A377520 The sum of the divisors of n that are terms in A207481.

1, 3, 4, 7, 6, 12, 8, 7, 13, 18, 12, 28, 14, 24, 24, 7, 18, 39, 20, 42, 32, 36, 24, 28, 31, 42, 40, 56, 30, 72, 32, 7, 48, 54, 48, 91, 38, 60, 56, 42, 42, 96, 44, 84, 78, 72, 48, 28, 57, 93, 72, 98, 54, 120, 72, 56, 80, 90, 60, 168, 62, 96, 104, 7, 84, 144, 68

Offset: 1

Author

Amiram Eldar, Oct 30 2024

Keywords

• nonn
• easy
• mult

Comments

First differs from A284341 at n = 81 = 3^4: a(81) = 40, while A284341(81) = 121.

The number of these divisors is A377519(n), and the largest of them is A377518(n).

Links

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

Crossrefs

Cf. A000203, A013661, A207481, A284341, A377517, A377518, A377519.

Programs

Mathematica

f[p_, e_] := (p^(Min[p, e] + 1) - 1)/(p-1); a[1] = 1; a[n_] := Times @@ f @@@ FactorInteger[n]; Array[a, 100]

PARI

a(n) = {my(f = factor(n)); prod(i = 1, #f~, (f[i,1]^(min(f[i,1], f[i,2]) + 1) - 1)/(f[i,1] - 1));}

Formula

a(n) = A000203(A377518(n)).

Multiplicative with a(p^e) = (p^(min(p, e)+1) - 1)/(p - 1).

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

Sum_{k=1..n} a(k) ~ c * n^2 / 2, where c = zeta(2) * Product_{p prime} (1 - 1/p^(p+1)) = 1.42145673335960701365... .