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The number of infinitary divisors of n that are terms of A366242.
4

%I #7 Oct 07 2023 08:47:38

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

%T 4,1,2,4,4,4,2,8,2,2,2,4,2,4,1,2,4,2,2,4,4,4,4,4,2,4,2,4,2,2,4,8,2,2,

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

%N The number of infinitary divisors of n that are terms of A366242.

%H Amiram Eldar, <a href="/A366308/b366308.txt">Table of n, a(n) for n = 1..10000</a>

%F Multiplicative with a(p^e) = 2^A139351(e).

%F a(n) = 2^A366246(n).

%F a(n) = A037445(n)/A366309(n).

%F a(n) = A037445(A366244(n)).

%F a(n) >= 1, with equality if and only if n is in A366243.

%F a(n) <= A037445(n), with equality if and only if n is in A366242.

%t s[0] = 0; s[n_] := s[n] = s[Floor[n/4]] + If[OddQ[Mod[n, 4]], 1, 0]; f[p_, e_] := 2^s[e]; a[1] = 1; a[n_] := Times @@ f @@@ FactorInteger[n]; Array[a, 100]

%o (PARI) s(e) = if(e > 3, s(e\4)) + e%2 \\ after _Charles R Greathouse IV_ at A139351

%o a(n) = vecprod(apply(x -> 2^s(x), factor(n)[, 2]));

%Y Cf. A037445, A139351, A366242, A366243, A366244, A366246, A366309.

%K nonn,easy,mult

%O 1,2

%A _Amiram Eldar_, Oct 06 2023