%I #14 Mar 26 2019 11:46:44

%S 9,11,30,82,6131,26735,29430,76172,151439,227416,771341,2712159,

%T 4490404

%N Greatest number m such that the fractional part of (10/9)^A153693(n) <= 1/m.

%F a(n) = floor(1/fract((10/9)^A153693(n))), where fract(x) = x-floor(x).

%e a(2)=11 since 1/12 < fract((10/9)^A153693(2)) = fract((10/9)^7) = 0.09075... <= 1/11.

%t A153693 = {1, 7, 50, 62, 324, 3566, 66877, 108201, 123956, 132891,

%t 182098, 566593, 3501843};

%t Table[fp = FractionalPart[(10/9)^A153693[[n]]]; m = Floor[1/fp];

%t While[fp <= 1/m, m++]; m - 1, {n, 1, Length[A153693]}] (* _Robert Price_, Mar 25 2019 *)

%Y Cf. A081464, A153669, A153677, A153685, A153693, A154130, A153705, A153713, A153721.

%K nonn,more

%O 1,1

%A _Hieronymus Fischer_, Jan 06 2009

%E a(12)-a(13) from _Robert Price_, Mar 25 2019