No!
Claim: the bends in concentric circle n=1,2... are the values 4(x^2+xy+y^2)-1 for x+y=n with x=0..n-1 repeated six times.
The spheres with x=0 have one outer neighbour and 3 inner neighbours. Other spheres have two of each.
Proof by induction on n:
The bend for a sphere touching 2,2,K1,K2 is a specialization of Soddy's formula:
K(K1,K2) = (4+K1+K2 + sqrt(3*(4+K1+K2)^2 - 6*(8+K1^2+K2^2))) / 2.
Define (x,y) = 4(x^2+xy+y^2)-1.
Take neighbouring spheres in circle n. By induction hypothesis K1 = (x,y), K2 = (x+1,y-1).
It is safe to use x=n-1 because K2=(n,0)=(0,n) is the first sphere of the next repetition.
The bend of the inner sphere touching 2,2,K1,K2 works out to be K(K1,K2) = (x+1,y).
There is one extra sphere in ring n+1 which touches (0,n) and (1,n), as described in my previous post.
The bend of this sphere works out to be K((0,n), (1,n)) = (0,n+1).
Corollary: Spheres on the six radial lines have bends given by (0,n) = 4n^2-1 = A000466(n).
A270248 starts with a(1)=0, which corresponds to the surrounding unit sphere with bend -1.