A342933 a(n) = n! * Sum_{k=1..n} k^2/floor(n/k).

1, 9, 80, 654, 6264, 59760, 665160, 7565040, 96929280, 1286046720, 18976083840, 286363123200, 4775047200000, 81792956044800, 1515077749785600, 28898470215014400, 594066352700620800, 12467555729620992000, 280797084422959104000, 6460327992512249856000, 157769680941941612544000

Offset

1,2

Comments

In general, for m>=0, Sum_{k=1..n} k^m / floor(n/k) ~ n^(m+1) * (-1 + Sum_{j=2..m+2} zeta(j) / (m+1)).

Links

Table of n, a(n) for n=1..21.

Formula

a(n) ~ c * n^3 * n!, where c = Sum_{j>=1} (1 + 3*j*(j+1)) / (3*j^4*(j+1)^3) = (zeta(4) + zeta(3) + zeta(2))/3 - 1 = Pi^2/18 + Pi^4/270 + zeta(3)/3 - 1.

Mathematica

Table[n!*Sum[k^2/Floor[n/k], {k, 1, n}], {n, 1, 25}]
(* Alternative: *)
Table[n!*Sum[(Floor[n/j]*(1 + Floor[n/j])*(1 + 2*Floor[n/j]) - Floor[n/(1 + j)]*(1 + Floor[n/(1 + j)])*(1 + 2*Floor[n/(1 + j)]))/6/j, {j, 1, n}], {n, 1, 25}]

Prog

(Python)
from math import factorial
def A342933(n):
    c, j, f = 0, 1, factorial(n)
    while j <= n:
        k = n//j
        m = n//k
        c += f*(m*(m+1)*(2*m+1)-(j-1)*j*(2*j-1))//k
        j = m+1
    return c//6 # Chai Wah Wu, May 14 2026

Crossrefs

Cf. A345683, A345684.

Sequence in context: A293916 A293731 A254833 * A275497 A171314 A370039

Adjacent sequences: A342930 A342931 A342932 * A342934 A342935 A342936

Keyword

nonn

Author

Vaclav Kotesovec, Jun 23 2021

Status

approved