A270137 - OEIS

%I #19 May 14 2019 21:46:54

%S 0,8,6,6,0,7,3,9,0,8,7,3,0,1,5,9,2,9,9,7,1,2,6,4,1,4,0,6,8,5,8,4,8,0,

%T 6,4,2,8,6,6,3,1,1,5,2,3,8,6,2,7,3,2,1,1,6,0,0,9,7,3,3,8,6,5,9,3,2,8,

%U 1,9,3,5,3,8,1,8,9,1,4,0,6,7,4,4,5,4,6

%N Decimal 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 the continued fraction expansion of this number, which is transcendental.

%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.

%H <a href="/index/Tra#transcendental">Index entries for transcendental numbers</a>

%F The continued fraction expansion takes the form

%F [0; 1, 6, A270124(0), A270121(1), ..., n*A270124(n-1), A270121(n), (n+1)*A270124(n), A270121(n+1), ...].

%e 0.86607390873015929971... = 6/A270121(1) + Sum_{n>=2} 1/A270121(n) = 6/7 + 1/112 + 1/403200 + 1/1755760043520000 + ... = [0; 1, 6, 2, 7, 32, 112, 10800, 403200, 17418254400, ...] = [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,cons

%O 1,2

%A _Andrew Hone_, Mar 11 2016

%E More terms from _Jon E. Schoenfield_, Nov 12 2016