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A050386
Exponential reversion of Moebius function A008683.
3
1, 1, 4, 25, 221, 2505, 34707, 568177, 10731571, 229706718, 5495040882, 145285035974, 4206973447847, 132410823640004, 4500857134998016, 164322352411837139, 6412953180173688644, 266421162165751276297, 11738996224580074369000, 546795310668016410175089, 26846006952196883028139188
OFFSET
1,3
FORMULA
E.g.f. A(x) satisfies: A(x) = x - Sum_{k>=2} mu(k) * A(x)^k / k!. - Ilya Gutkovskiy, Apr 22 2020
MATHEMATICA
length = 40; Range[length]! InverseSeries[Sum[MoebiusMu[n] x^n/n!, {n, 1, length}] + O[x]^(length+1)][[3]] (* Vladimir Reshetnikov, Nov 07 2015 *)
PROG
(PARI) seq(n)= Vec(serlaplace(serreverse(sum(k=1, n, moebius(k)*x^k/k!) + O(x*x^n)))); \\ Michel Marcus, Apr 21 2020
(PARI) \\ using function inverse_bell_matrix_row from A354794
a(n) = inverse_bell_matrix_row(n, x->moebius(x+1))[1] \\ Mikhail Kurkov, May 09 2026
(SageMath)
def A050386(n):
mu = [moebius(i) for i in range(n + 1)]
v2 = [ZZ(0)] * (n + 1); v2[1] = ZZ(1)
for i in range(2, n + 1):
s = ZZ(0); A = ZZ(1)
for j in range(1, i):
A = -(A * (i - n - j)) // j
term = (1 - n) * A - ((i - n) * A) // (j + 1)
if mu[j + 1] != 0:
s += term * v2[i - j] * mu[j + 1]
v2[i] = s // (i - 1)
return v2[n]
L = [A050386(i) for i in range(1, 22)]; print(L)
# after Mikhail Kurkov, Peter Luschny, May 04 2026
CROSSREFS
KEYWORD
nonn,changed
AUTHOR
Christian G. Bower, Nov 15 1999
EXTENSIONS
Typo in name corrected by Sean A. Irvine, Aug 15 2021
STATUS
approved