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%I #13 Jul 29 2022 09:53:16
%S 61,131,181,461,491,541,571,701,811,821,941,971,1021,1051,1091,1171,
%T 1181,1291,1301,1349,1381,1531,1571,1621,1741,1811,1829,1861,2141,
%U 2221,2251,2341,2371,2411,2621,2731,2741,2851,2861,2971,3011,3221,3251,3301
%N Primitive numbers whose decimal expansion of 1/n is equidistributed in base 10.
%C Usually once a number has the desired property, so do all its multiples. However there are exceptions. 61*7 in base 10 is not equidistributed. Multiples of earlier numbers are not included here.
%C From _Jianing Song_, Jul 29 2022: (Start)
%C There are 58 composite terms below 100000, 2 among which being even: a(239) = 25064 = 2^3 * 13 * 241, and a(613) = 72728 = 2^3 * 9091.
%C Conjecture 1: let p be a prime such that ord(10,p) is a multiple of 10, where ord(a,m) denotes the multiplicative order of a modulo m. Then p is a term if and only if 10 is a primitive root modulo p.
%C Conjecture 2: suppose that m is a term with bigomega(m) = 2, then m = p*q, where p == 1 (mod 10), q == 9 (mod 10), gcd(p-1,q-1) = 2, ord(10,p) = (p-1)/2, and ord(10,q) = q-1. Note that the converse is not true, though.
%C There are no counterexamples to the conjectures above below 100000.
%C Is there any odd term m such that bigomega(m) > 2? (End)
%D David Wells, The Penguin Dictionary of Curious and Interesting Numbers, revised edition, London, England, 1997, entry 61, page 110.
%H Jianing Song, <a href="/A073761/b073761.txt">Table of n, a(n) for n = 1..808</a> (all terms <= 100000)
%e 61 is a term because 1/61 = .016393... (period 60 digits, 6 of each 0,1,..9).
%t a = {}; Do[d = RealDigits[1/n][[1, 1]]; If[ !IntegerQ[d] && Count[d, 0] == Count[d, 1] == Count[d, 2] == Count[d, 3] == Count[d, 4] == Count[d, 5] == Count[d, 6] == Count[d, 7] == Count[d, 8] == Count[d, 9], If[ Select[n/a, IntegerQ] == {}, a = Append[a, n]]], {n, 11, 3330}]; a
%Y Cf. A074709.
%K nonn,base
%O 1,1
%A _Donald S. McDonald_, Sep 02 2002
%E Edited by _Robert G. Wilson v_, Sep 06 2002
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