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Second term of continued fraction for log(n).
1

%I #10 Jul 18 2023 17:43:58

%S 1,10,2,1,1,1,12,5,3,2,2,1,1,1,1,1,1,1,1,22,10,7,5,4,3,3,3,2,2,2,2,2,

%T 1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,136,39,23,16,12,10,9,7,6,6,

%U 5,5,4,4,4,4,3,3,3,3,3,3,2,2,2,2,2,2,2,2,2,2,2,2,2,2,1,1,1,1,1,1

%N Second term of continued fraction for log(n).

%F a(n) = floor(1/(log(n) - floor(log(n)))).

%t Table[ContinuedFraction[Log[n],10][[2]],{n,2,120}] (* _Harvey P. Dale_, Jul 18 2023 *)

%K easy,nonn

%O 2,2

%A _Leroy Quet_, Oct 07 2000