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%I #16 Mar 23 2019 17:57:43
%S 1,8,19,76,166,178,209,1907,20926,22925,32653,119136
%N Fractional part of e^a(n) is the largest yet.
%C Recursive definition: a(1)=1, a(n) = least number m>a(n-1) such that the fractional part of e^m is greater than the fractional part of e^k for all k, 1<=k<m.
%C The next such number must be greater than 100000. [_Hieronymus Fischer_, Jan 06 2009]
%C a(13) > 300,000. _Robert Price_, Mar 23 2019
%F Recursion: a(1):=1, a(k):=min{ m>1 | fract(e^m) > fract(e^a(k-1))}, where fract(x) = x-floor(x). [_Hieronymus Fischer_, Jan 06 2009]
%e a(2)=8, since fract(e^8)= 0.9579870417..., but fract(e^k)<=0.7182818... for 1<=k<=7;
%e thus fract(e^8)>fract(e^k) for 1<=k<8 and 8 is the minimal exponent > 1 with this property. [_Hieronymus Fischer_, Jan 06 2009]
%t a = 0; Do[b = N[ FractionalPart[ N[ E, 2^12]^n], 24]; If[b > a, Print[n]; a = b], {n, 1, 9400}] (* _Robert G. Wilson v_, Mar 16 2004 *)
%o (PARI) E=exp(1); /* use sufficient precision! */
%o ym=0;for(i=1,1000,x=E^i;y=x-floor(x);if(y>ym,print1(","i);ym=y))
%Y Cf. A153663, A153671, A153679, A153687, A153695, A153707, A154130, A153711, A153719, A000149. [_Hieronymus Fischer_, Jan 06 2009]
%K nonn,more
%O 1,2
%A _Jon Perry_, Mar 04 2004
%E a(8) from _Robert G. Wilson v_, Mar 16 2004
%E a(9)-a(11) from _Hieronymus Fischer_, Jan 06 2009
%E a(12) from _Robert Price_, Mar 23 2019
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