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 A360078 Moebius function for the floor quotient poset. 2
 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 (list; graph; refs; listen; history; text; internal format)
 OFFSET 1,30 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. 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. EXAMPLE 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. MATHEMATICA LinearSolve[Table[If[Floor[i/j] > Floor[i/(j + 1)], 1, 0], {i, n}, {j, n}], UnitVector[n, 1]] PROG (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 CROSSREFS Cf. A002321, A008683, A360079. Sequence in context: A163529 A283655 A262742 * A027354 A192227 A102673 Adjacent sequences: A360075 A360076 A360077 * A360079 A360080 A360081 KEYWORD sign,look AUTHOR Harry Richman, Jan 24 2023 STATUS approved

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Last modified September 8 08:25 EDT 2024. Contains 375752 sequences. (Running on oeis4.)