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A369184
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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.
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2
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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
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OFFSET
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0,2
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COMMENTS
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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.
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LINKS
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EXAMPLE
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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.
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MAPLE
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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);
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PROG
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(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
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CROSSREFS
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KEYWORD
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nonn,base
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AUTHOR
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STATUS
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approved
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