A319085 a(n) = Sum_{k=1..n} k^2*tau(k), where tau is A000005.

1, 9, 27, 75, 125, 269, 367, 623, 866, 1266, 1508, 2372, 2710, 3494, 4394, 5674, 6252, 8196, 8918, 11318, 13082, 15018, 16076, 20684, 22559, 25263, 28179, 32883, 34565, 41765, 43687, 49831, 54187, 58811, 63711, 75375, 78113, 83889, 89973, 102773, 106135

Offset: 1

Author

Vaclav Kotesovec, Sep 10 2018

Keywords

• nonn

Comments

In general, for m>=1, Sum_{k=1..n} k^m * tau(k) = Sum_{k=1..n} k^m * (Bernoulli(m+1, floor(1 + n/k)) - Bernoulli(m+1, 0)) / (m+1), where Bernoulli(n,x) are the Bernoulli polynomials. - Daniel Suteu, Nov 08 2018

Links

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

Crossrefs

Cf. A000005, A006218, A034714, A143127.

Programs

Mathematica

Accumulate[Table[k^2*DivisorSigma[0, k], {k, 1, 50}]]

PARI

a(n) = sum(k=1, n, k^2*numdiv(k)); \\ Michel Marcus, Sep 12 2018

PARI

f(n) = n*(n+1)*(2*n+1)/6; \\ A000330
a(n) = 2*sum(k=1, sqrtint(n), k^2 * f(n\k)) - f(sqrtint(n))^2; \\ Daniel Suteu, Nov 26 2020

Python

from math import isqrt
def A319085(n): return (-((s:=isqrt(n))*(s+1)*(2*s+1))**2//12 + sum(k**2*(q:=n//k)*(q+1)*(2*q+1) for k in range(1,s+1)))//3 # Chai Wah Wu, Oct 21 2023

Formula

a(n) ~ n^3 * (log(n) + 2*gamma - 1/3)/3, where gamma is the Euler-Mascheroni constant A001620.

a(n) = Sum_{k=1..n} k^2 * Bernoulli(3, floor(1 + n/k)) / 3, where Bernoulli(n,x) are the Bernoulli polynomials. - Daniel Suteu, Nov 08 2018

a(n) = Sum_{k=1..n} Sum_{i=1..floor(n/k)} i^2 * k^2. - Wesley Ivan Hurt, Nov 26 2020