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Greatest number m such that the fractional part of (11/10)^A153686(n) <= 1/m.
2

%I #6 Mar 03 2020 02:48:13

%S 10,4,3,18,253,58,618,484,6009,6767,21386,697723,634293,189959,

%T 4186162,31102351

%N Greatest number m such that the fractional part of (11/10)^A153686(n) <= 1/m.

%F a(n) = floor(1/fract((11/10)^A153686(n))), where fract(x) = x - floor(x).

%e a(4) = 18 since 1/19 < fract((11/10)^A153686(4)) = fract((11/10)^17) = 0.05447... <= 1/18.

%Y Cf. A153662, A153670, A153678, A153686, A154130, A153698, A153706, A153714, A153722.

%K nonn,more

%O 1,1

%A _Hieronymus Fischer_, Jan 06 2009

%E a(14)-a(16) from _Jinyuan Wang_, Mar 03 2020