OFFSET
1,2
COMMENTS
In general, for m > 1, Sum_{k=1..n} floor(n^m/k^m) ~ zeta(m)*n^m + zeta(1/m)*n.
LINKS
Vaclav Kotesovec, Table of n, a(n) for n = 1..10000
Vaclav Kotesovec, Plot of (a(n) - zeta(3)*n^3)/n for n = 1..100000
FORMULA
a(n) ~ zeta(3)*n^3 + zeta(1/3)*n.
MATHEMATICA
Table[Sum[Floor[n^3/k^3], {k, 1, n}], {n, 1, 50}]
PROG
(Python)
from sympy import integer_nthroot
def A344675(n):
c, j, n2 = 0, 1, n**3
while j <= n:
k = n2//j**3
m = integer_nthroot(n2//k, 3)[0]
c += k*(m-j+1)
j = m+1
return c # Chai Wah Wu, May 21 2026
CROSSREFS
KEYWORD
nonn
AUTHOR
Vaclav Kotesovec, May 26 2021
STATUS
approved
