OFFSET
0,1
COMMENTS
This number is transcendental - this follows from a result of Baker (1968) on linear forms of algebraic numbers.
REFERENCES
Jolley, Summation of Series, Dover (1961), eq (79) page 16.
Konrad Knopp, Theory and application of infinite series, Blackie & Son Limited, London and Glasgow, 1954. See pp. 236, 269.
Murray R. Spiegel, Seymour Lipschutz, John Liu. Mathematical Handbook of Formulas and Tables, 3rd Ed. Schaum's Outline Series. New York: McGraw-Hill (2009): p. 135, equation 21.16
LINKS
Ivan Panchenko, Table of n, a(n) for n = 0..1000
A. Baker, Linear forms in the logarithms of algebraic numbers (IV). Mathematika, 15 (1968) pp. 204-216
L. Euler, De fractionibus continuis observationes, The Euler Archive, Index Number 123, Section 5
Eric W. Weisstein, Euler's Series Transformation.
FORMULA
Equals Integral_{x = 0..1} dx/(1+x^3) = Sum_{k >= 0} (-1)^k/(3*k+1) = 1 - 1/4 + 1/7 - 1/10 + 1/13 - 1/16 + ... (see A016777). - Benoit Cloitre, Alonso del Arte, Jul 29 2011
Generalized continued fraction: 1/(1 + 1^2/(3 + 4^2/(3 + 7^2/(3 + 10^2/(3 + ... ))))) due to Euler. For a sketch proof see A024217. - Peter Bala, Feb 22 2015
Equals (1/2)*Sum_{n >= 0} n!*(3/2)^n/(Product_{k = 0..n} 3*k + 1) = (1/2)*Sum_{n >= 0} n!*(3/2)^n/A007559(n+1) (apply Euler's series transformation to Sum_{k >= 0} (-1)^k/(3*k + 1)). - Peter Bala, Dec 01 2021
From Peter Bala, Mar 03 2024: (Start)
Equals hypergeom([1/3, 1], [4/3], -1).
Gauss's continued fraction: 1/(1 + 1/(4 + 3^2/(7 + 4^2/(10 + 6^2/(13 + 7^2/(16 + 9^2/(19 + 10^2/(22 + 12^2/(25 + 13^2/(28 + ... )))))))))). (End)
Equals (1/12) * Sum_{n >= 0} (-1/2)^n * (9*n + 7)/((3*n + 2)*(n + 1)*binomial(2*n+1/3, n+1)). - Peter Bala, Mar 05 2025
MATHEMATICA
RealDigits[(Log[2]+\[Pi]/Sqrt[3])/3, 10, 120][[1]] (* Harvey P. Dale, Mar 26 2011 *)
PROG
(PARI) 1/3*(log(2)+Pi/sqrt(3))
CROSSREFS
KEYWORD
cons,nonn
AUTHOR
Benoit Cloitre, Jan 08 2006
STATUS
approved
