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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.
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%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