A365174 The sum of divisors d of n such that gcd(d, n/d) is an exponentially odd number (A268335).

1, 3, 4, 7, 6, 12, 8, 15, 13, 18, 12, 28, 14, 24, 24, 27, 18, 39, 20, 42, 32, 36, 24, 60, 31, 42, 40, 56, 30, 72, 32, 51, 48, 54, 48, 91, 38, 60, 56, 90, 42, 96, 44, 84, 78, 72, 48, 108, 57, 93, 72, 98, 54, 120, 72, 120, 80, 90, 60, 168, 62, 96, 104, 107, 84, 144

Offset: 1

Author

Amiram Eldar, Aug 25 2023

Keywords

• nonn
• easy
• mult

Comments

The number of these divisors is A365173(n).

Links

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

Crossrefs

Cf. A000203, A033634, A034448, A268335, A365172, A365173.

Cf. A002117, A013661, A013664.

Programs

Mathematica

f[p_, e_] := 1 + p^e + If[EvenQ[e], (p^(e + 1) - p)/(p^2 - 1), (1 + p^(2*Floor[e/4] + 1))*(p^(2*Floor[(e + 1)/4] + 1) - p)/(p^2 - 1)]; a[1] = 1; a[n_] := Times @@ f @@@ FactorInteger[n]; Array[a, 100]

PARI

a(n) = {my(f = factor(n), p, e); prod(i = 1, #f~, p = f[i,1]; e = f[i,2]; 1 + p^e + if(e%2, (1 + p^(2*(e\4) + 1))*(p^(2*((e+1)\4) + 1) - p)/(p^2 - 1), (p^(e+1)-p)/(p^2-1)));}

Formula

Multiplicative with 1 + p^e + (p^(e + 1) - p)/(p^2 - 1) if e is even, and 1 + p^e + (1 + p^(2*floor(e/4)+1))*(p^(2*floor((e+1)/4)+1) - p)/(p^2 - 1) if e is odd.

a(n) <= A000203(n), with equality if and only if n is not a biquadrateful number (A046101).

a(n) >= A034448(n), with equality if and only if n is squarefree (A005117).

Sum_{k=1..n} a(k) ~ c * n^2, where c = (zeta(2)*zeta(6)/(2*zeta(3))) * Product_{p prime} (1 + 1/p^3 - 1/p^6) = 0.809912096042... .