Complex numbers product and ratio, prove this relation.

Question

Define a table $T$ as follows:

$$\text{If}\; n\geq k \; \text{then} \; T(n,k) = (2+3 i) \sum _{i=1}^{n-1} T(n-i,k-1)+(5+7 i) \sum _{i=1}^{n-1} T(n-i,k) \; \text{else} \; T(n,k) = 0$$

Then take rows sums of that table $T$. This a sequence of complex numbers, call it $A$:

$$A = 1, 3 + 3 I, -11 + 47 I, -563 + 259 I, -7099 - 3565 I, -21147 - 99517 I ...$$

Define another table $t$ as follows:

$$\text{If}\; n\geq k \; \text{then} \; t(n,k) = (2+3 i) \sum _{i=1}^{k-1} t(n-i,k-1)+(5+7 i) \sum _{i=1}^{k-1} t(n-i,k) \; \text{else} \; t(n,k) = 0$$

Then take rows sums of that table $t$. This a sequence of complex numbers, call it $a$: $$a = 1, 3 + 3 I, -13 + 44 I, -540 + 183 I, -5883 - 4055 I, -4263 - 89973 I ...$$

Prove that: $$\prod _{n=2}^{\infty} \frac{1}{1-\frac{1}{A_n}} = \lim_{n\to \infty } \, \frac{A_n}{a_n}$$

which is approximately:

$$ = 1.141394903149-0.252933757966 i$$

Mathematica:

Clear[t, n, k, a, b, x, y, b1, at, T, AT, at];
nn = 200;
x = 2 + 3 I;
y = 5 + 7*I;
T[n_, 1] = 1;
T[n_, k_] := 
  T[n, k] = 
   If[n >= k, 
    x*Sum[T[n - i, k - 1], {i, 1, n - 1}] + 
     y*Sum[T[n - i, k], {i, 1, n - 1}], 0];
a = Table[Table[Expand[T[n, k]], {k, 1, nn}], {n, 1, nn}];
(*TableForm[a]*)
AT = Total[Transpose[a]];
AT[[1 ;; 20]];
Table[Limit[((s + 1)^(n - 1) + s - 1)/s, s -> y + 1], {n, 1, 12}];
c = Product[1/(1 - 1/AT[[i]]), {i, 2, nn}];
N[c, 12]

Clear[t];
t[n_, 1] = 1;
t[n_, k_] := 
  t[n, k] = 
   If[n >= k, 
    x*Sum[t[n - i, k - 1], {i, 1, k - 1}] + 
     y*Sum[t[n - i, k], {i, 1, k - 1}], 0];
a = Table[Table[Expand[t[n, k]], {k, 1, nn}], {n, 1, nn}];
aa = Table[
   Total[Table[t[n - k + 1, k], {k, 1, nn}]], {n, 1 + 1, nn + 1}];
(*TableForm[a]*)
at = Total[Transpose[a]];
at[[1 ;; 20]];
N[AT[[nn]]/at[[nn]], 12]