Sum Over Coprime Elements: Inversion Formula

[Leroy Quet]

Let us say we have the arbitrary sequence {a(k)}.

We define {b(k)}, based on {a(k)}, by:

b(n) = sum{1<=k<=n, GCD(k,n)=1} a(k).

(So, for example, b(12) = a(1) + a(5) + a(7) + a(11). And b(1) = b(2) = 
a(1), obviously.)


So, given {b(k)} (which must have b(1) = b(2)), how do we get the 
sequence {a(k)}?

If a(n) =

sum{k>=2} b(k) * c(n,k),

then there is a triangular array {c(n,k)} which begins as follows
(unless I made an error):

 1
-1, 1
-1, 0, 1
 1,-1,-1, 1
-1, 0, 0, 0, 1
 1, 0, 0,-1,-1, 1
 1, 0,-1, 0,-1, 0, 1
-1, 0, 2,-1, 0, 0,-1, 1

(n is from the top then down, and starts with n=1; and k is from left to 
right, and starts with k=2.)


Now the terms of {c(n,k)} are easily found by recursion, but I wonder 
what the closed form formula is that gets this triangle. (Something to do 
with the Moebius function, perhaps?)

I don't believe this triangle is in the EIS either.

But I believe the left-most column is sequence A096433.

Anything more that can be said about this triangular array?
I bet there is something somewhere in the literature about it.
 
thanks,
Leroy Quet