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A081750 Simple continued fraction of 2*e.

%I A081750 #27 Sep 01 2023 04:10:42
%S A081750 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,
%T A081750 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,
%U A081750 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
%N A081750 Simple continued fraction of 2*e.
%C A081750 Decimal expansion is A019762. - _Michael Somos_, May 07 2012
%H A081750 Antti Karttunen, <a href="/A081750/b081750.txt">Table of n, a(n) for n = 1..20000</a>
%H A081750 <a href="/index/Rec#order_09">Index entries for linear recurrences with constant coefficients</a>, signature (0,0,1,0,0,1,0,0,-1).
%F A081750 First 5 terms are 5, 2, 3, 2, 3.
%F A081750 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.
%e A081750 2*e = 5 + 1 / (2 + 1 / (3 + 1 / (2 + 1 / (3 + 1 / (1 + ...))))). - _Michael Somos_, May 07 2012
%t A081750 ContinuedFraction[ 2E, 94] (* _Robert G. Wilson v_, May 07 2012 *)
%o A081750 (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
%Y A081750 Cf. A006083, A006084, A006085, A019762.
%K A081750 cofr,easy,nonn
%O A081750 1,1
%A A081750 _Benoit Cloitre_, Apr 08 2003