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A153725 Least number m such that floor((3^n-m)/(2^n-m)) > floor(3^n/2^n).

%I A153725 #22 Jul 01 2019 01:49:17
%S A153725 1,2,2,3,2,4,7,4,8,7,12,9,17,4,8,16,99,20,39,235,49,97,194,885,1106,
%T A153725 439,2059,968,4034,5268,3070,1163,2325,4649,9297,18593,16210,4452,
%U A153725 8903,67524,68757,49124,98248,39360,288234,17763,35526,567677,1135354
%N A153725 Least number m such that floor((3^n-m)/(2^n-m)) > floor(3^n/2^n).
%C A153725 Provided A002379(n) = floor((3^n-1)/(2^n-1)) holds (which is proved only for 1 < n <= 305000), then a(n) > 1.
%H A153725 David A. Corneth, <a href="/A153725/b153725.txt">Table of n, a(n) for n = 1..8005</a>
%F A153725 a(n) = ceiling(((f + 1)*(2^n) - 3^n)/f) where f = floor(3^n/2^n). - _David A. Corneth_, Mar 27 2019
%e A153725 a(5)=2, since floor((3^5-1)/(2^5-1)) = floor(242/31) = 7 = floor(243/32) = floor(3^5/2^5), but floor((3^5-2)/(2^5-2)) = floor(241/30) = 8 > 7.
%t A153725 Table[n3 = 3^n; n2 = 2^n; m = 1;
%t A153725 While[Floor[(n3 - m)/(n2 - m)] <= Floor[n3/n2], m++]; m, {n,1,50}] (* _Robert Price_, Mar 27 2019 *)
%o A153725 (PARI) a(n) = my(f = floor(3^n/2^n)); ceil(((f + 1)*(2^n) - 3^n)/f) \\ _David A. Corneth_, Mar 27 2019
%Y A153725 Cf. A002379, A081464, A153701, A137994, A153717, A154130.
%K A153725 nonn,easy
%O A153725 1,2
%A A153725 _Hieronymus Fischer_, Jan 06 2009