4. Given
$$
A=17^{2012 n}+4 \times 17^{4 n}+7 \times 19^{7 n}(n \in \mathbf{N})
$$

can be expressed as the product of $k(k \in \mathbf{N}, k>1)$ consecutive integers. Then $n+k=$ $\qquad$ .
4. 2 .

If $k \geqslant 4$, then $81 A$.
$$
\begin{array}{l}
\text { But } A=17^{2012 n}+4 \times 17^{4 n}+7 \times 19^{7 n} \\
\equiv 1+4-3^{7 n} \equiv 5-3^{n} \not \equiv 0(\bmod 8),
\end{array}
$$

Contradiction.
If $k=3$, let $A=m\left(m^{2}-1\right)(m \in \mathbf{Z})$.
But $A \equiv 2^{2012 n}-2^{4 n}+2 \times(-1)^{7 n}$
$$
\equiv 2 \times(-1)^{n} \equiv \pm 2(\bmod 5),
$$

This contradicts $m\left(m^{2}-1\right) \equiv 0$ or $\pm 1(\bmod 5)$.
If $k=2$, let $A=m(m+1)$. Then
$$
(2 m+1)^{2}=4 A+1 \text {. }
$$

But when $n \geqslant 1$, we have
$$
\left(2 \times 17^{100 \sigma_{n}}\right)^{2}<4 A+1<\left(2 \times 17^{100 \sigma_{n}}+1\right)^{2} \text {, }
$$

Contradiction.
Therefore, $n=0$. In this case, $A=12=3 \times 4$ satisfies the condition.
Thus, $(n, k)=(0,2)$. Hence $n+k=2$.