%I #26 Feb 14 2024 20:17:56
%S 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,
%T -2,-2,-2,-2,-2,-2,-1,-1,-1,-1,0,0,-1,-1,-1,-2,-2,-2,-2,-3,-3,-3,-3,
%U -3,-1,-1,-1,-1,-1,-1,1,1,1,0,-1,-1,-1,-1,-1,-1,-1
%N Moebius function for the floor quotient poset.
%C 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.
%H Andrew Howroyd, <a href="/A360078/b360078.txt">Table of n, a(n) for n = 1..10000</a>
%H J.-P. Cardinal, <a href="https://arxiv.org/abs/0811.3701">Symmetric matrices related to the Mertens function</a>, arXiv:0811.3701 [math.NT], 2008-2009.
%H J. C. Lagarias and D. H. Richman, <a href="https://doi.org/10.1016/j.aam.2023.102615">The floor quotient partial order</a>, Adv. Appl. Math., 153 (2024); arXiv:<a href="https://arxiv.org/abs/2212.11689">2212.11689</a> [math.NT], 2022-2023.
%e 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.
%t LinearSolve[Table[If[Floor[i/j] > Floor[i/(j + 1)], 1, 0], {i, n}, {j, n}], UnitVector[n, 1]]
%o (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
%Y Cf. A002321, A008683, A360079.
%K sign,look
%O 1,30
%A _Harry Richman_, Jan 24 2023