OFFSET
2,1
COMMENTS
As with A146025, it is a plausible conjecture that there are no more terms, but this has not been proved. - Daniel Mondot, Dec 16 2016
From Devansh Singh, Jul 14 2026: (Start)
Suppose a(n) exists, and let N = a(n). For each base b <= n, write N = Sum_{i>=0} A_b(i)*b^i, where each A_b(i) is 0 or 1.
For any c with 1 <= c < b, we have N - Sum_{i>=0} A_b(i)*c^i = Sum_{i>=1} A_b(i)*(b^i-c^i). Since b-c divides b^i-c^i, it follows that b-c divides N - Sum_{i>=0} A_b(i)*c^i.
In particular, for b = 6, 4 divides N - Sum A_6(i)*2^i. Since N also has only digits 0 and 1 in base 4, N == 0 or 1 (mod 4).
But N == A_6(0) + 2*A_6(1) (mod 4), because 2^i == 0 (mod 4) for i >= 2. Therefore A_6(1) = 0.
Hence the last two base-6 digits of N are 00 or 01, so N == 0 or 1 (mod 36). Thus, if a(6) exists, it must be of the form 36*q or 36*q+1, where q >= 1. (End)
LINKS
Thomas Oléron Evans, Solution: Covering all the bases
Richard Green, A Curious Property of 82000
James Grime and Brady Haran, Why 82,000 is an extraordinary number, Numberphile video, 2015.
Devansh Singh, Some necessary congruence conditions for the possible existence of a(6) in OEIS A258107, Jun 24 2026.
EXAMPLE
a(4) = 4 because it is 100 in base 2, 11 in base 3 and 10 in base 4. No smaller number, except 1, can be expressed in such bases with only 0's and 1's.
a(5) = 82000: 82000 in bases 2 through 5 is 10100000001010000, 11011111001, 110001100, 10111000, containing only 0's and 1's, while all smaller numbers have a larger digit in one of those bases. For example, 12345 is 11000000111001, 121221020, 3000321, 343340. - N. J. A. Sloane, Feb 01 2016
MATHEMATICA
Table[k = 2; While[Total[Total@ Drop[RotateRight[DigitCount[k, #]], 2] & /@ Range[3, n]] > 0, k++]; k, {n, 2, 5}] (* Michael De Vlieger, Aug 29 2015 *)
CROSSREFS
KEYWORD
nonn,base,more
AUTHOR
Bernardo Boncompagni, May 20 2015
STATUS
approved