A119813 Partial quotients of the continued fraction of the constant A119812 defined by binary sums involving Beatty sequences: c = Sum_{n>=1} A049472(n)/2^n = Sum_{n>=1} 1/2^A001951(n).

0, 1, 6, 18, 1032, 16777344, 288230376151842816, 1393796574908163946345982392042721617379328

Offset: 1

Author

Paul D. Hanna, May 26 2006

Keywords

• cofr
• nonn

Comments

Convergents begin: [0/1, 1/1, 6/7, 109/127, 112494/131071,...], where the denominators of the convergents are equal to [2^A001333(n-1)-1], where A001333 are numerators of continued fraction convergents to sqrt(2). The number of digits in these partial quotients are (beginning at n=2): [1,1,2,4,8,18,43,102,246,594,1432,3457,8345,20146,48636,117417,...].

Examples

			c = 0.858267656461002055792260308433375148664905190083506778667684867..
The partial quotients start:
a(1) = 0; a(2) = 1; a(3) = 4^1 + 2^1; a(4) = 4^2 + 2^1;
a(5) = 4^5 + 2^3; a(6) = 4^12 + 2^7; a(7) = 4^29 + 2^17;
and continue as a(n) = 4^A000129(n-2) + 2^A001333(n-3) where
A000129(n) = ( (1+sqrt(2))^n - (1-sqrt(2))^n )/(2*sqrt(2));
A001333(n) = ( (1+sqrt(2))^n + (1-sqrt(2))^n )/2.
		

Links

• W. W. Adams and J. L. Davison, A remarkable class of continued fractions, Proc. Amer. Math. Soc. 65 (1977), 194-198.
• P. G. Anderson, T. C. Brown, P. J.-S. Shiue, A simple proof of a remarkable continued fraction identity Proc. Amer. Math. Soc. 123 (1995), 2005-2009.

Crossrefs

Cf. A119812 (constant), A119814 (convergents); A119809 (dual constant).

Programs

PARI

{a(n)=if(n==1,0,if(n==2,1, 4^round(((1+sqrt(2))^(n-2)+(1-sqrt(2))^(n-2))/(2*sqrt(2))) +if(n==3,2,2^round(((1+sqrt(2))^(n-3)-(1-sqrt(2))^(n-3))/2))))}

Formula

a(n) = 4^A000129(n-2) + 2^A001333(n-3) for n>2, with a(1)=0, a(2)=1.