A131371 Number of anagrams of n that are semiprimes.
0, 0, 0, 1, 0, 1, 0, 0, 1, 1, 0, 1, 0, 1, 2, 0, 0, 0, 1, 0, 1, 1, 0, 0, 1, 2, 0, 1, 0, 0, 0, 0, 1, 1, 1, 0, 0, 1, 2, 0, 1, 0, 1, 0, 0, 1, 1, 0, 2, 0, 2, 1, 1, 0, 1, 1, 1, 2, 1, 0, 0, 2, 0, 1, 1, 0, 0, 1, 1, 0, 0, 0, 0, 1, 1, 0, 1, 1, 0, 0, 0, 1, 1, 0, 2, 1, 1, 0, 0, 0, 1, 0, 2, 2, 1, 1, 0, 0, 0, 0, 0, 1, 1, 0, 1
Offset: 1
Examples
a(123) = 3 because 123 = 3 * 41 is semiprime, 213 = 3 * 71 is semiprime, 321 = 3 * 107 is semiprime, while the other anagrams 132, 231 and 312 have respectively 3, 3 and 5 prime factors with multiplicity. a(129) = 4 because 129 = 3 * 43 is semiprime, 219 = 3 * 73 is semiprime, 291 = 3 * 97 is semiprime, 921 = 3 * 307 is semiprime, while 192 and 912 have 7 and 6 prime factors with multiplicity. a(134) = 5 because 134 = 2 * 67 and 143 = 11 * 13 and 314 = 2 * 157 and 341 = 11 * 31 and 413 = 7 * 59 are semiprimes, while 431 is prime.
Links
- Robert Israel, Table of n, a(n) for n = 1..10000
Programs
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Maple
f:= proc(n) local L,m,t,i; L:= convert(n,base,10); m:= nops(L); nops(select(t -> t[-1] <> 0 and numtheory:-bigomega(add(t[i]*10^(i-1), i=1..m))=2, combinat:-permute(L))); end proc: map(f, [$1..200]); # Robert Israel, Jun 11 2023
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Python
from sympy import factorint from sympy.utilities.iterables import multiset_permutations as mp def c(n): return sum(factorint(n).values()) == 2 def a(n): return sum(1 for p in mp(str(n)) if p[0]!="0" and c(t:=int("".join(p)))) print([a(n) for n in range(1, 106)]) # Michael S. Branicky, Jun 11 2023
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