cp's OEIS Frontend

A281384 Least integer with more than 1 digit in base n, such that the set of its base-n digits equals the set of its binary digits.

%I A281384 #16 Jan 22 2017 13:00:11
%S A281384 2,9,4,5,6,49,8,9,10,11,12,13,14,225,16,17,18,19,20,21,22,23,24,25,26,
%T A281384 27,28,29,30,961,32,33,34,35,36,37,38,39,40,41,42,43,44,45,46,47,48,
%U A281384 49,50,51,52,53,54,55,56,57,58,59,60,61,62,3969,64,65,66,67,68,69,70
%N A281384 Least integer with more than 1 digit in base n, such that the set of its base-n digits equals the set of its binary digits.
%C A281384 "More than 1 digit in base n" is equivalent to ">= n", which also implies "more than 1 digit in base 2", since n >= 2 by definition. If such a number written in binary is not a string of 1's, then its set of binary digits is {0, 1}, and the smallest number to have the same digits in base n is n = 10[n] itself. If n has all binary digits equal to 1, i.e., n = 2^k-1, then the smallest solution is the next larger number having the digits {0, 1} in base n (and also in base 2), which is, since 11[n] = n+1 is excluded, 100[n] = n^2 = (2^k-1)^2 = 2^{2k} + 2^{k+1} + 1 = 1...01[2]. - _M. F. Hasler_, Jan 22 2017
%F A281384 For any n>1: if n belongs to A000225, then a(n)=n^2, otherwise a(n)=n. - _Rémy Sigrist_, Jan 22 2017
%o A281384 (PARI) a(n) = my(m=n); while (Set(digits(m, n)) != Set(digits(m,2)), m++); m;
%o A281384 (PARI) A281384(n) = if(n+1==1<<logint(n),n^2,n) \\ _M. F. Hasler_, Jan 22 2017
%Y A281384 First column of triangular array A281383.
%K A281384 nonn
%O A281384 2,1
%A A281384 _Michel Marcus_, Jan 21 2017