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A153415
Decimal expansion of Sum_{n>=1} 1/A000032(2*n).
11
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
OFFSET
0,1
COMMENTS
From Peter Bala, Oct 15 2019: (Start)
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
REFERENCES
J. M. Borwein and P. B. Borwein, Pi and the AGM, Wiley, 1987, p. 91.
LINKS
Daniel Duverney, Keiji Nishioka, Kumiko Nishioka and Iekata Shiokawa, Transcendence of Rogers-Ramanujan continued fraction and reciprocal sums of Fibonacci numbers, Proceedings of the Japan Academy, Series A, Mathematical Sciences, Vol. 73, No. 7 (1997), pp. 140-142.
EXAMPLE
0.56617767581138455027...
MATHEMATICA
First[ RealDigits[ N[(EllipticTheta[3, 0, GoldenRatio^(-2)]^2 - 1)/4, 120], 10, 105]](* Jean-François Alcover, Jun 07 2012, after Eric W. Weisstein *)
PROG
(PARI) th3(x)=1 + 2*suminf(n=1, x^n^2)
phi=(sqrt(5)+1)/2
(th3(phi^-2)^2-1)/4 \\ Charles R Greathouse IV, Jun 06 2016
CROSSREFS
KEYWORD
nonn,cons
AUTHOR
Eric W. Weisstein, Dec 25 2008
STATUS
approved