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A242600
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Decimal expansion of -dilog(phi) = -polylog(2, 1-phi) with phi = (1 + sqrt(5))/2.
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2
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5, 4, 2, 1, 9, 1, 2, 1, 6, 4, 5, 0, 6, 9, 3, 3, 7, 8, 3, 4, 0, 5, 0, 1, 5, 3, 1, 0, 4, 2, 6, 4, 3, 6, 9, 5, 6, 7, 9, 3, 7, 6, 7, 8, 5, 4, 5, 8, 0, 6, 9, 9, 3, 9, 6, 8, 6, 5, 7, 2, 6, 7, 7, 4, 0, 3, 1, 0, 5, 3, 1, 5, 3, 7, 7, 7, 9, 9, 4, 4, 3, 0, 4, 0, 9, 2, 4, 2, 8, 6, 7, 0, 4, 7, 0, 9, 2, 8, 4, 5, 9, 3, 7, 3, 0, 1, 3
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OFFSET
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1,1
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COMMENTS
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-dilog(phi) = -polylog(2, 1-phi) = -Sum_{k>=1} ((1-phi)^k/k^2) = Pi^2/15 - (log(phi-1)^2)/2. See the Abramowitz-Stegun link, p. 1004, eqs. 27.7.3 - 27.7.6 with x = phi-1, solving for -dilog(x+1) = -f(1+x), using log(2-phi) = 2*log(phi-1).
This solution for -Sum)k>=1} (-2*sin(Pi/10)^k/k^2) should also have been mentioned in the Jolley reference pp. 66-69 under (360).
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REFERENCES
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L. B. W. Jolley, Summation of Series, Dover, 1961.
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LINKS
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M. Abramowitz and I. A. Stegun, eds., Handbook of Mathematical Functions, National Bureau of Standards, Applied Math. Series 55, Tenth Printing, 1972 [alternative scanned copy].
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FORMULA
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-dilog(phi) = -polylog(2, 1-phi) = -sum(((1-phi)^k)/k^2, k=1..infinity) = Pi^2/15 - (log(phi-1)^2)/2 = 0.542191216450693..., with the golden section phi = (1 + sqrt(5))/2.
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MATHEMATICA
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RealDigits[PolyLog[2, 1 - GoldenRatio], 10, 120][[1]] (* Amiram Eldar, May 30 2023 *)
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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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