A068337 a(n) = n!*Sum_{k=1..n} mu(k)/k, where mu(k) is the Möbius function.

1, 1, 1, 4, -4, 96, -48, -384, -3456, 328320, -17280, -207360, -481697280, -516741120, 79427174400, 1270834790400, 681401548800, 12265227878400, -6169334376038400, -123386687520768000, -158218429759488000, 47610136717000704000

Offset

1,4

Links

Amiram Eldar, Table of n, a(n) for n = 1..450

Friedrich Roesler, Riemann's hypothesis as an eigenvalue problem, Linear Algebra and its Applications, Vol. 81 (1986), pp. 153-198.

Friedrich Roesler, Riemann's hypothesis as an eigenvalue problem. II, Linear Algebra and its Applications, Vol. 92 (1987), pp. 45-73.

Formula

a(n) = (-1)^(n-1)*{determinant of the n X n matrix m(i,j) = i+(j (mod i))} - Benoit Cloitre, May 28 2004

From Amiram Eldar, Oct 22 2020: (Start)

a(n) = A000142(n)*A070888(n)/A070889(n).

a(n) ~ O(n! * n^(-1/2 + eps)), for every eps>0, if and only if Riemann's hypothesis is true (Roesler, 1986). (End)

Mathematica

n = 25; Accumulate[Table[MoebiusMu[k]/k, {k, 1, n}]] * Range[n]! (* Amiram Eldar, Oct 22 2020 *)

Prog

(Python)
from math import factorial
from functools import lru_cache
from sympy import harmonic
@lru_cache(maxsize=None)
def f(n):
    if n <= 1:
        return 1
    c, j = 1, 2
    k1 = n//j
    while k1 > 1:
        j2 = n//k1 + 1
        c += (harmonic(j-1)-harmonic(j2-1))*f(k1)
        j, k1 = j2, n//j2
    return c+harmonic(j-1)-harmonic(n)
def A068337(n): return factorial(n)*f(n) # Chai Wah Wu, Nov 03 2023

Crossrefs

Cf. A000142, A008683, A070888, A070889.

Sequence in context: A224092 A217188 A092209 * A009534 A009559 A163196

Adjacent sequences: A068334 A068335 A068336 * A068338 A068339 A068340

Keyword

sign

Author

Leroy Quet, Feb 27 2002

Status

approved