[seqfan] Sequences Related to the Riemann Zeta Zeros

[Paul D Hanna]

        Instead of the sequences of least *increasing* integers that
satisfy (*), another set of very interesting sequences would be the least
positive integers not used previously that satisfy (*).  These would form
a permutation of the natural numbers, one for each Riemann Zeta zero. 
Anyone like to isolate these?

        Of course, there are probably many other complex numbers for
which one could select integers to satisfy (*), but the Zeta zeros are
simply more interesting.

        Thanks Much,
                Paul


On Fri, 23 May 2003 02:31:18 -0400 Paul D Hanna <pauldhanna at juno.com>
writes:
        I would like to describe a set of sequences that could be
generated from the non-trivial zeros of the Riemann Zeta function.

        While it is obvious that the following sum does not converge:
           sum(n>=1, 1/n^(1/2 + i*y) )
where (1/2 + i*y) is a Riemann Zeta Zero, perhaps certain subsets of the
integers would allow this to converge to zero.

        Suppose we define such an integer sequence {a(n)} such that

(*)        sum(n>=1, 1/a(n)^(1/2 + i*y) ) = 0    

by requiring that the modulus of the partial sums be always decreasing in
magnitude, so that the sum approaches zero as a limit.  The n-th term
a(n) is to be the least positive integer that causes the n-th partial sum
of (*) to be less in magnitude than the (n-1)-th partial sum.
        
        Further, it would be interesting to derive a table of related row
sequences satisfying (*) for the same Zeta zero, such that they
collectively would form a permutation of the natural numbers.  There
would then be a separate table defined for each Riemann Zeta zero.  

        Perhaps someone would like to derive some of these tables?  I can
generate the first row sequences, but the subsequent rows are beyond my
calculating capability at present.

        Here are a few sequences that satisfy (*) for six different Zeta
zeros, and would form the first row of the above tables.

        Thanks,
                Paul
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