OFFSET
1,1
COMMENTS
a(n) is the first number that has n anagrams k such that A001222(k) = n.
Does 9 divide a(n) for n > 6? - David A. Corneth, Jan 16 2024
LINKS
David A. Corneth, Table of n, a(n) for n = 1..61
Robert Israel, Anagrams of a(n) with n prime divisors, for n = 1 to 18
EXAMPLE
a(4) = 135 is a term because 135 has 4 anagrams having 4 prime divisors, counted by multiplicity: 135 = 3^3 * 5, 315 = 3^2 * 5 * 7, 351 = 3^3 * 13 and 513 == 3^3 * 19, and no number < 135 works.
a(6) != 2367 because 2367 has exactly 7 anagrams with each having exactly 6 prime divisors (namely 2673, 3276, 3726, 6237, 6372, 6732, 7236). - David A. Corneth, Jan 16 2024
MAPLE
f:= proc(n) # numbers k such that n has k anagrams with Omega = k
local L, W, WS, V, 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);
W:= map(t -> numtheory:-bigomega(t), L);
WS:= convert(W, set);
for x in WS do V[x]:= 0 od;
for x in W do V[x]:= V[x]+1 od;
select(x -> V[x] = x, WS);
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:
V:= 'V': m:= 0:
for d from 1 to 9 do
for x in h(d) do
for y in f(x) do
if not assigned(V[y]) then V[y]:= x: m:= max(m, y) fi
od od od:
seq(V[y], y=1..m);
PROG
(Python)
from collections import Counter
from sympy import primeomega as W
from sympy.utilities.iterables import multiset_permutations as MP
from itertools import combinations_with_replacement, count, islice
def counteq(n):
c = Counter(W(int("".join(p))) for p in MP(str(n)) if p[0]!='0')
return [i for i in c if c[i] == i]
def agen(): # generator of terms
adict, n = dict(), 1
for d in count(len(str(2**n))):
for f in "123456789":
for r in combinations_with_replacement("0123456789", d-1):
k = int(f+"".join(r))
for v in counteq(k):
if v not in adict:
adict[v] = k
while n in adict: yield adict[n]; n += 1
print(list(islice(agen(), 8))) # Michael S. Branicky, Jan 16 2024
CROSSREFS
KEYWORD
nonn,base
AUTHOR
Robert Israel, Jan 15 2024
EXTENSIONS
More terms from David A. Corneth, Jan 16 2024
STATUS
approved