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Numbers m such that v^n - u^m >= u^(m+1) - v^n, where u=2, v=3, and u^m < v^n < u^(m+1).
1

%I #10 May 31 2024 14:26:04

%S 1,3,5,8,10,13,15,17,20,22,25,27,29,32,34,37,39,41,44,46,49,51,54,56,

%T 58,61,63,66,68,70,73,75,78,80,82,85,87,90,92,94,97,99,102,104,107,

%U 109,111,114,116,119,121,123,126,128,131,133,135,138,140,143,145

%N Numbers m such that v^n - u^m >= u^(m+1) - v^n, where u=2, v=3, and u^m < v^n < u^(m+1).

%e The condition u^m < v^n < u^(m + 1) implies m = floor(n*log(v)/log(u)). With u=2 and v=3, for n = 1, we have m = 1 and 3 - 2 >= 4 - 3, so 1 is in this sequence. For n = 2, we have m = 3 and 9 - 8 < 16 - 9, so 2 is in A372779.

%t z = 200; {u, v} = {2, 3};

%t m[n_] := Floor[n*Log[v]/Log[u]];

%t Table[m[n], {n, 0, z}];

%t s = Select[Range[z], v^# - u^m[#] < u^(m[#] + 1) - v^# &] (* A372779 *)

%t Complement[Range[Max[s]], s] (* this sequence *)

%Y Cf. A000079, A000244, A056576, A372779 (complement).

%K nonn

%O 1,2

%A _Clark Kimberling_, May 18 2024