

A037207


Smallest natural number k such that periodic part of 1/k is n, or 0 if no such k exists.


3



1, 9, 45, 3, 144, 18, 6, 36, 72, 1, 99, 9, 825, 1584, 12672, 66, 61875, 193359375, 55, 405504, 495, 825, 45, 309375, 4125, 396, 792, 44, 6336, 7734375, 33, 1584, 309375, 3, 103809024, 50688, 88, 25952256, 202752, 528, 2475, 12672, 4125, 103809024, 144, 22, 1546875, 12976128
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OFFSET

0,2


COMMENTS

Suppose n is an mdigit number and write n/(10^m1) 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 3digit 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



EXAMPLE

1/1584 = 0.00063131313..., a(13)=1584.


CROSSREFS



KEYWORD

nonn,base,nice


AUTHOR



EXTENSIONS



STATUS

approved



