A235063 Continued fraction expansion of Sum(i=1..inf, 1/2^(2^i+1) ).

2, 2, 4, 2, 8, 3, 8, 1, 8, 3, 4, 2, 12, 2, 8, 1, 8, 3, 4, 2, 8, 3, 8, 1, 12, 2, 4, 2, 12, 2, 8, 1, 8, 3, 4, 2, 8, 3, 8, 1, 8, 3, 4, 2, 12, 2, 8, 1, 12, 2, 4, 2, 8, 3, 8, 1, 12, 2, 4, 2, 12, 2, 8, 1, 8, 3, 4, 2, 8, 3, 8, 1, 8, 3, 4, 2, 12, 2, 8, 1, 8, 3, 4, 2, 8, 3, 8, 1, 12, 2, 4, 2, 12, 2, 8, 1, 12, 2, 4

Offset: 0

Author

Ralf Stephan, Jan 03 2014

Keywords

• cofr
• nonn

Examples

			0.40821075451094657... = 2/(2+1/(2+1/(4+1/(2+1/(8+1/(3+1/8...
		

Links

• H. Cohn, Symmetry and specializability in continued fractions. arXiv preprint math/0008221 (2000), published in Acta Arithmetica 75 (1996): 4.
• Jeffrey Shallit, Simple continued fractions for some irrational numbers, J. Number Theory 11 (1979), no. 2, 209-217.
• Index entries for continued fractions for constants

Crossrefs

Cf. A007400.

Programs

PARI

a(n)=contfrac(suminf(i=0, 1/2^(2^i+1)))[n+1]

Formula

0.40821075451094657... = (1/2) A007400.

Recurrence: a(8n)=1, a(8n+4)=a(16n+14)=a(32n+26)=2, a(16n+6)=a(32n+10)=3, a(8n+3)=4, a(8n+7)=a(16n+5)=a(32n+9)=8, a(16n+13)=a(32n+25)=12, a(8n+1)=a(4n+1), a(8n+2)=a(4n+2), starting 2,2,4 (conjectured).