9
\$\begingroup\$

Introduction

According to the Riemann Hypothesis, all zeroes of the Riemann zeta function are either negative even integers (called trivial zeroes) or complex numbers of the form 1/2 ± i*t for some real t value (called non-trivial zeroes). For this challenge, we will be considering only the non-trivial zeroes whose imaginary part is positive, and we will be assuming the Riemann Hypothesis is true. These non-trivial zeroes can be ordered by the magnitude of their imaginary parts. The first few are approximately 0.5 + 14.1347251i, 0.5 + 21.0220396i, 0.5 + 25.0108576i, 0.5 + 30.4248761i, 0.5 + 32.9350616i.

The Challenge

Given an integer N, output the imaginary part of the Nth non-trivial zero of the Riemann zeta function, rounded to the nearest integer (rounded half-up, so 13.5 would round to 14).

Rules

  • The input and output will be within the representable range of integers for your language.
  • As previously stated, for the purposes of this challenge, the Riemann Hypothesis is assumed to be true.
  • You may choose whether the input is zero-indexed or one-indexed.

Test Cases

The following test cases are one-indexed.

1       14
2       21
3       25
4       30
5       33
6       38
7       41
8       43
9       48
10      50
50      143
100     237

OEIS Entry

This is OEIS sequence A002410.

\$\endgroup\$
5
\$\begingroup\$

Mathematica, 23 bytes

⌊Im@ZetaZero@#+.5⌋&

Unfortunately, Round rounds .5 to the nearest even number, so we have to implement rounding by adding .5 and flooring.

\$\endgroup\$
1
\$\begingroup\$

PARI/GP, 25 bytes

There's not much support in GP for analytic number theory (it's mostly algebraic), but just enough for this challenge.

n->lfunzeros(1,15*n)[n]\/1
\$\endgroup\$
1
\$\begingroup\$

Sage, 34 bytes

lambda n:round(lcalc.zeros(n)[-1])

Try it online

This solution is a golfed form of the program found on the OEIS page.

lcalc.zeros is a function (which is thankfully spelled the shorter way, rather than zeroes for an extra byte) that returns the imaginary parts of the first n non-trivial Riemann zeta zeros. Taking the -1st index returns the nth zero (1-indexed), and round rounds it to the nearest integer. In Python 3, round uses banker's rounding (half-to-nearest-even), but thankfully Sage runs on Python 2, where round uses half-up rounding.

\$\endgroup\$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy