Finding $\frac{I_1}{I_2}$ from recurrence relation

Question

Consider following recurrence relation:

$$2\left(\sum_{i = 0}^k I_{n-i}\right) + I_{n-k} = I_{n-k-1}$$ $k = 0 , 1 , \dots , n-1$

Also $I_0 = \frac{V}{R}$. Assuming that $n \ge2$ is a fixed integer, I'm looking for $\frac{I_1}{I_2}$ in terms of $n$. After that I will take the limit as $n \to \infty$. This problem has come up when solving a circuit with infinite number of resistors. I don't know how to tackle this problem. Setting $S(k) =\sum_{i = 0}^k I_{n-i}$ isn't helpful. Also I tried to find a pattern in value of $\frac{I_1}{I_2}$ as $n$ increases but it was so complicated. The first terms are $3, \frac{11}{3} $ and $ \frac{41}{11}$. Using equivalent resistor we find that $\frac{I_1}{I_2} = \sqrt{3} + 2$.

Edit:

Here is the circuit:

I added a voltage source between a-b, so we have $I_0 = \frac{V}{R}$. Assuming there are $n$ blocks, we calculate $\frac{I_1}{I_2}$ and then taking the limit yields the result. According to Cesareo's answer, the answer doesn't depend on $n$. How this is plausible? Increasing $n$ should change the value of $\frac{I_1}{I_2}$.

Answer 1

Hint.

Making

$$ R_{n,k} = 2\sum_{i=0}^{i=k}I_{n-i}+I_{n-k}-I_{n-k-1}=0 $$

we have

$$ R_{n,k}-R_{n,k-1} = I_{n-k+1}-4I_{n-k}+I_{n-k-1}=0 $$

In this recurrence calling $n-k=m$ or

$$ I_{m+1}-4I_{m}+I_{m-1}=0 $$

and this recurrence has the general solution

$$ I_{m} = (2-\sqrt{3})^m C_1+(2+\sqrt{3})^m C_2 $$

now depending on the initial/boundary conditions for the determination of $C_1, C_2$ we have

$$ \frac{I_2}{I_1}=\frac{(2-\sqrt{3})^2 C_1+(2+\sqrt{3})^2 C_2}{(2-\sqrt{3}) C_1+(2+\sqrt{3}) C_2} $$

for instance: if $I_n$ is limited then necessarily $C_2=0$ but if $I_n$ is unlimited then we need $C_1, C_2$

NOTE

Included a MATHEMATICA script for the calculus of $\frac{I_2}{I_1}$ in the ladder resistance circuit included in the OP.

R[1, r] = 3 r;
R[n_, r_] := R[n, r] = 2 r + r R[n - 1, r]/(r + R[n - 1, r])

n = 10;
For[j = 1; rel = {}, j <= n, j++,
  equ1 = Table[i[k] - i0[k + 1] + i0[k], {k, 1, n}];
  equ2 = Table[r i[k] - i0[k] R[k, r], {k, 1, n}];
  vars = Union[Variables[equ1], Variables[equ2]];
  equs = Join[Join[equ1, equ2], {i0[n + 1] - in}];
  solIJ = Solve[Thread[equs == 0], Take[vars, {2, Length[vars]}]][[1]];
  AppendTo[rel, i[n - 1]/i[n - 2] /. solIJ]
]

gr1 = ListPlot[rel, PlotStyle -> Red];
gr2 = Plot[2 + Sqrt[3], {k, 0, n + 2}];
Show[gr1, gr2]

tr = Table[R[k, r], {k, 1, n}] /. {r -> 1}
gr3 = ListPlot[tr, PlotStyle -> Red];
gr4 = Plot[(1 + Sqrt[3]), {k, 0, n}];
Show[gr3, gr4, PlotRange -> All]

Answer 2

This isn't an answer, but it may be a start. It's too big to go in a comment box.

For a given $n$, we can write this as a system of $n$ linear equations in $n+1$ unknowns. Then we can set $I_0$ arbitrarily in order to get a $n\times n$ system. I did this in a python script, setting the value of $I_0$ to the determinant of the coefficient matrix in order to get integer values.

import numpy as np

def test(n):
    A = 2*np.ones((n,n))
    A = np.tril(A)
    for k in range(n):
        A[k,k] = 3
    for k in range(n-1):
        A[k,k+1] = -1
    X = np.zeros(n)
    X[n-1] = np.linalg.det(A)
    return np.linalg.inv(A)@X

for n in range(2,10):
    print(n, test(n))

This produced the output,

2 [1. 3.]
3 [ 1.  3. 11.]
4 [ 1.  3. 11. 41.]
5 [  1.   3.  11.  41. 153.]
6 [  1.   3.  11.  41. 153. 571.]
7 [1.000e+00 3.000e+00 1.100e+01 4.100e+01 1.530e+02 5.710e+02 2.131e+03]
8 [1.000e+00 3.000e+00 1.100e+01 4.100e+01 1.530e+02 5.710e+02 2.131e+03
 7.953e+03]
9 [1.0000e+00 3.0000e+00 1.1000e+01 4.1000e+01 1.5300e+02 5.7100e+02
 2.1310e+03 7.9530e+03 2.9681e+04]

The answers are "backwards". That is for $n=4$, for example, the answer $[1,3,11,41]$ means $I_1=41, I_2=11,I_3=3,I_4=1$.

The sequence $$1,3,11,41,153,571,2131,7953,29681,\dots$$ does not appear in OEIS. If we could figure out a formula for this sequence, at least we'd have something concrete to try to prove. I don't know a way to proceed absent such a formula.

EDIT

The sequence seems to satisfy the recurrence, $$a_n=4a_{n-1}-a_{n-2},\ n\geq2$$ with initial conditions $a_0=a_1=1.$ This gives the solution $$a_n=\left(\frac{3-\sqrt3}6\right)(2+\sqrt3)^n+\left(\frac{3+\sqrt3}6\right)(2-\sqrt3)^n$$ by standard methods. It's clear that $$\frac{a_{n+1}}{a_n}\to2+\sqrt3\text{ as }n\to\infty,$$ but it remains to prove that the sequence satisfies the original equations. This should be just a matter of subsituting the values into the equations, and confirming that the latter hold.