TITLE: The 1-step vanishing polyplets on Conway's game of life
QUESTION [22 upvotes]: A $n$-polyplet is a collection of $n$ cells on a grid which are orthogonally or diagonally connected.
The number of $n$-polyplets is given by the OEIS sequence A030222: $1, 2, 5, 22, 94, 524, 3031, \dots$    
See below the five $3$-polyplets:

A polyplet will be called $1$-step vanishing on Conway's game of life, if every cell dies after one step.
We observed that for $n\le 4$, a $n$-polyplet is $1$-step vanishing iff $n \le 2$.  
We found $1$-step vanishing polyplets with $n=9, 12$, see below:

Question:  Is there an other $1$-step vanishing $n$-polyplet with $n \le 12$?
If yes, what are they? (note that there are exactly $37963911$ $n$-polyplets with $n \le 12$).
We can build infinitely many $1$-step vanishing polyplets with such pattern, see below with $n=142$:

Bonus question: Are there $1$-step vanishing polyplets of an other kind?

REPLY [5 votes]: Theorem: The set of positive integers $n$ such that there is a $1$-step vanishing polyplet with $n$ cells is $$ \{1,2,9,10,12,14 \} \cup  \mathbb{N}_{\ge 15}$$ It can be represented by $1$-step vanishing polyominoes without hole and a non-trivial symmetry group.   
Proof: The finite part is given by:

For the infinite part, we will do three steps.
Firstly, the family beginning as follows:  

provides the integers $6a+b$ with $b \in \{ 3 , \dots , a+2 \}$. Now, $$6a + (a+2) +1 \le 6(a+1)+3$$ if and only if $a \ge 6$. But $6 \times 6+3  = 39$, so it is ok for $\mathbb{N}_{\ge 39}$.    
Next, the gaps of the previous family for $14<n<39$ are given by the set $$\{17,18,19,20,24,25,26,31,32,38\}$$ and can be filled ($n=19,20$ excepted) by the family beginning as follows:     

providing the integers $7a+b$ with $b \in \{ 3 , \dots , a+2 \}$.    
Finally, $n=19,20$ are given by:   

The result follows. $\square$  
Bonus question: Can we classify (in some sense) the $1$-step vanishing polyominoes without hole?  
There are $21$ such polyominoes with $n$ cells and $n \le 18$.