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A228893 Numbers n such that the concatenation of the frequency of each decimal digit of n is divisible by n. 0

%I

%S 1,2,4,5,10,14,20,25,32,40,50,52,73,100,112,120,125,150,152,188,200,

%T 226,240,250,300,305,320,335,400,420,500,600,674,700,1000,1100,1144,

%U 1210,1250,1430,2000,2020,2500,2660,3250,4000,4015,5000,5500,6500,8629,10000

%N Numbers n such that the concatenation of the frequency of each decimal digit of n is divisible by n.

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

%C The corresponding m are:

%C 100000000, 5000000, 25000, 2000, 110000000, 7150000, 50500000, 400400, 343750, 25002500, 20000200, 192500, 13700, 21000000, 1875000, 9250000, 880080, 7333400, 723750, 531915,...

%e 226 is in the sequence because the table of the frequencies is:

%e frequency of 0 = 0;

%e frequency of 1 = 0;

%e frequency of 2 = 2;

%e frequency of 3 = 0;

%e frequency of 4 = 0;

%e frequency of 5 = 0;

%e frequency of 6 = 1;

%e frequency of 7 = 0;

%e frequency of 8 = 0;

%e frequency of 9 = 0.

%e The concatenation of the frequencies is 20001000, and 20001000/226 = 88500.

%p with(numtheory): T:=array(0..9):

%p for n from 1 to 200000 do:

%p x:=convert(n,base,10):n1:=nops(x):(1..n1):

%p for k from 0 to 9 do:

%p T[k]:=0:od:jj:=0:

%p for i from 0 to 9 do:

%p for j from 1 to n1 do:

%p if x[j]=i

%p then

%p T[i]:=T[i]+1:jj:=jj+1:

%p else

%p fi:

%p od:

%p od:

%p s:= sum('T[9-i]*10^i', 'i'=0..9):s1:=s/n:

%p if floor(s1)=s1

%p then

%p printf(`%d, `,n):

%p else

%p fi:

%p od:

%K nonn,base,less

%O 1,2

%A _Michel Lagneau_, Sep 07 2013

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Last modified January 23 07:56 EST 2022. Contains 350510 sequences. (Running on oeis4.)