

A278439


Numbers n such that n  A047842(n).


3



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, 1123316, 31123317
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OFFSET

1,2


COMMENTS

The sequence is bounded. Let us consider a kdigit 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.


LINKS

Table of n, a(n) for n=1..33.
Paolo P. Lava, List of terms and corresponding ratios for n <= 10^9.


EXAMPLE

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


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);


CROSSREFS

Cf. A047842, A237605, A278440, A278441.
Sequence in context: A321608 A241345 A272821 * A126797 A101206 A294467
Adjacent sequences: A278436 A278437 A278438 * A278440 A278441 A278442


KEYWORD

nonn,base,easy,fini


AUTHOR

Paolo P. Lava, Nov 22 2016


STATUS

approved



