A308388 - OEIS

%I #20 Jan 20 2020 21:43:49

%S 2,3,5,7,10,14,15,21,26,34,35,38,39,46,51,57,58,62,65,69,74,82,85,86,

%T 87,91,93,94,95,102,105,116,117,130,138,154,165,170,171,174,182,186,

%U 188,190,195,212,230,231,238,242,244,246,258,273,285,286,290,292,310

%N Let N(k, d_i) be the number of digits d_i in the decimal expansion of k, and let M(k, p_j) be the multiplicity of a prime factor p_j of k. The sequence lists the numbers k such that, for any digit d_i, there exists at least a prime factor p_j where N(k,d_i) = M(k,p_j), and for any prime factor p_j there exists at least a digit d_i where M(k,p_j) = N(k,d_i).

%C a(n) is composite for n > 4.

%C The sequence contains two subsequences:

%C Subsequence 1: numbers with distinct digits. This finite subsequence begins with the numbers 2, 3, 5, 7, 10, 14, 15, 21, 26, 34, ...

%C Subsequence 2: numbers with non-distinct digits. This subsequence begins with the numbers 116, 117, 171, 188, 212, 242, 244, ... with at least a prime factor of multiplicity greater than 1.

%C We can find a subsequence E where there exists a bijection between the set of the distinct digits D = {d_1, d_2, ..., d_r} and the set of the distinct prime factors P = {p_1, p_2, ..., p_r}. In other words, for each digit d_i there exists a unique prime factor p_j where N(k,d_i) = M(k,p_j), and for each prime factor p_j there exists a unique digit d_i where M(k,p_j) = N(k,d_i). This subsequence begins with the numbers 2, 3, 5, 7, 116, 117, 171, 188, ...

%C Remark: the numbers of the sequence of the form p_1^a * p_2^b * ... * p_i^m, where p_1, p_2, ..., p_i are prime factors, are in E if the integers a, b, ..., m are distinct. For example, the number 2001200 = 2^4 * 5^2 * 5003 is in E but 2007972 = 2^2 * 3^2 * 17^2 * 193 is not in E.

%C Numbers k such that the multisets of prime exponents and digit frequencies of k are identical. - _Charlie Neder_, Jul 05 2019

%e 188 = 2^2*47 is in the sequence because this number contains one digit "1" and two digits "8". The number 188 has a prime factor of multiplicity 1 and a prime factor of multiplicity 2. So, N(188, 1) = M(188, 47) = 1 and N(188, 8) = M(188, 2) = 2.

%e 2007972 = 2^2 * 3^2 * 17^2 * 193 is in the sequence.

%p with(numtheory):nn:=400:

%p for n from 2 to nn do:

%p   x:=convert(n,base,10):n0:=length(n):

%p   T:=Array(1..10,[0$n0]):W:=Array(1..10):

%p    for i from 0 to 9 do:

%p     for j from 1 to n0 do:

%p      if x[j]=i

%p       then

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

%p        else

%p       fi:

%p      od:

%p     od:

%p      W:=sort(T):s1:=sum(‘W[i]*10^(10-i)’, ‘i’=1..10):

%p      d:=factorset(n):n1:=nops(d):

%p      U:=Array(1..n1,[0$n1]):V:=Array(1..n1):f:=ifactors(n):

%p       for p from 1 to n1 do:

%p        U[p]:=f[2][p][2]:

%p       od:

%p         V:=sort(U):s2:=sum(‘V[i]*10^(10-i)’, ‘i’=1..n1):

%p         s2:=s2/10^(10-n1):

%p          if s1=s2

%p           then

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

%p           else

%p    od:

%t Select[Range[2, 310], Sort[ Tally[IntegerDigits[#]][[;; , 2]] ] == Sort[ FactorInteger[#][[;; , 2]] ] &] (* _Amiram Eldar_, Jul 05 2019 *)

%Y Cf. A001222.

%K nonn,base

%O 1,1

%A _Michel Lagneau_, May 23 2019