A352495 Decimal expansion of the pearl of the Riemann zeta function.

1, 0, 0, 0, 0, 2, 7, 8, 5, 7, 6, 3, 3, 0, 6, 6, 4, 4, 0, 7, 3, 0, 2, 1, 5, 0, 9, 1, 8, 5, 7, 3, 6, 2, 1, 7, 7, 8, 2, 9, 7, 1, 0, 0, 9, 1, 4, 0, 5, 3, 3, 3, 0, 4, 7, 8, 7, 9, 7, 3, 1, 9, 2, 8, 4, 5, 8, 6, 4, 7, 3, 5, 4, 1, 6, 6, 6, 1, 2, 9, 3, 5, 2, 6, 5, 0, 0

Offset: 1

Author

Eduard Roure Perdices, Mar 18 2022

Keywords

• nonn
• cons

Comments

Let Z be the Riemann zeta function, and consider its sequence of nontrivial zeros with nonnegative imaginary part, {r(m)}, so that for every m >= 1, Z(r(m)) = 0, 0 <= Re(r(m)) <= 1, and 0 <= Im(r(m)), and for every k > m, Im(r(m)) < Im(r(k)), or Im(r(m)) = Im(r(k)) and Re(r(m)) < Re(r(k)).

Let i be the imaginary unit, and define the sequence {b(m)} as follows: b(1) = Z((r(1)-1/2)/i), b(2) = Z((r(1)-1/2)/i + Z((r(2)-1/2)/i)), b(3) = Z((r(1)-1/2)/i + Z((r(2)-1/2)/i + Z((r(3)-1/2)/i))), and so on. If this sequence converges, we call its limit the pearl of Z.

Suppose that the Riemann Hypothesis is true. Then the sequence {b(m)} is real. On the interval [2,oo), Z is decreasing, positive, and bounded above by 2, so {b(2*m-1)} is decreasing and bounded below by 0, and hence, it converges to a real value, say A. Moreover, {b(2*m)} is increasing and b(2*m) <= b(2*m+1), and by repeated application of the mean value theorem, b(2*m+1) - b(2*m) <= Z(Im(r(2*m+1))) * |Z'(Im(r(1)))|^(2*m) <= 2*(4/100000)^(2*m), so {b(2*m)} also converges to A, and {a(n)} is the decimal expansion of this value.

We don't know if the existence of a real pearl of Z implies the Riemann Hypothesis.

More generally, the definition of pearl works for Dirichlet L-functions, giving rise to analogous constants, not necessarily real.

Examples

			1.00002785763306644073021509185736217782971009140533304787973192845864...
		

Links

• Eduard Roure Perdices, Table of n, a(n) for n = 1..5000

Crossrefs

Cf. A072449, A099874, A099876, A099877, A105546, A105817, A151558, A239349, A277313, A278812.

Programs

Mathematica

RealDigits[Re[res = Fold[Zeta[#1 + #2] &, 0, Reverse[(ZetaZero[Range[10]] - 1/2)/I]]], 10, 100][[1]]