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A366309
The number of infinitary divisors of n that are terms of A366243.
3
1, 1, 1, 2, 1, 1, 1, 2, 2, 1, 1, 2, 1, 1, 1, 1, 1, 2, 1, 2, 1, 1, 1, 2, 2, 1, 2, 2, 1, 1, 1, 1, 1, 1, 1, 4, 1, 1, 1, 2, 1, 1, 1, 2, 2, 1, 1, 1, 2, 2, 1, 2, 1, 2, 1, 2, 1, 1, 1, 2, 1, 1, 2, 2, 1, 1, 1, 2, 1, 1, 1, 4, 1, 1, 2, 2, 1, 1, 1, 1, 1, 1, 1, 2, 1, 1, 1
OFFSET
1,4
LINKS
FORMULA
Multiplicative with a(p^e) = 2^A139352(e).
a(n) = 2^A366247(n).
a(n) = A037445(n)/A366308(n).
a(n) = A037445(A366245(n)).
a(n) >= 1, with equality if and only if n is in A366242.
a(n) <= A037445(n), with equality if and only if n is in A366243.
Asymptotic mean: Limit_{m->oo} (1/m) * Sum_{k=1..m} a(k) = Product_{p prime} (1 - 1/p)*(1 + Sum_{k>=1} 2^A139352(k)/p^k) = 1.44736831993091923328... .
MATHEMATICA
s[0] = 0; s[n_] := s[n] = s[Floor[n/4]] + If[Mod[n, 4] > 1, 1, 0]; f[p_, e_] := 2^s[e]; a[1] = 1; a[n_] := Times @@ f @@@ FactorInteger[n]; Array[a, 100]
PROG
(PARI) s(e) = if(e > 3, s(e\4)) + e%4\2 \\ after Charles R Greathouse IV at A139352
a(n) = vecprod(apply(x -> 2^s(x), factor(n)[, 2]));
KEYWORD
nonn,easy,mult
AUTHOR
Amiram Eldar, Oct 06 2023
STATUS
approved