Finding where I am in a linear recurrence relation
First of all, your linear recurrence relation is not exactly linear, it is affine. However, in general, it is not difficult to get rid of the constant term $d$.

To do this, define $A(n)=a(n)+t$, and choose $t$ to make sure that the constant term vanishes. This requires $t=(c_1+...+c_k)t+d$, clearly, this can be done unless $c_1+...+c_k=1$.

From now on, I assume that we have no constant term $d$. You want to recover a possible $n$ from a single value $a(n)$. As far as I know, there is no direct relation between this and discrete logarithms. 

However, there is a direct relation to the discrete logarithm problem for a simple variant of your question, i.e. given $a(n), a(n-1), ..., a(n-k+1)$ recover $n$.

This variant is a classical question about LFSR (linear feedback shift register). Usually, people consider LFSRs modulo 2, but it is also possible to consider them modulo $p$ and this is precisely what we are doing here.

To the recurrence formula, you can associate a polynomial $P(x)=x^n-(c_n\cdot x^{(n-1)}+...+c_2\cdot x+c_1)$. If you associate to a state $a(1),...,a(n)$ the polynomial $a(1)+a(2)\cdot x+...+a(n)\cdot x^{n-1}$ then updating a state is simply multiplication by $x$ modulo $p$. If this polynomial is irreducible, then finding the correct power of $x$ is simply a discrete logarithm problem in $GF(p^n)$.

When $P(x)$ is not irreducible, you are in fact simultaneously computing the values modulo several irreducible polynomials (and the results are pasted together using the Chinese Remainder Theorem), so you can solve it by computing discrete logarithms in smaller fields.

Coming back to your original question, a equivalent formulation in terms of discrete logarithms is: given $x$ an integer modulo $p$ and an extension field $GF(p^n)$ find an arbitrary element $X$ in $GF(p^n)$ represented by a polynomial whose constant term is $x$, together with a logarithm of $X$.