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%I #34 Apr 17 2023 10:18:09
%S 142857,588235294117647,52631578947368421,434782608695652173913,
%T 344827586206896551724137931,
%U 212765957446808510638297872340425531914893617,169491525423728813559322033898305084745762711864406779661
%N Numbers with x digits such that the first x multiples are cyclic permutations of the number, leading 0's omitted (or cyclic numbers).
%C Periodic part of decimal expansion of 1/A001913(n). The number of digits in each term (including leading zeros), plus one, makes the sequence A001913.
%H Ray Chandler, <a href="/A180340/b180340.txt">Table of n, a(n) for n = 1..60</a>
%H Edwin E. Freed, <a href="https://archive.org/details/dr_dobbs_journal_vol_08/page/176/mode/2up">Binary Magic Numbers</a>, Dr. Dobb's Journal, Vol. 78 (April 1983), pp. 24-37.
%H OEIS Wiki, <a href="/wiki/Cyclic_numbers">Cyclic numbers</a>
%H Eric Weisstein's World of Mathematics, <a href="http://mathworld.wolfram.com/CyclicNumber.html">Cyclic number</a>
%H Wikipedia, <a href="http://en.wikipedia.org/wiki/Cyclic_number">Cyclic number</a>
%F a(n) = (10^(A001913(n)-1) - 1) / A001913(n).
%e 142857 is in the sequence because it has 6 digits and the first 6 multiples of 142857 are 142857, 285714, 428571, 571428, 714285, and 857142, all cyclic permutations of the number. Also the first term of A001913 is 7, and 1/7 = 0.142857142857... .
%e 588235294117647 is the next number because 0588235294117647 has 16 digits and the first 16 multiples are cyclic permutations of the number; the second term of A001913 is 17, and 1/17 = 0.05882352941176470588235294117647... .
%t Map[(10^(# - 1) - 1)/# &, Select[Prime@ Range@ 17, MultiplicativeOrder[10, #] == # - 1 &]] (* _Michael De Vlieger_, Apr 03 2017 *)
%Y A006883 starting from the second term of A006883, omitting ending 0's.
%Y The n-th terms of A060284 where n is a member of A001913.
%K base,nonn
%O 1,1
%A Ralph Kerchner (daxkerchner(AT)hotmail.com), Aug 28 2010
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