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A352495
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Decimal expansion of the pearl of the Riemann zeta function.
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1
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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
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OFFSET
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1,6
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COMMENTS
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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.
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LINKS
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EXAMPLE
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1.00002785763306644073021509185736217782971009140533304787973192845864...
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MATHEMATICA
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RealDigits[Re[res = Fold[Zeta[#1 + #2] &, 0, Reverse[(ZetaZero[Range[10]] - 1/2)/I]]], 10, 100][[1]]
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CROSSREFS
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KEYWORD
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AUTHOR
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STATUS
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approved
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