The given equation is a functional equation with a recursive definition of function f.

To solve this equation, we can use the method of generating functions. We define the generating function F(x) as:

F(x) = f(0) + f(1)x + f(2)x^2/2! + f(3)x^3/3! + ...

Multiplying both sides of the original equation by x^k/k! and summing over all k, we get:

∑(k>=0) f(k) x^k/k! = ∑(k>=0) f(k-1) x^k/k! + an ∑(k>=0) x^k/k!

Using the generating function F(x) and its derivatives, we can simplify this expression as:

F'(x) = F(x) + an * e^x

Solving this differential equation, we get:

F(x) = c * e^x - an

where c is a constant of integration. To determine c, we use the initial condition f(0) = 1, which gives:

F(0) = f(0) = 1

c = 1 + an

Therefore, the generating function F(x) is:

F(x) = (1 + an) * e^x - an

We can then use the generating function to find the coefficients of the function f. Taking the k-th derivative of F(x) and evaluating it at x=0, we get:

f(k) = [x^k/k!] F(x) |x=0
= [(1+an) * x^k/k! e^x - an * x^k/k!] |x=0
= (1+an) * f(k-1) - an * δ(k,0)

where δ(k,0) is the Kronecker delta, which equals 1 if k=0 and 0 otherwise.

Therefore, the first derivative of f is:

f'(k) = (1+an) * f(k-1) - an * δ(k,0)

And the second derivative of f is:

f''(k) = (1+an) * f'(k-1) = (1+an)^2 * f(k-2) - an * (1+an) * δ(k-1,0)
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