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A153415
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Decimal expansion of Sum_{n>=1} 1/A000032(2*n).
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11
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5, 6, 6, 1, 7, 7, 6, 7, 5, 8, 1, 1, 3, 8, 4, 5, 5, 0, 2, 7, 5, 9, 2, 9, 3, 2, 1, 2, 1, 2, 0, 6, 2, 0, 0, 3, 7, 3, 6, 1, 4, 4, 1, 9, 7, 8, 6, 5, 9, 0, 5, 5, 7, 0, 4, 9, 2, 3, 4, 4, 4, 1, 3, 2, 5, 4, 5, 7, 5, 5, 5, 4, 5, 3, 0, 2, 0, 8, 6, 8, 5, 6, 1, 4, 8, 5, 5, 6, 7, 8, 4, 2, 1, 8, 1, 8, 3, 2, 6, 6, 4, 6, 1, 5, 3
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OFFSET
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0,1
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COMMENTS
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c = (1/4)*(theta_3( (3-sqrt(5))/2 )^2 - 1 ), where theta_3(q) = 1 + 2*Sum_{n >= 1} q^n^2. See Borwein and Borwein, Proposition 3.5 (i), p. 91. Cf. A056854.
Series acceleration formulas (L(n) = A000032(n)):
c = 1 - 5*Sum_{n >= 1} 1/( L(2*n)*(L(2*n)^2 - 5) ).
c = (1/6) + 15*Sum_{n >= 1} 1/( L(2*n)*(L(2*n)^2 + 5) ).
c = (11/16) - 10*Sum_{n >= 1} (L(2*n)^2 - 10)/( L(2*n)*(L(2*n)^2 - 5)*(L(2*n)^2 - 20) ). (End)
Compare with Sum_{n >= 1} 1/(L(2*n) - sqrt(5)) = phi and Sum_{n >= 1} 1/(L(2*n) + sqrt(5)) = 2 - phi, where phi = (sqrt(5) + 1)/2. - Peter Bala, Nov 23 2019
This constant is transcendental (Duverney et al., 1997). - Amiram Eldar, Oct 30 2020
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REFERENCES
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J. M. Borwein and P. B. Borwein, Pi and the AGM, Wiley, 1987, p. 91.
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LINKS
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EXAMPLE
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0.56617767581138455027...
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MATHEMATICA
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PROG
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(PARI) th3(x)=1 + 2*suminf(n=1, x^n^2)
phi=(sqrt(5)+1)/2
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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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