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The number of divisors of the largest divisor of n whose number of divisors is a power of 2.
2

%I #13 Apr 30 2024 11:43:48

%S 1,2,2,2,2,4,2,4,2,4,2,4,2,4,4,4,2,4,2,4,4,4,2,8,2,4,4,4,2,8,2,4,4,4,

%T 4,4,2,4,4,8,2,8,2,4,4,4,2,8,2,4,4,4,2,8,4,8,4,4,2,8,2,4,4,4,4,8,2,4,

%U 4,8,2,8,2,4,4,4,4,8,2,8,4,4,2,8,4,4,4

%N The number of divisors of the largest divisor of n whose number of divisors is a power of 2.

%C First differs from A286324 at n = 32, and from A331109 at n = 64.

%C Also, the number of infinitary divisors of the largest divisor of n whose number of divisors is a power of 2.

%H Amiram Eldar, <a href="/A372381/b372381.txt">Table of n, a(n) for n = 1..10000</a>

%F Multiplicative with a(p^e) = 2^floor(log_2(e+1)).

%F a(n) = A000005(A372379(n)).

%F a(n) = A037445(A372379(n)).

%F a(n) = A000005(n) if and only if n is in A036537.

%F a(n) <= A372380(n), with equality if and only if n is cubefree (A004709).

%t f[p_, e_] := 2^Floor[Log2[e + 1]]; a[1] = 1; a[n_] := Times @@ f @@@ FactorInteger[n]; Array[a, 100]

%o (PARI) a(n) = vecprod(apply(x -> 2^exponent(x+1), factor(n)[, 2]));

%o (Python)

%o from math import prod

%o from sympy import factorint

%o def A372381(n): return prod(1<<(e+1).bit_length()-1 for e in factorint(n).values()) # _Chai Wah Wu_, Apr 30 2024

%Y Cf. A000005, A004709, A036537, A037445, A372379, A372380.

%Y Cf. A286324, A331109.

%K nonn,easy,mult

%O 1,2

%A _Amiram Eldar_, Apr 29 2024