A058281 Continued fraction for square root of e.
1, 1, 1, 1, 5, 1, 1, 9, 1, 1, 13, 1, 1, 17, 1, 1, 21, 1, 1, 25, 1, 1, 29, 1, 1, 33, 1, 1, 37, 1, 1, 41, 1, 1, 45, 1, 1, 49, 1, 1, 53, 1, 1, 57, 1, 1, 61, 1, 1, 65, 1, 1, 69, 1, 1, 73, 1, 1, 77, 1, 1, 81, 1, 1, 85, 1, 1, 89, 1, 1, 93, 1, 1, 97, 1, 1, 101, 1, 1, 105, 1, 1, 109, 1, 1, 113, 1, 1
Offset: 0
Examples
sqrt(e) = 1 + 1/(1 + 1/(1 + 1/(1 + 1/(5 + ...)))). - _Harry J. Smith_, May 01 2009
Links
- Harry J. Smith, Table of n, a(n) for n = 0..20000
- D. H. Lehmer, Continued fractions containing arithmetic progressions, Scripta Mathematica, 29 (1973): 17-24. [Annotated copy of offprint]
- Keith Matthews, Finding the continued fraction of e^(l/m).
- Thomas J. Osler, A proof of the continued fraction expansion of e^(1/M), Amer. Math. Monthly, 113 (No. 1, 2006), 62-66.
- Gang Xiao, Contfrac.
- Index entries for continued fractions for constants.
- Index entries for linear recurrences with constant coefficients, signature (0,0,2,0,0,-1).
Programs
-
Mathematica
ContinuedFraction[ Sqrt[E], 100] LinearRecurrence[{0,0,2,0,0,-1},{1,1,1,1,5,1},100] (* Harvey P. Dale, Aug 05 2025 *) -
PARI
contfrac(sqrt(exp(1)))
-
PARI
{ allocatemem(932245000); default(realprecision, 60000); x=contfrac(sqrt(exp(1))); for (n=1, 20001, write("b058281.txt", n-1, " ", x[n])); } \\ Harry J. Smith, May 01 2009
Formula
a(3k+1) = 4k+1, a(i) = 1 otherwise.
G.f.: -(x^2-x+1)*(x^3-2*x^2-2*x-1) / ((x-1)^2*(x^2+x+1)^2). - Colin Barker, Jun 24 2013
E.g.f.: exp(-x/2)*(exp(3*x/2)*(5 + 4*x) + (4 + 8*x)*cos(sqrt(3)*x/2) - 4*sqrt(3)*sin(sqrt(3)*x/2))/9. - Stefano Spezia, May 05 2022
Sum_{n>=1} (-1)^(n+1)/a(n) = (Pi + 2*log(sqrt(2)+1)) / (4*sqrt(2)). - Amiram Eldar, May 03 2025
Extensions
More terms from Jason Earls, Jul 10 2001