A372331 The number of infinitary divisors of the smallest number k such that k*n is a number whose number of divisors is a power of 2 (A036537).

1, 1, 1, 2, 1, 1, 1, 1, 2, 1, 1, 2, 1, 1, 1, 4, 1, 2, 1, 2, 1, 1, 1, 1, 2, 1, 1, 2, 1, 1, 1, 2, 1, 1, 1, 4, 1, 1, 1, 1, 1, 1, 1, 2, 2, 1, 1, 4, 2, 2, 1, 2, 1, 1, 1, 1, 1, 1, 1, 2, 1, 1, 2, 2, 1, 1, 1, 2, 1, 1, 1, 2, 1, 1, 2, 2, 1, 1, 1, 4, 4, 1, 1, 2, 1, 1, 1

Offset

1,4

Comments

First differs from A370077 and A370080 at n = 32.

The number of divisors d of n that are infinitarily relatively prime to n (see A064379), i.e., d have no common infinitary divisors with n.

Equivalently, the number of divisors d of n such that for each prime divisor p of d, bitand(v_p(n), v_p(d)) = 0, where v_p(k) is the highest power of p that divides k. Note that for infinitary divisors d of n (A077609), bitand(v_p(n), v_p(d)) = v_p(d).

Links

Amiram Eldar, Table of n, a(n) for n = 1..10000

Formula

a(n) = A037445(A372328(n)).

Multiplicative with a(p^e) = 2^A023416(e) = A080100(e).

a(n) = 1 if and only if n is in A036537.

Asymptotic mean: Limit_{m->oo} (1/m) * Sum_{k=1..m} a(k) = Product_{p prime} (1 + Sum_{k>=1} A080100(k)/p^k) = 1.51209151045338102469... .

Mathematica

f[p_, e_] := 2^DigitCount[e, 2, 0]; a[1] = 1; a[n_] := Times @@ f @@@ FactorInteger[n]; Array[a, 100]

Prog

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

Crossrefs

Cf. A036537, A037445, A064379, A077609, A080100, A372328.

Cf. A370077, A370080.

Sequence in context: A367987 A370077 A370080 * A318498 A093997 A157196

Adjacent sequences: A372328 A372329 A372330 * A372332 A372333 A372334

Keyword

nonn,easy,mult

Author

Amiram Eldar, Apr 28 2024

Status

approved