|
%I #8 Jan 08 2019 09:12:41
%S 1,2,5,11,573,71081,506860777
%N 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.
%e 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.
%o (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,","))
%K hard,nonn
%O 0,2
%A _Benoit Cloitre_, Jun 11 2002
%E One more term from _Michel ten Voorde_ Jun 13 2003
|
