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Search: id:A037207
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| A037207 |
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Smallest natural number k such that periodic part of 1/k is n, or 0 if no such k exists. |
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+0
3
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| 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
(list; graph; listen)
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OFFSET
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0,2
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COMMENT
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Comment from N. J. A. Sloane (njas(AT)research.att.com): 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?
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
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LINKS
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T. D. Noe, Table of n, a(n) for n=0..1000
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EXAMPLE
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1/1584 = 0.00063131313..., a(13)=1584.
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CROSSREFS
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Cf. A037211.
Sequence in context: A104470 A084016 A125679 this_sequence A096688 A124983 A087969
Adjacent sequences: A037204 A037205 A037206 this_sequence A037208 A037209 A037210
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KEYWORD
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nonn,base,nice
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AUTHOR
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Jean Marc Rebert (jm.rebert(AT)rmcnet.fr)
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EXTENSIONS
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More terms from N. J. A. Sloane (njas(AT)research.att.com) 6/13/98. I believe a(n)=0 will eventually occur.
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