A228893 Numbers k such that the concatenation of the frequency of each decimal digit of k is divisible by k.

1, 2, 4, 5, 10, 14, 20, 25, 32, 40, 50, 52, 73, 100, 112, 120, 125, 150, 152, 188, 200, 226, 240, 250, 300, 305, 320, 335, 400, 420, 500, 600, 674, 700, 1000, 1100, 1144, 1210, 1250, 1430, 2000, 2020, 2500, 2660, 3250, 4000, 4015, 5000, 5500, 6500, 8629, 10000

Offset

1,2

Comments

Start with any number k and denote by (f0, f1, ..., f9) the number of the digits 0, 1, 2,..., 9 of k. The sequence lists the numbers k such that m = f0f1f2f3f4f5f6f7f8f9/k is an integer.

The corresponding m are: 100000000, 5000000, 25000, 2000, 110000000, 7150000, 50500000, 400400, 343750, 25002500, 20000200, 192500, 13700, 21000000, 1875000, 9250000, 880080, 7333400, 723750, 531915, ...

Links

Harvey P. Dale, Table of n, a(n) for n = 1..79 (all terms up to 1 million)

Example

226 is in the sequence because the table of the frequencies is:
  frequency of 0 = 0;
  frequency of 1 = 0;
  frequency of 2 = 2;
  frequency of 3 = 0;
  frequency of 4 = 0;
  frequency of 5 = 0;
  frequency of 6 = 1;
  frequency of 7 = 0;
  frequency of 8 = 0;
  frequency of 9 = 0.
The concatenation of the frequencies is 20001000, and 20001000/226 = 88500.

Maple

with(numtheory): T:=array(0..9):
  for n from 1 to 200000 do:
     x:=convert(n, base, 10):n1:=nops(x):(1..n1):
       for k from 0 to 9 do:
          T[k]:=0:od:jj:=0:
            for i from 0 to 9 do:
              for j from 1 to n1 do:
                if x[j]=i
                then
                T[i]:=T[i]+1:jj:=jj+1:
                else
                fi:
              od:
            od:
               s:= sum('T[9-i]*10^i', 'i'=0..9):s1:=s/n:
               if floor(s1)=s1
               then
               printf(`%d, `, n):
               else
               fi:
       od:
# second Maple program:
q:= n-> (p-> is(irem(parse(cat(seq(coeff(p, x, i), i=0..9)
         )), n)=0))(add(x^i, i=convert(n, base, 10))):
select(q, [$1..10000])[];  # Alois P. Heinz, Aug 29 2022

Mathematica

Select[Range[10000], Divisible[FromDigits[RotateRight[DigitCount[#]]], #]&] (* Harvey P. Dale, Aug 29 2022 *)

Prog

(Python)
def ok(n): s = str(n); return n > 0 and int("".join(str(s.count(d)) for d in "0123456789"))%n == 0
print([k for k in range(10001) if ok(k)]) # Michael S. Branicky, Aug 29 2022

Crossrefs

Sequence in context: A002237 A329136 A067935 * A272622 A097133 A023165

Adjacent sequences: A228890 A228891 A228892 * A228894 A228895 A228896

Keyword

nonn,base,less

Author

Michel Lagneau, Sep 07 2013

Status

approved