A270137 Decimal expansion of the constant 6/A270121(1) + Sum_{n>=2} 1/A270121(n).

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

Offset: 1

Author

Andrew Hone, Mar 11 2016

Keywords

• nonn
• cons

Comments

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.

Examples

			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).
		

Links

• A. N. W. Hone, Curious continued fractions, nonlinear recurrences and transcendental numbers, Journal of Integer Sequences, Vol. 18 (2015), Article 15.8.4.
• A. N. W. Hone, Continued fractions for some transcendental numbers, arXiv:1509.05019 [math.NT], 2015-2016, Monatsh. Math. DOI: 10.1007/s00605-015-0844-2.
• Index entries for transcendental numbers

Crossrefs

Cf. A112373, A114550, A114551, A114552.

Formula

The continued fraction expansion takes the form

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

Extensions

More terms from Jon E. Schoenfield, Nov 12 2016