A360078 Moebius function for the floor quotient poset.

1, -1, -1, 0, 0, 1, 1, 0, 1, 1, 1, 0, 0, 0, 0, 1, 1, -1, -1, -1, -1, -1, -1, 0, 0, 0, -1, -1, -1, -2, -2, -2, -2, -2, -2, -1, -1, -1, -1, 0, 0, -1, -1, -1, -2, -2, -2, -2, -3, -3, -3, -3, -3, -1, -1, -1, -1, -1, -1, 1, 1, 1, 0, -1, -1, -1, -1, -1, -1, -1

Offset: 1

Author

Harry Richman, Jan 24 2023

Keywords

• sign
• look

Comments

Say d is a "floor quotient" of n if d = [n/k] for some positive integer k. This defines a partial order relation on the positive integers. This sequence records the Moebius function values of this poset.

Examples

			For n = 9, the set of floor quotients of 9 are Q(9) = {1, 2, 3, 4, 9} with Moebius values a(1) = 1, a(2) = -1, a(3) = -1, and a(4) = 0. The Moebius recursion requires that the Moebius values summed over Q(9) must equal zero, so a(9) = 1.
		

Links

• Andrew Howroyd, Table of n, a(n) for n = 1..10000
• J.-P. Cardinal, Symmetric matrices related to the Mertens function, arXiv:0811.3701 [math.NT], 2008-2009.
• J. C. Lagarias and D. H. Richman, The floor quotient partial order, Adv. Appl. Math., 153 (2024); arXiv:2212.11689 [math.NT], 2022-2023.

Crossrefs

Cf. A002321, A008683, A360079.

Programs

Haskell

isFQ d n = (n `div` (n `div` d)) == d
a360078 1 = 1
a360078 n = - sum [a360078 d | d <- [1..(n-1)], d `isFQ` n]
-- Harry Richman, Jun 13 2025

Mathematica

LinearSolve[Table[If[Floor[i/j] > Floor[i/(j + 1)], 1, 0], {i, n}, {j, n}], UnitVector[n, 1]]

PARI

seq(n)={my(v=vector(n)); v[1]=1; for(n=2, #v, my(S=Set(vector(n-1, k, n\(k+1)))); v[n]=-sum(i=1, #S, v[S[i]])); v} \\ Andrew Howroyd, Jan 24 2023