A360720 a(n) is the sum of unitary divisors of n that are powerful (A001694).

1, 1, 1, 5, 1, 1, 1, 9, 10, 1, 1, 5, 1, 1, 1, 17, 1, 10, 1, 5, 1, 1, 1, 9, 26, 1, 28, 5, 1, 1, 1, 33, 1, 1, 1, 50, 1, 1, 1, 9, 1, 1, 1, 5, 10, 1, 1, 17, 50, 26, 1, 5, 1, 28, 1, 9, 1, 1, 1, 5, 1, 1, 10, 65, 1, 1, 1, 5, 1, 1, 1, 90, 1, 1, 26, 5, 1, 1, 1, 17, 82

Offset

1,4

Comments

The number of these divisors is given by A323308.

Links

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

Vaclav Kotesovec, Plot of Sum_{k=1..n} a(k) / n^(3/2) for n = 1..10^8.

Index entries for sequences related to divisors of numbers.

Formula

Multiplicative with a(p) = 1 and a(p^e) = p^e + 1 for e > 1.

a(n) <= A034448(n), with equality if and only if n is powerful (A001694).

a(n) <= A183097(n), with equality if and only if n is cubefree (A004709).

Dirichlet g.f.: zeta(s)*zeta(s-1)*Product_{p prime} (1 - p^(1-s) + p^(2-2*s) - p^(2-3*s)).

From Vaclav Kotesovec, Feb 18 2023: (Start)

Dirichlet g.f.: zeta(s) * zeta(2*s-2) * Product_{primes p} (1 - p^(3-4*s) - p^(2-3*s) + p^(3-3*s)).

Sum_{k=1..n} a(k) ~ c * zeta(3/2) * n^(3/2) / 3, where c = Product_{primes p} (1 + 1/p^(3/2) - 1/p^(5/2) - 1/p^3) = 1.48039182258752809541724060173644... (End)

a(n) = A034448(A057521(n)) (the sum of unitary divisors of the powerful part of n). - Amiram Eldar, Dec 12 2023

Mathematica

f[p_, e_] := If[e == 1, 1, p^e + 1]; a[1] = 1; a[n_] := Times @@ f @@@ FactorInteger[n]; Array[a, 100]

Prog

(PARI) a(n) = {my(f = factor(n)); prod(i = 1, #f~, if(f[i, 2] == 1, 1, f[i, 1]^f[i, 2] + 1)); }
(PARI) for(n=1, 100, print1(direuler(p=2, n, (1 - p^3*X^4 - p^2*X^3 + p^3*X^3) / ((1 - X) * (1 - p^2*X^2)))[n], ", ")) \\ Vaclav Kotesovec, Feb 18 2023

Crossrefs

Cf. A001694, A004709, A034444, A034448, A057521, A077610, A323308.

Similar sequences: A183097, A360722.

Sequence in context: A345939 A140210 A380400 * A010130 A361063 A363972

Adjacent sequences: A360717 A360718 A360719 * A360721 A360722 A360723

Keyword

nonn,mult,easy

Author

Amiram Eldar, Feb 18 2023

Status

approved