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Greatest number m such that the fractional part of (1024/1000)^A153679(m) >= 1-(1/m).
3

%I #2 Mar 31 2012 13:21:05

%S 1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,2,2,2,2,3,3,4,5,6,9,17,93,123,1061,

%T 1360,4137,66910,571666,1192010

%N Greatest number m such that the fractional part of (1024/1000)^A153679(m) >= 1-(1/m).

%F a(n):=floor(1/(1-fract((1024/1000)^A153679(n)))), where fract(x) = x-floor(x).

%e a(5)=1, since 1-(1/2)=0.5>fract((1024/1000)^A153679(5))=fract((1024/1000)^5)=0.0510...>=1-(1/1).

%Y Cf. A153663, A153671, A153679, A154130, A153691, A153699, A153707, A153715, A153723.

%K nonn,more

%O 1,18

%A _Hieronymus Fischer_, Jan 06 2009