cp's OEIS Frontend

A360078 Moebius function for the floor quotient poset.

%I A360078 #29 Jun 18 2025 21:02:08
%S A360078 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 A360078 -2,-2,-2,-2,-2,-2,-1,-1,-1,-1,0,0,-1,-1,-1,-2,-2,-2,-2,-3,-3,-3,-3,
%U A360078 -3,-1,-1,-1,-1,-1,-1,1,1,1,0,-1,-1,-1,-1,-1,-1,-1
%N A360078 Moebius function for the floor quotient poset.
%C A360078 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 A360078 Andrew Howroyd, <a href="/A360078/b360078.txt">Table of n, a(n) for n = 1..10000</a>
%H A360078 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 A360078 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 A360078 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 A360078 LinearSolve[Table[If[Floor[i/j] > Floor[i/(j + 1)], 1, 0], {i, n}, {j, n}], UnitVector[n, 1]]
%o A360078 (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
%o A360078 (Haskell)
%o A360078 isFQ d n = (n `div` (n `div` d)) == d
%o A360078 a360078 1 = 1
%o A360078 a360078 n = - sum [a360078 d | d <- [1..(n-1)], d `isFQ` n]
%o A360078 -- _Harry Richman_, Jun 13 2025
%Y A360078 Cf. A002321, A008683, A360079.
%K A360078 sign,look
%O A360078 1,30
%A A360078 _Harry Richman_, Jan 24 2023