A153415 Decimal expansion of Sum_{n>=1} 1/A000032(2*n).

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

Author

Eric W. Weisstein, Dec 25 2008

Keywords

• nonn
• cons

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

Examples

			0.56617767581138455027...
		

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.
• Index entries for transcendental numbers

Crossrefs

Cf. A000032, A000122, A056854, A093540, A153416.

Programs

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 *)

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