A360079 Finite differences of Moebius function for the floor quotient poset.
1, -2, 0, 1, 0, 1, 0, -1, 1, 0, 0, -1, 0, 0, 0, 1, 0, -2, 0, 0, 0, 0, 0, 1, 0, 0, -1, 0, 0, -1, 0, 0, 0, 0, 0, 1, 0, 0, 0, 1, 0, -1, 0, 0, -1, 0, 0, 0, -1, 0, 0, 0, 0, 2, 0, 0, 0, 0, 0, 2, 0, 0, -1, -1, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, -2
Offset: 1
Keywords
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.
- J. C. Lagarias and D. H. Richman, The floor quotient partial order, arXiv:2212.11689 [math.NT], 2022.
Programs
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Haskell
isFQ d n = (n `div` (n `div` d)) == d fqMobius 1 = 1 fqMobius n = - sum [fqMobius d | d <- [1..(n-1)], d `isFQ` n] a360079 1 = 1 a360079 n = fqMobius n - fqMobius (n-1) -- Harry Richman, Jun 13 2025
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Mathematica
LinearSolve[Table[If[Floor[i/j] > Floor[i/(j + 1)], 1, 0], {i, n}, {j, n}] . Table[If[i >= j, 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]])); vector(#v, i, v[i]-if(i>1, v[i-1]))} \\ Andrew Howroyd, Jan 24 2023
Comments