A345683 a(n) = n! * Sum_{k=1..n} 1/floor(n/k).

1, 3, 14, 66, 444, 2880, 25080, 216720, 2247840, 24071040, 304335360, 3752179200, 54965433600, 810550540800, 13176376012800, 219079045785600, 4078723532083200, 75227891042304000, 1550619342784512000, 31871016307113984000, 710529031487987712000, 16180987966182014976000

Offset: 1

Author

Vaclav Kotesovec, Jun 23 2021

Keywords

• nonn

Links

• Seiichi Manyama, Table of n, a(n) for n = 1..448
• Vaclav Kotesovec, Plot of a(n) / (n*n!) for n = 1..1000000
• Mathoverflow, An asymptotic formula for a sum involving powers of floor functions, 2019.

Crossrefs

Cf. A006218, A010786, A092143, A117871, A345682, A345684, A345686, A355987.

Programs

Mathematica

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

PARI

a(n) = n!*sum(k=1, n, 1/(n\k)); \\ Michel Marcus, Jun 24 2021

PARI

my(N=30, x='x+O('x^N)); Vec(serlaplace(-sum(k=1, N, (1-x^k)*log(1-x^k))/(1-x))) \\ Seiichi Manyama, Jul 23 2022

Python

from math import factorial, isqrt
def A345683(n): return (m:=factorial(n))*(n-1)+m//n+sum((q:=n//k)*(m//k-m//(k-1))+m//q for k in range(2,isqrt(n)+1)) # Chai Wah Wu, Oct 27 2023

Formula

a(n) ~ c * n * n!, where c = Sum_{j>=1} 1/(j^2*(j+1)) = Pi^2/6 - 1 = 0.644934... [proved by Harry Richman, see Mathoverflow link]

E.g.f.: -(1/(1-x)) * Sum_{k>0} (1 - x^k) * log(1 - x^k). - Seiichi Manyama, Jul 23 2022