A081750 Simple continued fraction of 2*e.
5, 2, 3, 2, 3, 1, 2, 1, 3, 4, 3, 1, 4, 1, 3, 6, 3, 1, 6, 1, 3, 8, 3, 1, 8, 1, 3, 10, 3, 1, 10, 1, 3, 12, 3, 1, 12, 1, 3, 14, 3, 1, 14, 1, 3, 16, 3, 1, 16, 1, 3, 18, 3, 1, 18, 1, 3, 20, 3, 1, 20, 1, 3, 22, 3, 1, 22, 1, 3, 24, 3, 1, 24, 1, 3, 26, 3, 1, 26, 1, 3, 28, 3, 1, 28, 1, 3, 30, 3, 1, 30, 1, 3, 32
Offset: 1
Examples
2*e = 5 + 1 / (2 + 1 / (3 + 1 / (2 + 1 / (3 + 1 / (1 + ...))))). - _Michael Somos_, May 07 2012
Links
- Antti Karttunen, Table of n, a(n) for n = 1..20000
- Index entries for linear recurrences with constant coefficients, signature (0,0,1,0,0,1,0,0,-1).
Programs
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Mathematica
ContinuedFraction[ 2E, 94] (* Robert G. Wilson v, May 07 2012 *)
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PARI
A081750(n) = if(1==n,5,if(n<6,2+(n%2), my(k=n\6, r=n%6); if(0==r || 2==r, 1, if(1==r, 2*k, if(n%2, 3, 2*(k+1)))))); \\ Antti Karttunen, Feb 20 2023
Formula
First 5 terms are 5, 2, 3, 2, 3.
For k >= 1, a(6k)=1; a(6k+1)=2k; a(6k+2)=1; a(6k+3)=3; a(6k+4)=2k+2; a(6k+5)=3.
Comments