COMMENTS

a(n) might be called the "smallest trans-basic multiple of n." In order to be a valid binary number, the terms may contain only 0's and 1's. The number of 1's in a(n) is conjectured to be n; the number of 0's separating each one digit is usually A268336(n)-1 for small n. The number of trailing 0's is A051903(n). The first 21 terms have been found and verified to be minimal via an advanced search; the 13th term (see b-file) contains 146 binary digits, and in general for every prime n the corresponding a(n) is conjectured to contain 2+(n-1)^2 0's and 1's. A lower bound for a(n) is given by a(A032742(n)). Proof: If a(n) were smaller than a(A032742(n)), then a(A032742(n)) would not be the smallest trans-basic multiple of A032742(n); a(n) would be. By definition a(n) is the smallest trans-basic multiple of n, so we have a contradiction; QED. To verify a trans-basic multiple of n for n > 2, one must only: A) make sure the string has some multiple of n of '1' digits; B) make sure the string ends with at least one '0' digit; and C) check that, for all prime bases below n, the resulting number is divisible by n. If these three conditions are met, the string is a trans-basic multiple of n. While the formula given below is guaranteed to provide a trans-basic multiple of n, it does not always yield a(n) which by definition is the smallest such number. [Corrected by M. F. Hasler, Nov 14 2019] From N. J. A. Sloane, Nov 12 2019: (Start) For each n, the values of (string a(n) read in base b)/n for b = 1,2,3,... give a sequence of integers. For n=1 this is the all-1's sequence A000012. For n=2, a(2) = 110 which in base b is b+b^2. Divided by 2 we get (b+b^2)/2, which evaluated at b = 1,2,3,4,... is 1,3,6,10, ..., the triangular numbers A000217. For n=3, we get (b+b^3+b^5)/3, which is A220892. For n=4, we get A328994. (End)