ABSTRACT If we take the divisor matrix Dij = 1 if j|i, 0 otherwise, we can find it’s inverse Dij−1, the determinant of D is always 1. When we do this we find that the firsts column of D−1, takes the values of the Mobius μ(n), and the subsequent rows take the pattern D^{-1}_{ij}=\mu\left({j}\right) where we allow μ(x)=0, x ∉ ℕ. One thing we can do is to take then the absolute value of each element of Dij−1, we will call this matrix Hij = |Dij−1|, then we may find the inverse of H , i.e. H−1, we then find that the left most column of this matrix is Liouville’s function λ(n), where \lambda(n)=(-1)^{\Omega(n)} with Ω(n) the number of prime factors with multiplicity of n. In fact then the explicit formula for H−1 appears to be H^{-1}_{ij}=\lambda\left({j}\right) where again we allow that λ(x)=0, x ∉ ℕ. We then see that Dij = |Hij−1|, the absolute value of the matrix is just the divisor matrix again! We have the statement “The Liouville function’s Dirichlet inverse is the absolute value of the Möbius function.” (Wikipedia-Liouville Function). One question is what is the matrix H? It appears to be the divisor matrix, minus a few elements in numbers which have a square part. We can look at the matrix D − H, we see that the first column is the characteristic function of the numbers which are not square free 1 − χSF(n), and again we have the same pattern D_{ij}-H_{ij}=1-\left({j}\right) FURTHER EXPERIMENTATION We may play with this idea. Consider the matrix instead Pij = 1 if pj|i otherwise 0, with pj being the jth prime. Then we may take the inverse of P + I, with I the identity matrix, or trace element 1 matrix for the inversion procedure. We then get the first column of (P + I)−1 begin the sequence 1,-1,1,-1,-1,0,1,-1,1,-2,1,0,0,0,0,-1,-1,0,1,-2,2,0,-1,0,-1,-1,1,0,2,-1,\cdots we can define the function \kappa(n)=(-1)^{M(n)} where M(n) is Mertens function, and then define K^{-1}_{ij}=-\kappa\left({j}\right) then find the inverse of this, i.e. K. The sequence associated with K is 1,1,-1,0,1,-3,1,0,2,1,1,-2,-1,3,-3,0,1,8,-1,-2,-1,1,1,2,2,-3,-4,2,1,-9,\cdots it would appear to be 0 for 4, 8, 16, 32, 49, 56, 64, 96. We can try a sum over the divisors d of n, [\kappa](d) we get the sequence 1,2,0,2,2,-2,2,2,2,4,2,-4,0,6,-2,2,2,8,0,2,0,4,2\cdots We can define Q as Qij = isprime(i/j), and find (Q + I)−1. The sequence associated to this is 1,-1,-1,1,-1,2,-1,-1,1,2,-1,-3,-1,2,2,1,-1,-3,-1,-3,2,2,-1,4,1,2,-1,-3,-1,-6,-1,-1 \cdots without the sign we can say that this is the number of prime factors of n with multiplicity, which we will call b(n), then |Q^{-1}_{ij}|=b\left({j}\right) and in fact with sign we have Q^{-1}_{ij}=b\left({j}\right)\lambda\left({j}\right) If we let Sij = σ₀(i/j), then we get the sequence A007427, for whom σ₀(n) is the Dirichlet inverse, we can see a pattern forming with all these results... PRIME SEQUENCE AND 1 Let our matrix ρij = prime*(i/j), letting the function prime^*(n)=1, \; n=1\\ prime^*(n)=p_{n-1} \; n\in \\ prime^*(x)=0 \; x\notin then if we find the inverse ρ−1, we gain the following sequence g(n)\to1,-2,-3,-1,-7,1,-13,-5,-10,5,-29,7,-37,11,-1,2,-53,29,-61,11,7,43,-79,37,-40,51,\cdots if we look at prime indexed values only we get -2,-3,-7,-13,-29,-37,-53,-61,-79,-107,-113,\cdots which is the sequence −prime(prime(n)−1) A055003. We can the suspect the sign is negative for all primes in ρ−1, and that every prime index of the sequence is also prime, they also appear to be records of the sequence. Interestingly enough, this would mean the inversion process would preserve primality. This is then the Dirichlet inverse of the sequence f(n)→1, 2, 3, 5, 7, 11, 13, 17, 19, ⋯. Such that (f*h)(n)=f(d)h\left({d}\right)=1,\; \forall n such that when n = p, a prime, (f*h)(p)=f(1)h(p)+f(p)h(1)=1\\ h(p)+f(p)=1\\ h(p)=1-(p-1) for h(n)= g(d) \to 1,-1,-2,-2,-6,-3,-12,-7,-12,-3,-28,3,-36 so h(1)f(2)+h(2)f(1)= 1\cdot2 -1\cdot1 =1 \\ h(1)f(3)+h(3)f(1)= 1\cdot3 -2\cdot1 =1 \\ h(1)f(5)+h(5)f(1)= 1\cdot7 -6\cdot1 =1 and so on. whereas (h*g)(p)=0, \forall p MORE If we insert ϕ(n), the Euler totient function we get A023900, the Dirichlet inverse of the Euler totient function. If we insert n, we get A055615, which is nμ(n). For a general input function f(n), we find that the general inverse function g(n) is given by the relationship g(n)= ^{b(n)} \cdot n\cdot ^n -f(\omega_i) where the product is such that $^n \omega_i = n$, and then combination of the ω is used once. These begin -f(2)\\-f(3)\\f(2)^2-f(4)\\f(5)\\2f(2)f(3)-f(6)\\f(7)\\-f(2)^3+2f(2)f(4)-f(8) to clarify, a sum over all the different ways of making n with any factor, weighted by the number of distinct factors used times minus one to the power of the number of distinct factors used. for example, 8 can be made in three ways, 8, 2 ⋅ 4, 2 ⋅ 2 ⋅ 2, so we have \cdot(-f(2))(-f(2))(-f(2))+\cdot(-f(2))(-f(4))+\nu_8\cdot(-f(8)) and ν2, 2, 2 = 1 as only one distinct divisor is used, ν2, 4 = 2 as two distinct divisors are used, and ν₈ = 1 also. ADDITIVE ANALOGY? So we found the formula for the Dirichlet inverse of a general function above, can we do the same formula, but with the number of ways of partitioning a number additively rather than multiplicatively? So, 1 is easy, for an input f(n), we have g(1)=1\\ g(2)=(-f(2))+(-f(1))(-f(1))= f(1)^2-f(2)\\ g(3)=-f(3)+f(2)f(1)-f(1)f(1)f(1)=-f(3)+2f(2)f(1)-f(1)^3\\ g(4)=-f(4)+2f(1)f(3)+f(2)^2-2f(1)^2f(2)+f(1)^4\\ g(5)=-f(5)+2f(4)f(1)+2f(3)f(2)-2f(3)f(1)^2-2f(2)^2f(1)+2f(2)f(1)^3-f(1)^5 obviously these will grow very quickly in size.