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

1,30

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