1729 Wrote: (02/21/2023, 11:52 PM)tommy1729 Wrote: I did some thinking and calculations , and yeah it seems it is better to give up the original idea.
Some coincidences lead to wrong conclusions it seems.
But I think I can modify the idea to be more meaningful and get a new conjecture.
This might take a while though.
regards
tommy1729
Ok at first it seems if we add a factor (-1)^n
f(x) = 1 - 2/2^s + 3/3^s - 5/4^s + ...
, then we get no zeros with Re(s) > 5/2.
Not sure how (un)remarkable that is yet.
regards
tommy1729
The sequence \(p(n) \sim n\log(n)\); and therefore:
\[
\sum_{n=1}^\infty (-1)^n p(n)n^{-s}\\
\]
Converges for \(\Re(s) > 1\); by the conditional convergence theorem for Dirichlet series.
Posts: 1,924
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(02/23/2023, 06:52 AM)JmsNxn Wrote: (02/22/2023, 12:49 AM)tommy1729 Wrote: (02/21/2023, 11:52 PM)tommy1729 Wrote: I did some thinking and calculations , and yeah it seems it is better to give up the original idea.
Some coincidences lead to wrong conclusions it seems.
But I think I can modify the idea to be more meaningful and get a new conjecture.
This might take a while though.
regards
tommy1729
Ok at first it seems if we add a factor (-1)^n
f(x) = 1 - 2/2^s + 3/3^s - 5/4^s + ...
, then we get no zeros with Re(s) > 5/2.
Not sure how (un)remarkable that is yet.
regards
tommy1729
The sequence \(p(n) \sim n\log(n)\); and therefore:
\[
\sum_{n=1}^\infty (-1)^n p(n)n^{-s}\\
\]
Converges for \(\Re(s) > 1\); by the conditional convergence theorem for Dirichlet series.
I am well aware !
regards
tommy1729
