A349693 Dirichlet convolution of the ruler function (A001511) with itself.

1, 4, 2, 10, 2, 8, 2, 20, 3, 8, 2, 20, 2, 8, 4, 35, 2, 12, 2, 20, 4, 8, 2, 40, 3, 8, 4, 20, 2, 16, 2, 56, 4, 8, 4, 30, 2, 8, 4, 40, 2, 16, 2, 20, 6, 8, 2, 70, 3, 12, 4, 20, 2, 16, 4, 40, 4, 8, 2, 40, 2, 8, 6, 84, 4, 16, 2, 20, 4, 16, 2, 60, 2, 8, 6, 20, 4, 16, 2, 70

Offset: 1

Author

Ilya Gutkovskiy, Nov 25 2021

Keywords

• nonn
• mult

Comments

Dirichlet convolution of A000005 with A104117. - Ridouane Oudra, Jul 23 2025

Links

• Antti Karttunen, Table of n, a(n) for n = 1..20000

Crossrefs

Cf. A000217, A001511, A115364, A129628.

Cf. A000005, A104117, A007814, A085058, A090739, A099777, A372784, A001227, A000292,

Programs

Maple

a:= n-> (f-> add(f(d)*f(n/d), d=numtheory[divisors](n)))(k-> padic[ordp](2*k, 2)):
seq(a(n), n=1..80);  # Alois P. Heinz, Nov 25 2021

Mathematica

Table[Sum[IntegerExponent[2 d, 2] IntegerExponent[2 n/d, 2], {d, Divisors[n]}], {n, 1, 80}]
f[p_, e_] := If[p == 2, Binomial[e + 3, 3], e + 1]; a[1] = 1; a[n_] := Times @@ f @@@ FactorInteger[n]; Array[a, 80] (* Amiram Eldar, Nov 25 2021 *)

PARI

A001511(n) = (1+valuation(n,2));
A349693(n) = sumdiv(n,d,A001511(n/d)*A001511(d)); \\ Antti Karttunen, Nov 25 2021

Python

from sympy import divisor_count
def A349693(n): return divisor_count(n)*(m:=(n&-n).bit_length()+1)*(m+1)//6 # Chai Wah Wu, Jul 13 2022

Formula

Dirichlet g.f.: zeta(s)^2 * 4^s / (2^s-1)^2.

a(n) = Sum_{d|n} A001511(d) * A001511(n/d).

a(n) = Sum_{d|n} A000217(A001511(d)).

Multiplicative with a(p^e) = binomial(e+3,3) if p = 2 and e+1 otherwise. - Amiram Eldar, Nov 25 2021

Sum_{k=1..n} a(k) ~ 4*n*(log(n) - 1 + 2*gamma - 2*log(2)), where gamma is the Euler-Mascheroni constant A001620. - Vaclav Kotesovec, Nov 26 2021

From Ridouane Oudra, Jul 23 2025: (Start)

a(n) = Sum_{i=0..A007814(n)} (i+1)*tau(n/2^i).

a(n) = Sum_{d|n} A115364(d).

a(n) = (1/6)*A090739(n)*A085058(n-1)*A000005(n).

a(n) = (1/6)*A001511(n)*A090739(n)*A099777(n).

a(n) = (1/3)*A115364(n)*A372784(n).

a(n) = A001227(n)*A000292(A001511(n)).

a(2*n+1) = tau(2*n+1).

a(2^k*(2*n+1)) = binomial(k+3, 3)*tau(2*n+1), for k, n >= 0. (End)