A113476 Decimal expansion of (log(2) + Pi/sqrt(3))/3.
8, 3, 5, 6, 4, 8, 8, 4, 8, 2, 6, 4, 7, 2, 1, 0, 5, 3, 3, 3, 7, 1, 0, 3, 4, 5, 9, 7, 0, 0, 1, 1, 0, 7, 6, 6, 7, 8, 6, 5, 2, 2, 1, 2, 7, 4, 8, 4, 3, 3, 1, 9, 4, 3, 2, 3, 0, 1, 8, 8, 3, 1, 4, 9, 6, 0, 5, 0, 5, 6, 0, 1, 0, 3, 2, 0, 1, 6, 1, 9, 9, 7, 6, 3, 3, 2, 9, 4, 3, 8, 4, 0, 2, 8, 2, 6, 2, 8, 5, 4, 6, 6, 0, 7
Offset: 0
Examples
0.835648848264721053337... = A073010 + A193535.
References
- Jolley, Summation of Series, Dover (1961), eq (79) page 16.
- 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.
- Index entries for transcendental numbers
Programs
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Mathematica
RealDigits[(Log[2]+\[Pi]/Sqrt[3])/3,10,120][[1]] (* Harvey P. Dale, Mar 26 2011 *)
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PARI
1/3*(log(2)+Pi/sqrt(3))
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
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