cp's OEIS Frontend

A364089 a(n) is the greatest k such that the base-n digits of 2^k are all distinct.

%I A364089 #16 Jul 20 2024 10:53:48
%S A364089 1,1,3,4,5,8,5,10,29,19,19,19,16,18,7,43,41,37,45,39,55,33,43,60,35,
%T A364089 61,56,50,44,69,9,64,44,80,43,88,53,71,56,68,59,78,76,74,95,109,111,
%U A364089 81,86,136,117,75,98,83,84,99,104,116,95,118,60,81,11,119,119,172,140,97,105,113,93,122,92
%N A364089 a(n) is the greatest k such that the base-n digits of 2^k are all distinct.
%C A364089 a(n) <= log_2(A062813(n)).
%e A364089 a(10) = 29 because all decimal digits of 2^29 = 536870912 are distinct.
%p A364089 f:= proc(b) local M,k,L;
%p A364089   M:= b^b - (b^b-b)/(b-1)^2;
%p A364089   for k from ilog2(M) to 1 by -1 do
%p A364089     L:= convert(2^k,base,b);
%p A364089     if nops(L) = nops(convert(L,set)) then return k fi
%p A364089   od
%p A364089 end proc:
%p A364089 map(f, [$2..100]);
%o A364089 (Python)
%o A364089 from sympy.ntheory.factor_ import digits
%o A364089 def A364089(n):
%o A364089     m = 1<<(l:=((r:=n**n)-(r-n)//(n-1)**2).bit_length()-1)
%o A364089     while len(d:=digits(m,n)[1:]) > len(set(d)):
%o A364089         l -= 1
%o A364089         m >>= 1
%o A364089     return l # _Chai Wah Wu_, Jul 07 2023
%Y A364089 Cf. A004642, A004643, A000866, A004645, A004646, A004647, A001357, A000079, A364049.
%K A364089 nonn,base
%O A364089 2,3
%A A364089 _Robert Israel_, Jul 04 2023