%I #7 Mar 13 2016 10:08:06
%S 0,1,6,2,7,32,112,10800,403200,17418254400,1755760043520000
%N Continued fraction expansion of the constant 6/A270121(1)+Sum_{n>=2}1/A270121(n).
%C A270121 is defined by the following recurrence: if A270121(n)=x(n) then x(n+1)*x(n-1)=x(n)^2*(1+n*x(n)) for n>=1, with x(1)=7, x(2)=112; and for A270124, if A270124(n)=y(n) then y(0)=2 and y(n)=x(n+1)/x(n) for n>=1. Both of these sequences appear in this continued fraction expansion, which defines a transcendental number.
%H A. N. W. Hone, <a href="https://cs.uwaterloo.ca/journals/JIS/VOL18/Hone/hone3.html">Curious continued fractions, nonlinear recurrences and transcendental numbers</a>, Journal of Integer Sequences, Vol. 18 (2015), Article 15.8.4.
%H A. N. W. Hone, <a href="http://arxiv.org/abs/1509.05019">Continued fractions for some transcendental numbers</a>, arXiv:1509.05019 [math.NT], 2015-2016, Monatsh. Math. DOI: 10.1007/s00605-015-0844-2.
%F a(2*n+1) = n*A270124(n-1), a(2*n+2) = A270121(n) for n>=1.
%e 6/A270121(1)+Sum_{n>=2}1/A270121(n)=6/7+1/112+1/403200+1/1755760043520000+...
%e =[0;1,6,2,7,32,112,10800,403200,17418254400,...]
%e =[0;1,6,A270124(0),A270121(1),2*A270124(1),A270121(2),3*A270124(2),A270121(3),4*A270124(3),...] (continued fraction).
%Y Cf. A112373, A114550, A114551, A114552.
%K nonn,cofr
%O 0,3
%A _Andrew Hone_, Mar 11 2016