$$\sum_{n=0}^{\infty} \frac{n^3((2n)!) + (2n-1)(n!)}{(n!)(2n)!} = ae + \frac{b}{e} + c$$ where $$a, b, c \in \mathbb{Z}$$ and $$e = \sum_{n=0}^{\infty} \frac{1}{n!}$$. Then $$a^2 - b + c$$ is equal to ______.

Step 1: Break down the general term of the series

Let the general term of the summation be $$T_n$$:

$$T_n = \frac{n^3((2n)!) + (2n-1)(n!)}{(n!)(2n)!}$$

Splitting the fraction into two independent parts gives:

$$T_n = \frac{n^3((2n)!)}{(n!)(2n)!} + \frac{(2n-1)(n!)}{(n!)(2n)!}$$

$$T_n = \frac{n^3}{n!} + \frac{2n-1}{(2n)!}$$

Now, we can split the infinite summation into two separate series, $$S_1$$ and $$S_2$$:

$$\sum_{n=0}^{\infty} T_n = \sum_{n=0}^{\infty} \frac{n^3}{n!} + \sum_{n=0}^{\infty} \frac{2n-1}{(2n)!} = S_1 + S_2$$

Step 2: Evaluate the first series $$S_1$$

To evaluate $$S_1 = \sum_{n=0}^{\infty} \frac{n^3}{n!}$$, we express $$n^3$$ in terms of falling factorials:

$$n^3 = n(n-1)(n-2) + 3n(n-1) + n$$

Substituting this into the summation:

$$S_1 = \sum_{n=3}^{\infty} \frac{n(n-1)(n-2)}{n!} + \sum_{n=2}^{\infty} \frac{3n(n-1)}{n!} + \sum_{n=1}^{\infty} \frac{n}{n!}$$

$$S_1 = \sum_{n=3}^{\infty} \frac{1}{(n-3)!} + 3\sum_{n=2}^{\infty} \frac{1}{(n-2)!} + \sum_{n=1}^{\infty} \frac{1}{(n-1)!}$$

Using the definition $$e = \sum_{k=0}^{\infty} \frac{1}{k!}$$, each of these transformed parts sums to $$e$$:

$$S_1 = e + 3e + e = 5e$$

Step 3: Evaluate the second series $$S_2$$

To evaluate $$S_2 = \sum_{n=0}^{\infty} \frac{2n-1}{(2n)!}$$, we separate the terms in the numerator:

$$S_2 = \sum_{n=0}^{\infty} \frac{2n}{(2n)!} - \sum_{n=0}^{\infty} \frac{1}{(2n)!}$$

For the first part, the $$n=0$$ term is 0, so the summation starts from $$n=1$$:

$$\sum_{n=1}^{\infty} \frac{2n}{(2n)!} = \sum_{n=1}^{\infty} \frac{1}{(2n-1)!} = \frac{1}{1!} + \frac{1}{3!} + \frac{1}{5!} + \dots = \sinh(1)$$

For the second part:

$$\sum_{n=0}^{\infty} \frac{1}{(2n)!} = \frac{1}{0!} + \frac{1}{2!} + \frac{1}{4!} + \dots = \cosh(1)$$

Combining the two parts gives:

$$S_2 = \sinh(1) - \cosh(1)$$

Using exponential definitions $$\sinh(1) = \frac{e - e^{-1}}{2}$$ and $$\cosh(1) = \frac{e + e^{-1}}{2}$$:

$$S_2 = \frac{e - e^{-1}}{2} - \frac{e + e^{-1}}{2} = -e^{-1} = -\frac{1}{e}$$

Step 4: Combine both series and compare coefficients

Combining $$S_1$$ and $$S_2$$:

$$\sum_{n=0}^{\infty} T_n = 5e - \frac{1}{e}$$

Comparing this result with the given equation $$ae + \frac{b}{e} + c$$:

- $$a = 5$$

- $$b = -1$$

- $$c = 0$$

Step 5: Calculate the final required expression

$$a^2 - b + c = (5)^2 - (-1) + 0$$

$$a^2 - b + c = 25 + 1 + 0 = 26$$

Conclusion:

The value of $$a^2 - b + c$$ is equal to 26.