OFFSET
0,5
COMMENTS
C arises when looking for a sequence b(n) such that b(1)=0 and b(n+1) is the smallest integer > b(n) such that the continued fraction for 1/2^b(1) + 1/2^b(2) + ... + 1/2^b(n+1) contains only 1's or 2's. It arises because b(n) = 2^n - 1 and C = Sum_{k>=0} 1/2^b(k). - Benoit Cloitre, Nov 03 2002
REFERENCES
N. J. A. Sloane and Simon Plouffe, The Encyclopedia of Integer Sequences, Academic Press, 1995 (includes this sequence).
LINKS
Harry J. Smith, Table of n, a(n) for n = 0..20000
Boris Adamczewski, The Many Faces of the Kempner Number, Journal of Integer Sequences, Vol. 16 (2013), #13.2.15.
Jeffrey Shallit, Simple continued fractions for some irrational numbers. J. Number Theory 11 (1979), no. 2, 209-217 [DOI]
FORMULA
Recurrence: a(5n) = a(5n+1) = a(2) = a(5n+3) = a(20n+14) = a(40n+9) = 1, a(20n+4) = a(40n+29) = 2, a(5n+2) = 3 - a(5n-1), a(20n+19) = a(10n+9). - Ralf Stephan, May 17 2005
EXAMPLE
1.632843018043786287416159475... = 1 + 1/(1 + 1/(1 + 1/(1 + 1/(2 + ...)))). - Harry J. Smith, May 09 2009
PROG
(PARI) { allocatemem(932245000); default(realprecision, 10000); x=suminf(n=0, 1/2^(2^n)); x=contfrac(2*x); for (n=1, 20001, write("b006466.txt", n-1, " ", x[n])); } \\ Harry J. Smith, May 09 2009
CROSSREFS
KEYWORD
nonn,cofr
AUTHOR
EXTENSIONS
Better description and more terms from Antonio G. Astudillo (afg_astudillo(AT)hotmail.com), Jun 19 2001
STATUS
approved