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%I #13 Apr 09 2014 10:13:01
%S 1,3,21,861,1275141,2551762438701,9546380157472159016030421,
%T 126857284256055227389078067834858327568823447932861
%N Define b by b(1) = 1 and for n>1, b(n) = b(n-1)+1/(1+1/b(n-1)); sequence gives numerator of b(n).
%D Suggested by _Leroy Quet_, Feb 14 2003.
%F Conjecture (Quet): a(m+1) = a(m)^2 + a(m)^3 /(2a(m-1)^2) - a(m)a(m-1)^2/2 for m >= 2.
%e The b sequence begins 1, 3/2, 21/10, 861/310, 1275141/363010, 2551762438701/594665194510, ... = A079268/A079269.
%p b := proc(n) option remember; if n=1 then 1 else b(n-1)+1/(1+1/b(n-1)); fi; end;
%t nxt[n_]:=n+1/(1+1/n); Numerator/@Nest[Append[#,nxt[Last[#]]]&,{1},10] (* _Harvey P. Dale_, Apr 21 2011 *)
%Y Cf. A079278, A080581, A080582.
%K nonn,frac
%O 1,2
%A _N. J. A. Sloane_, Feb 16 2003
%E The next term is too large to include.
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