The expression $16^n+4^n+1$ is equiavalent to the expression $(2^{p(n)}-1)/(2^{q(n)}-1)$ for all positive integers $n>1$ where $p(n)$ and $q(n)$ are functions and $\tfrac{p(n)}{q(n)}$ is constant.  Find $p(2006)-q(2006)$.
1. We start with the given expression \(16^n + 4^n + 1\). We need to show that this expression can be written in the form \(\frac{2^{p(n)} - 1}{2^{q(n)} - 1}\) for some functions \(p(n)\) and \(q(n)\).

2. Notice that \(16^n = (2^4)^n = 2^{4n}\) and \(4^n = (2^2)^n = 2^{2n}\). Therefore, the given expression can be rewritten as:
   \[
   16^n + 4^n + 1 = 2^{4n} + 2^{2n} + 1
   \]

3. We recognize that \(2^{4n} + 2^{2n} + 1\) can be seen as a sum of terms in a geometric series with the common ratio \(2^{2n}\). Specifically, we can write:
   \[
   2^{4n} + 2^{2n} + 1 = 2^{2n \cdot 2} + 2^{2n \cdot 1} + 2^{2n \cdot 0}
   \]

4. The sum of a geometric series \(a + ar + ar^2 + \cdots + ar^{k-1}\) is given by:
   \[
   S = \frac{a(r^k - 1)}{r - 1}
   \]
   Here, \(a = 1\), \(r = 2^{2n}\), and the number of terms \(k = 3\). Thus, we have:
   \[
   1 + 2^{2n} + 2^{4n} = \frac{2^{6n} - 1}{2^{2n} - 1}
   \]

5. From the above, we identify \(p(n) = 6n\) and \(q(n) = 2n\). We need to verify that \(\frac{p(n)}{q(n)}\) is constant:
   \[
   \frac{p(n)}{q(n)} = \frac{6n}{2n} = 3
   \]
   This confirms that \(\frac{p(n)}{q(n)}\) is indeed constant.

6. Finally, we need to find \(p(2006) - q(2006)\):
   \[
   p(2006) = 6 \cdot 2006 = 12036
   \]
   \[
   q(2006) = 2 \cdot 2006 = 4012
   \]
   \[
   p(2006) - q(2006) = 12036 - 4012 = 8024
   \]

The final answer is \(\boxed{8024}\).