1.4 Use the digits $2,4,6$ to construct a six-digit number, but do not allow two consecutive 2s to appear in the six-digit number (for example, 626442 is allowed, 226426 is not allowed), how many such six-digit numbers are there?
[Solution 1] (1) If there is no 2 in the six-digit number, then each digit can only be chosen from 4 or 6, in this case there are $2^{6}=64$.
(2) If there is only 1 two in the six-digit number, then the 2 has $C_{6}^{1}=6$ position choices, and the remaining 5 positions are chosen from 4 or 6, so there are $6 \cdot 2^{5}=192$.
(3) If there are 2 twos in the six-digit number, at this time there are $2^{4} \cdot C_{5}^{2}=160$.
(4) If there are 3 twos in the six-digit number, at this time there are $2^{3} \cdot C_{4}^{3}=32$.
By the problem, it is impossible to have 4 or more twos in the six-digit number.
Thus, there are
$$
64+192+160+32=448 \text { in total. }
$$
[Solution 2] Let $a_{n}$ be the number of $n$-digit numbers that can be constructed using the digits $2,4,6$ and meet the conditions, then
$$
a_{1}=3, \quad a_{2}=8\binom{24,42,26,62,}{46,64,44,66} .
$$

If the first digit $Y$ is 2, then the second digit $Y$ can only be 4 or 6, so there are $2 a_{n-2}$ such $n$-digit numbers.

If the first digit $Y$ is not 2, then the first digit $Y$ is 4 or 6, and there are $2 a_{n-1}$ such $n$-digit numbers.
Thus, we have the recurrence relation
$$
a_{n}=2 a_{n \sim 1}+2 a_{n-2} \quad(n=3,4, \cdots) .
$$

From this, we derive
$$
\begin{array}{l}
a_{3}=2\left(a_{1}+a_{2}\right)=22, a_{4}=2\left(a_{2}+a_{3}\right)=60, \\
a_{5}=2\left(a_{3}+a_{4}\right)=164, a_{6}=2\left(a_{4}+a_{5}\right)=448,
\end{array}
$$

That is, there are 448 such six-digit numbers.