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%I #17 Jun 05 2022 08:29:47
%S 1,2,3,11,16,21,27,35,42,51,55,63,75,89,350,364,385,385,416,450,453,
%T 468,476,483,526,604,617,780,1125,1157,1263,1935,7000,7028,7774,8928,
%U 9378,62628,865117,17731648
%N a(1)=1, a(n) is the smallest number >= a(n-1) such that the simple continued fraction for S(n) = 1/a(1) + 1/a(2) + ... + 1/a(n) contains exactly n elements.
%e The continued fraction for S(6) = 1+1/2+1/3+1/11+1/16+1/21 is [2, 29, 9, 1, 3, 3] which contains 6 elements. The continued fraction for 1+1/2+1/3+1/11+1/16+1/21+1/27 is [2, 14, 169, 1, 1, 1, 4] which contains 7 elements and 27 is the smallest number >21 with this property, hence a(7) = 27.
%t seq[len_] := Module[{s = {}, sum = 1, t = 1}, Do[sum += 1/t; While[Length[ContinuedFraction[sum + 1/t]] != n, t++]; AppendTo[s, t], {n, 1, len}]; s]; seq[39] (* _Amiram Eldar_, Jun 05 2022 *)
%o (PARI) s=1; t=1; for(n=1,38,s=s+1/t; while(abs(n-length(contfrac(s+1/t)))>0,t++); print1(t,","))
%Y Cf. A201267, A354742.
%K nonn,more
%O 1,2
%A _Benoit Cloitre_, May 19 2002
%E One more term from _Thomas Baruchel_, Nov 16 2003
%E Name corrected and a(40) added by _Amiram Eldar_, Jun 05 2022
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