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The following Mathematica program converges to most of the riemann zeta zeros, by using an approximation as a starting point.

Clear[n, k, a]
Do[a = 1/2 + N[I*2*Pi*(n - 11/8)/LambertW[(n - 11/8)/Exp[1]], 100];
 Do[a = a - Im[Zeta[a]]*I;,
   {k, 1, 80}] 
  Print[a],
 {n, 1, 12}]

Where the key statement is the iteration of:

a = a - Im[Zeta[a]]*I
a = a - Im[Zeta[a]]*I
a = a - Im[Zeta[a]]*I
a = a - Im[Zeta[a]]*I

80 times.

The approximation is the one by Andre LeClair in a paper on arXiv.

Out comes the values of the zeta zeros, but it misses a few and doubles some due to the starting values in the LambertW approximation.

$$1/2+14.134725141734693790457251983562 I$$ $$1/2+21.022039638771554992628479593896902777334 I$$ $$1/2+25.0108575801456887632137909925628218186595499 I$$ $$1/2+30.42487612585951321031189753058409132018156002372 I$$

Is this accurate enough to approach to the Riemann hypothesis?

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  • $\begingroup$ I'm afraid that, yes, it almost certainly is useless. The first ten trillion some-odd zeroes have been found and computed to reasonable accuracy, using various other schemes, and it's hard to see how any approximation scheme could lead to reasonable progress on the conjecture itself. $\endgroup$ – Steven Stadnicki Nov 16 '13 at 20:04
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    $\begingroup$ It is still interesting that such a simple recursion leads to an accuracy of over 15 digits. In subsequent work, we derived an exact equation satisfied by the n-th zero that depends only on n. You can solve it numerically with FindRoot, to whatever accuracy you like. We went up to thousands of digits. I think this later formula is an important step toward the RH. $\endgroup$ – André LeClair Nov 26 '13 at 20:38
  • $\begingroup$ @AndréLeClair I used a word with a negative connotation regarding the approximation. It is still what experts would call a partial progress towards the Riemann Hypothesis, I agree. $\endgroup$ – Mats Granvik Nov 27 '13 at 12:31
  • $\begingroup$ I like your question...+1 $\endgroup$ – draks ... Nov 28 '13 at 10:23

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