OFFSET
0,2
COMMENTS
Comment from N. J. A. Sloane: (Start)
Suppose n is an m-digit number and write n/(10^m-1) in lowest terms as r/s.
A necessary condition for k to exist is that 2^w*5^x=r (mod s) be solvable for w and x. This certainly fails for s=1353 and some r, since 2 and 5 have order 20 mod s and 20*20<phi(1353)=800.
So certainly a(n)=0 for some numbers n with 800 digits. When is the first time a(n)=0? I believe a(n)=0 will eventually occur. (End)
After finding a(n) for n up to 100, it is easy to verify that n=101 is the first time a(n)=0. Because 101 is a 3-digit number, m=3. In lowest terms, r/s = 101/999 because 101 is prime. Modulo 999, both 2^w and 5^x have period 36. Checking the 36^2 values of 2^w*5^x mod 999, we find none equal to 101 (or 110 or 011). - T. D. Noe, Mar 21 2007
LINKS
T. D. Noe, Table of n, a(n) for n = 0..1000
EXAMPLE
1/1584 = 0.00063131313..., a(13)=1584.
CROSSREFS
KEYWORD
nonn,base,nice
AUTHOR
EXTENSIONS
More terms from N. J. A. Sloane, Jun 13 1998
STATUS
approved