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Smallest natural number k such that periodic part of 1/k is n, or 0 if no such k exists.
3

%I #21 Feb 07 2019 02:11:17

%S 1,9,45,3,144,18,6,36,72,1,99,9,825,1584,12672,66,61875,193359375,55,

%T 405504,495,825,45,309375,4125,396,792,44,6336,7734375,33,1584,309375,

%U 3,103809024,50688,88,25952256,202752,528,2475,12672,4125,103809024,144,22,1546875,12976128

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

%C Comment from _N. J. A. Sloane_: (Start)

%C Suppose n is an m-digit number and write n/(10^m-1) in lowest terms as r/s.

%C 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.

%C 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)

%C 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

%H T. D. Noe, <a href="/A037207/b037207.txt">Table of n, a(n) for n = 0..1000</a>

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

%Y Cf. A037211.

%K nonn,base,nice

%O 0,2

%A _Jean-Marc Rebert_

%E More terms from _N. J. A. Sloane_, Jun 13 1998