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%I #12 Feb 25 2022 09:44:35
%S 1,1,1,3,1,1,5,1,1,7,1,1,9,1,1,11,1,1,13,1,1,15,1,1,17,1,1,19,1,1,21,
%T 1,1,23,1,1,25,1,1,27,1,1,29,1,1,31,1,1,33,1,1,35,1,1,37,1,1,39,1,1,
%U 41,1,1,43,1,1,45,1,1,47,1,1,49,1,1,51,1,1,53,1,1,55,1,1,57,1,1,59,1,1,61,1
%N Continued fraction expansion of (I_0(1/2)/I_1(1/2)-1)/2 = 1.56185896... (where I_n is the modified Bessel function of the first kind).
%H <a href="/index/Rec#order_06">Index entries for linear recurrences with constant coefficients</a>, signature (0,0,2,0,0,-1).
%F G.f.: 1 + x*U(0) where U(k)= 1 + x/(1 - x*(2*k+2)/(1+x*(2*k+2) - 1/((2*k+2) + 1 - (2*k+2)*x/(x + 1/U(k+1))))) ; (continued fraction, 5-step). - _Sergei N. Gladkovskii_, Oct 07 2012
%t LinearRecurrence[{0, 0, 2, 0, 0, -1}, {1, 1, 1, 3, 1, 1}, 92] (* _Georg Fischer_, Feb 25 2022 *)
%o (PARI) /* first version */ contfrac((besseli(0,1/2)/besseli(1,1/2)-1)/2)[n+1] /* alternate version */ 2/3*n*!(n%3)+1
%K cofr,easy,nonn
%O 0,4
%A _Thomas Baruchel_, Jan 18 2005
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