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%I #45 Jul 12 2022 20:55:10
%S 1,2,10,310,363010,594665194510,1871071000515058250871610,
%T 21362861761506953021644584296874581450310229239910
%N Define a rational sequence {b(n)} as b(1) = 1, b(n) = b(n-1) + 1/(1 + 1/b(n-1)) for n > 1; a(n) is the denominator of b(n).
%C The next term is too large to include.
%C The same sequence of denominators is produced by c(1) = 1 and for n > 1, c(n) = c(n-1) + 1/(n + 1 - c(n-1)). In that case, the sequence begins 1, 3/2, 19/10, 689/310, 902919/363010, 1610893922869/594665194510, ... . - _Leonid Broukhis_, Jul 09 2022
%D Suggested by _Leroy Quet_, Feb 14 2003.
%F Conjecture (Quet): a(m+1) = a(m)^2 + a(m)^3 / a(m-1)^2 - a(m)*a(m-1)^2 for m >= 2.
%e The b sequence begins 1, 3/2, 21/10, 861/310, 1275141/363010, 2551762438701/594665194510, ...
%p b := proc(n) option remember; if n=1 then 1 else b(n-1)+1/(1+1/b(n-1)); fi; end;
%t Denominator[NestList[#+1/(1+1/#)&,1,10]] (* _Harvey P. Dale_, Oct 07 2012 *)
%Y Cf. A079269 (numerators), A355615 (other numerators).
%Y Cf. A080581, A080582.
%K nonn,frac
%O 1,2
%A _N. J. A. Sloane_, Feb 16 2003
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