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A071293
a(0)=1, a(n) is the smallest integer > a(n-1) such that the continued fraction for 1/a(0)+1/a(1)+1/a(2)+...+1/a(n) contains exactly 2^n elements.
0
1, 2, 5, 11, 573, 71081, 506860777
OFFSET
0,2
EXAMPLE
The continued fraction for 1/a(0)+1/a(1)+1/5 = 1+1/2+1/5 is {1, 1, 2, 3} which contains 2^2 elements. 5 is the smallest integer > 2 with this property, hence a(2)=5.
PROG
(PARI) s=1; t=1; for(n=1, 5, s=s+1/t; while(abs(2^n-length(contfrac(s+1/t)))>0, t++); print1(t, ", "))
CROSSREFS
Sequence in context: A069504 A158997 A101828 * A234572 A341804 A109623
KEYWORD
hard,nonn
AUTHOR
Benoit Cloitre, Jun 11 2002
EXTENSIONS
One more term from Michel ten Voorde Jun 13 2003
STATUS
approved