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Number of semiprime divisors of the n-th squarefree number (A005117).
1

%I #15 Aug 13 2024 01:56:33

%S 0,0,0,0,1,0,1,0,0,1,1,0,0,1,1,0,1,0,3,0,1,1,1,0,1,1,0,3,0,1,0,1,0,1,

%T 1,1,0,0,1,1,3,0,1,3,0,0,1,1,3,0,1,0,1,1,1,0,1,1,1,1,0,0,3,0,3,1,0,0,

%U 3,1,0,3,1,1,1,1,1,0,1,3,0,1,1,0,3,0,1,1,1,1,1,0,0,3,1,0,1

%N Number of semiprime divisors of the n-th squarefree number (A005117).

%H Wikipedia, <a href="http://en.wikipedia.org/wiki/Squarefree_integer">Squarefree integer</a>.

%F a(n) = A086971(A005117(n)).

%F a(n) = c*(c-1)/2, where c = A001221(A005117(n)).

%e a(19) = 3 since A005117(19) = 30 and 30 has 3 semiprime divisors, namely {6, 10, 15}.

%t Table[If[PrimeNu[n] == PrimeOmega[n], PrimeNu[n] (PrimeNu[n] - 1)/2, {}], {n, 200}] // Flatten

%o (PARI) apply(x->(sumdiv(x, d, bigomega(d)==2)), select(issquarefree, [1..300])) \\ _Michel Marcus_, Nov 22 2023

%o (Python)

%o from math import isqrt

%o from sympy import mobius, primenu

%o def A367452(n):

%o def f(x): return n+x-sum(mobius(k)*(x//k**2) for k in range(1, isqrt(x)+1))

%o m, k = n, f(n)

%o while m != k:

%o m, k = k, f(k)

%o return (c:=primenu(m))*(c-1)>>1 # _Chai Wah Wu_, Aug 12 2024

%Y Cf. A001221 (omega), A001358 (semiprimes), A005117 (squarefree numbers), A086971.

%K nonn

%O 1,19

%A _Wesley Ivan Hurt_, Nov 18 2023