A369203 a(n) is the first number that has exactly n anagrams that each have exactly n prime divisors, counted by multiplicity.
2, 15, 117, 135, 1224, 10023, 10026, 50688, 104445, 100368, 1012257, 1002258, 1034568, 10027899, 10024569, 100002789, 100234566, 100236789, 1000024569, 1012566789, 10000224468, 10002367899, 10002345678, 100012344588, 100012234689, 100223456778, 1000012457889, 1002345566778
Offset: 1
Examples
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
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
Programs
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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); -
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
Extensions
More terms from David A. Corneth, Jan 16 2024
Comments