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A278439 Numbers k such that k | A047842(k).

%I A278439 #20 May 27 2025 10:33:50
%S A278439 1,2,5,22,105,188,258,327,663,15425,16654,27848,40324,80328,481263,
%T A278439 10213223,10311233,10313314,10313315,10313316,10313317,10313318,
%U A278439 10313319,21322314,21322315,21322316,21322317,21322318,21322319,31123314,31123315,31123316,31123317
%N A278439 Numbers k such that k | A047842(k).
%C A278439 The sequence is bounded. Let us consider a k-digit number n in which all 10 numerals from 0 to 9 are equally distributed: there are k/10 0's, k/10 1's, etc. This is the best case in order to have a number with the greatest number of digits under the transform n -> A047842(n). The number of digits we get is 10 + 10*floor(log_10(k/10) + 1), which must be >= k. The inequality becomes log_10(k/10) >= k/10 - 2, which is solved by k <= 23.75... This means that no term of the sequence can have more than 23 digits.
%H A278439 Paolo P. Lava, <a href="/A278439/a278439.txt"> List of terms and corresponding ratios for n <= 10^9.</a>
%e A278439 A237605(258) = 121518 and 121518/258 = 471.
%p A278439 with(numtheory): P:=proc(q) local a,b,c,d,j,k,n; for n from 1 to q do
%p A278439 a:=sort(convert(n,base,10)); k:=1; b:=a[1]; c:=0; for j from 2 to nops(a) do
%p A278439 if a[j]=b then k:=k+1; else d:=10*k+b; c:=c*10^(ilog10(d)+1)+d; k:=1; b:=a[j]; fi; od;
%p A278439 d:=10*k+b; c:=c*10^(ilog10(d)+1)+d; if type(c/n,integer) then print(n); fi; od; end: P(10^10);
%Y A278439 Cf. A047842, A237605, A278440, A278441.
%K A278439 nonn,base,easy,fini
%O A278439 1,2
%A A278439 _Paolo P. Lava_, Nov 22 2016
%E A278439 a(32) corrected by _Sean A. Irvine_, May 27 2025