Missing Numbers - 5

$4\times2-2=6$ $6\times3-2=16$ $16\times4-2=62$ $62\times5-2=308$ $308\times6-2=1846$

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Generally, it follows $a_n=na_{n-1}-2$

Explicit formula:

$a_n = 4 \cdot n! - 2 \left( 1+ \displaystyle \sum_{k=2}^{n-1} \frac {n!}{k!} \right), n > 1$

$\Rightarrow a_{\color{#D61F06}6} = 4 \cdot {\color{#D61F06}6}! - 2 \left( \frac{{\color{#D61F06}6}!}{2!} + \frac{{\color{#D61F06}6}!}{3!} + \frac{{\color{#D61F06}6}!}{4!} + \frac{{\color{#D61F06}6}!}{5!}\right) = \boxed{1846}$

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I found this by multiplying out the terms, but I didn't simplify anything

$\begin{aligned} a_1 & = 4 = a \\ a_2 & = a\cdot2-2 \\ a_3 & = (a\cdot2-2)\cdot3-2=a\cdot2\cdot3-2\cdot3 \\ a_4 & = ((a\cdot2-2)\cdot3-2)\cdot4-2=a\cdot2\cdot3\cdot4-2\cdot3\cdot4-2 \\ a_5 & = (((a\cdot2-2)\cdot3-2)\cdot4-2)\cdot5-2=a\cdot2\cdot3\cdot4\cdot5-2\cdot3\cdot4\cdot5-2\cdot5 \\ a_6 & = ((((a\cdot2-2)\cdot3-2)\cdot4-2)\cdot5-2)\cdot6+2=a\cdot2\cdot3\cdot4\cdot5\cdot6-2\cdot3\cdot4\cdot5\cdot6-2\cdot5\cdot6-2\\ a_7 & = (((((a\cdot2-2)\cdot3-2)\cdot4-2)\cdot5-2)\cdot6-2)\cdot7-2=a\cdot2\cdot3\cdot4\cdot5\cdot6\cdot7-2\cdot3\cdot4\cdot5\cdot6\cdot7-2\cdot5\cdot6\cdot7-2\cdot7-2 \\ \vdots & \end{aligned}$