A369184 a(n) is the first positive number that has exactly n anagrams which have 3 prime divisors, counted by multiplicity, or 0 if there is no such number.

1, 8, 103, 117, 156, 268, 1038, 1027, 1059, 1246, 1245, 1347, 1578, 3789, 10136, 10126, 10234, 10355, 10157, 10236, 10158, 11456, 10247, 10245, 10289, 10237, 10235, 10347, 10256, 10257, 10246, 10789, 10239, 10579, 12567, 10578, 13457, 12369, 14559, 12458, 12579, 23789, 24789, 12459, 100258, 12345

Offset: 0

Author

Robert Israel, Jan 15 2024

Keywords

• nonn
• base

Comments

An anagram of a number k is a permutation of the base-10 digits of k with no leading 0's.

a(n) is the first k such that A175854(k) = n, or 0 if n is not a term of A175854.

Conjecture: all a(n) > 0.

Examples

			a(5) = 268 because 268 has 5 anagrams that have 3 prime divisors, counted by multiplicity, and is the first number that does that:
268 = 2^2 * 67, 286 = 2 * 11 * 13, 628 = 2^2 * 157, 682 = 2 * 11 * 31, 826 = 2 * 7 * 59.
		

Links

• Robert Israel, Table of n, a(n) for n = 0..1000

Crossrefs

Cf. A014612, A175854. All terms are in A179239.

Programs

Maple

f:= proc(n) local L, d, w, x, i;
      L:= convert(n, base, 10); d:= nops(L);
      L:= select(t -> t[-1] <> 0, combinat:-permute(L));
      L:= map(t-> add(t[i]*10^(i-1), i=1..d), L);
      nops(select(t -> numtheory:-bigomega(t)=3, L))
end proc:
g:= proc(xin,d,n) # first anagrams with n digits starting xin, all other digits >= d
  option remember;
  local i;
  if 1 + ilog10(xin) = n then return xin fi;
  seq(procname(10*xin+i,i,n), i=d..9)
end proc:
h:= proc(n) # first anagrams with n digits
  local i,j;
  seq(seq(g(i*10^j,i,n),j=n-1..0,-1),i=1..9)
end proc:
N:= 100: # for a(0) .. a(N)
V:= Array(0..N): count:= 0:
for i from 1 while count < N+1 do
  for x in [h(i)] while count < N+1 do
    v:= f(x);
    if v <= N and V[v] = 0 then V[v]:= x; count:= count+1; fi
  od
od:
convert(V,list);

Python

from sympy import primeomega
from sympy.utilities.iterables import multiset_permutations
from itertools import combinations_with_replacement, count, islice
def func(n): return sum(1 for p in multiset_permutations(str(n)) if p[0]!='0' and primeomega(int("".join(p)))==3)
def agen(): # generator of terms
    adict, n = dict(), 0
    for d in count(1):
        for f in "123456789":
            for r in combinations_with_replacement("0123456789", d-1):
                k = int(f+"".join(r))
                v = func(k)
                if v not in adict:
                    adict[v] = k
                    while n in adict: yield adict[n]; n += 1
print(list(islice(agen(), 44))) # Michael S. Branicky, Jan 15 2024