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 A308388 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). 0
 2, 3, 5, 7, 10, 14, 15, 21, 26, 34, 35, 38, 39, 46, 51, 57, 58, 62, 65, 69, 74, 82, 85, 86, 87, 91, 93, 94, 95, 102, 105, 116, 117, 130, 138, 154, 165, 170, 171, 174, 182, 186, 188, 190, 195, 212, 230, 231, 238, 242, 244, 246, 258, 273, 285, 286, 290, 292, 310 (list; graph; refs; listen; history; text; internal format)
 OFFSET 1,1 COMMENTS a(n) is composite for n > 4. The sequence contains two subsequences: Subsequence 1: numbers with distinct digits. This finite subsequence begins with the numbers 2, 3, 5, 7, 10, 14, 15, 21, 26, 34, ... 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. 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, ... 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. Numbers k such that the multisets of prime exponents and digit frequencies of k are identical. - Charlie Neder, Jul 05 2019 LINKS EXAMPLE 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. 2007972 = 2^2 * 3^2 * 17^2 * 193 is in the sequence. MAPLE with(numtheory):nn:=400: for n from 2 to nn do:   x:=convert(n, base, 10):n0:=length(n):   T:=Array(1..10, [0\$n0]):W:=Array(1..10):    for i from 0 to 9 do:     for j from 1 to n0 do:      if x[j]=i       then        T[i+1]:=T[i+1]+1:        else       fi:      od:     od:      W:=sort(T):s1:=sum(‘W[i]*10^(10-i)’, ‘i’=1..10):      d:=factorset(n):n1:=nops(d):      U:=Array(1..n1, [0\$n1]):V:=Array(1..n1):f:=ifactors(n):       for p from 1 to n1 do:        U[p]:=f[2][p][2]:       od:         V:=sort(U):s2:=sum(‘V[i]*10^(10-i)’, ‘i’=1..n1):         s2:=s2/10^(10-n1):          if s1=s2           then            printf(`%d, `, n):           else    od: MATHEMATICA Select[Range[2, 310], Sort[ Tally[IntegerDigits[#]][[;; , 2]] ] == Sort[ FactorInteger[#][[;; , 2]] ] &] (* Amiram Eldar, Jul 05 2019 *) CROSSREFS Cf. A001222. Sequence in context: A209000 A115024 A167050 * A019529 A194242 A173538 Adjacent sequences:  A308385 A308386 A308387 * A308389 A308390 A308391 KEYWORD nonn,base AUTHOR Michel Lagneau, May 23 2019 STATUS approved

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Last modified September 20 09:04 EDT 2021. Contains 347579 sequences. (Running on oeis4.)