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%I #50 Jan 05 2026 14:05:15
%S 3,6,47,56,920483,46341,888667668,13207922282067
%N Smallest number k such that log_b(k) - floor(log_b(k)) >= 1/2 for b = 2..n, and not for b = n+1.
%C Probably infinite.
%C All the terms are of the form b^(y/2), where b is any of 2..n and where y is any odd number greater than 1. (See the link for proof.)
%C It is suspected that terms of the form (n+1)^z could exist, where z is any integer greater than 0.
%H Jakub Buczak, <a href="/A392007/a392007_3.pdf">A guide to A392007</a>
%e a(4) = 47, because 47 is the smallest number k for which the inequality log_b(47) - ceiling(log_b(47)) >= 1/2 holds for b = 2,3,4 and not for b=5.
%Y Cf. A020857, A153459.
%Y Cf. A002193, A002194, A002163, A010464.
%K nonn,more
%O 2,1
%A _Jakub Buczak_, Dec 26 2025