Zig-zag resistance and OEIS sequence A029907
David Broadhurst
11 April 2024

Definitions: Let Z[n] be the zig-zag graph with vertices
v[k], for k = 0 to n, and 2*n-1 edges, comprising
n diagonal edges d[k], connecting v[k-1] to v[k], and
n-1 horizontal edges h[k], connecting v[k-1] to v[k+1].
Let each edge have unit resistance.
Let R[n] be the resistance of Z[n] between v[0] and v[n].

Example: Remove the curved edge from the figure in
https://arxiv.org/pdf/1208.1890.pdf
to obtain Z[5] with resistance R[5] = 15/11.

Lucas numbers: Let F[n] be the n-th Fibonacci number.
Then the n-th Lucas number is
L[n] = F[n+1] + F[n-1] = F[n] + 2*F[n-1].
With s = sqrt(5) and r = (s + 1)/2, Binet gives
L[n] = r^n + (1-r)^n, F[n] = (r^n - (1-r)^n)/s.

Theorem: R[n] = (n + 4*F[n]/L[n])/5.

Proof: It is easy to verify that R[1] = 1 and R[2] = 2/3.
For n > 2, Kirchhoff's laws are satisfied for a current
L[n] entering at v[0] and leaving at v[n], with currents
D[k] = F[n+1-k]*F[k] - F[n-k]*F[k-1] ................. (1)
H[k] = 2*F[n-k]*F[k] ..................................(2)
on the diagonal and horizontal edges d[k] and h[k].
Kirchhoff's second law, for closed loops, requires that
H[k] = D[k] + D[k+1], for each of the n-1 triangles.
This is proved by applying to Eqs (1,2) the recursion
F[m+1] = F[m] + F[m-1], with m = k and m = n-k.
To conserve charge, Kirchhoff's first law requires that
L[n] = D[1] + H[1], at vertex v[0],
D[1] = D[2] + H[2], at vertex v[1], and
D[k+1] + H[k] = D[k+2] + H[k+2] ...................... (3)
at vertices v[k+1], for 0 < k < n - 2.
Note that we need not consider vertices v[n-1] and v[n],
since the vectors D and H in Eqs (1,2) are palindromic.
The condition at v[0] is verified by using
D[1] = F[n] and H[1] = 2*F[n-1].
The condition at v[1] is verified by using
D[2] = F[n-3] and H[2] = 2*F[n-2].
To prove the remaining relation (3), use the identity
5*F[a]*F[b] = L[a+b] - (-1)^b*L[a-b] ................. (4)
which results from the Binet representation.
Applying (4) to Eqs (1,2), we obtain
D[k] = (1/5)*L[n] - (2/5)*(-1)^k*L[n-2*k+1] .......... (5)
H[k] = (2/5)*L[n] - (2/5)*(-1)^k*L[n-2*k] ............ (6)
Then condition (3) follows, by using the recursion
L[m+1] = L[m] + L[m-1]. The voltage at v[k] is given by
V[k] = (k/5)*L[n] + (2/5)*(F[n] - (-1)^k*F[n-2*k]) ... (7)
which solves Ohm's law, with
D[k] = V[k] - V[k-1] verified by Eqs (5,7) using
L[m] = F[m+1] + F[m-1]. The proof is completed by setting
k = n in formula (7) to obtain the resistance
R[n] = V[n]/L[n] = (n + 4*F[n]/L[n])/5 as claimed.

Comment: For n > 0, we obtain the integer sequence
V[n] = 1,2,4,8,15,28,51,92,164,290,509,888,1541,2662,...
which occurs in a different context in the OEIS entry
https://oeis.org/A029907

David Broadhurst