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

1, 2, 5, 22, 105, 188, 258, 327, 663, 15425, 16654, 27848, 40324, 80328, 481263, 10213223, 10311233, 10313314, 10313315, 10313316, 10313317, 10313318, 10313319, 21322314, 21322315, 21322316, 21322317, 21322318, 21322319, 31123314, 31123315, 31123316, 31123317

Offset: 1

Author

Paolo P. Lava, Nov 22 2016

Keywords

• nonn
• base
• easy
• fini

Comments

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.

Examples

			A237605(258) = 121518 and 121518/258 = 471.
		

Links

• Paolo P. Lava, List of terms and corresponding ratios for n <= 10^9.

Crossrefs

Cf. A047842, A237605, A278440, A278441.

Programs

Maple

with(numtheory): P:=proc(q) local a,b,c,d,j,k,n; for n from 1 to q do
a:=sort(convert(n,base,10)); k:=1; b:=a[1]; c:=0; for j from 2 to nops(a) do
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;
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);

Extensions

a(32) corrected by Sean A. Irvine, May 27 2025