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A267795 Integers n such that n, 2n, 3n ... 10n contain almost equally many copies of each base 10 digit.

%I A267795 #17 Mar 19 2023 15:32:31
%S A267795 1,9,109,909,10909,90909,1090909,9090909,13431958,25834963,32973507,
%T A267795 38296415,45096237,51546969,94845303,96237045,109090909,113431958,
%U A267795 126084879,132868745,132875488,133595248,134319558,134755956,134758658,137584878,143865844,153584878
%N A267795 Integers n such that n, 2n, 3n ... 10n contain almost equally many copies of each base 10 digit.
%C A267795 Here 'almost equally many' means that the most common digit appears only once more than the least common.
%H A267795 Lars Blomberg, <a href="/A267795/b267795.txt">Table of n, a(n) for n = 1..4000</a>
%e A267795 The first 10 multiples of 109 are 109, 218, 327, 436, 545, 654, 763, 872, 981, 1090. Every digit appears 3 times except for '1' which appears 4 times. It is clear that all numbers of the form 10909..0909 and 90909..0909 appear in the list, and it seems likely that these are the only members.
%o A267795 (Python)
%o A267795 def f(n):
%o A267795   """ This returns True iff n is in the sequence """
%o A267795   l = [ n * i for i in range(1, 11) ]
%o A267795   s = "".join(str(i) for i in l)
%o A267795   c = [ s.count(str(j)) for j in range(10) ]
%o A267795   return min(c) >= max(c) - 1
%o A267795 for n in range(1, 10000000):
%o A267795   if f(n):
%o A267795     print(n, end=', ')
%Y A267795 Cf. A038365.
%K A267795 nonn,easy,base
%O A267795 1,2
%A A267795 _Jack W Grahl_, Jan 20 2016
%E A267795 a(7)-a(28) from _Lars Blomberg_, Aug 11 2016