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\end{array}
\end{equation}
It can be seen that some numbers lead to previous numbers and therefore arrive at the same prime number to terminate the sequence. The steps till the resulting number is prime has the sequence
$A(n)$=1,1,1,3,1,2,1,14,3,5,1,2,1,7,5,5,1,2,1,16,2,2,1,3,4,5,5,2,1,3,1,3,2,17,4,3,1,3,2,10... Although this is really $A(n)$=0,0,0,2,0,1,0,13,2,4,0,1,0,6,4,4,0,1,0,15,1,1,0,2,3,4,4,1,0,2,0,2,1,16,3,2,0,2,1,9... $A(n)$=0,0,0,2,0,1,0,13,2,4,0,1,0,5,4,4,0,1,0,15,1,1,0,2,3,4,4,1,0,2,0,2,1,16,3,2,0,2,1,9,0,2,0,9,6,1,0,0,-,2,1,1,0,1,2,3,2,1,0,2,... as the prime numbers require $0$ transforms to be prime.
We could call this measure $A(n)$ the anti-primality of $n$, although, at the moment just because all prime numbers express anti-primality of $0$. But does the index even mean anything like this...
Comparing the divisors sequence $\sigma_0(n)$ and the $A(n)$ sequence, we observe that:
1, 2, 2, 3, 2, 4, 2, 4, 3, 4, 2, 6, 2, 4, 4, 5, 2, 6, 2, 6, 4, 4, 2, 8, 3, 4, 4, 6, 2, 8, \\
0, 0, 0, 2, 0, 1, 0, 13,2, 4, 0, 1, 0,
6, 5, 4, 4, 0, 1, 0,
15,1, 1, 0, 15,1 ,1 ,0 ,2 ,3 ,4 ,4 ,1 ,0, 2,
3, 4, 4, 1, 0, 2, ... \\
2, 6, 4, 4, 4, 9, 2, 4, 4, 8,
0, 2, 1, 16,3, 2, 0, 2, 1, 9
2, 6, 4, 4, 4, 9, 2, 4, 4, 8, 2, 8, 2, 6, 6, 4, 2, 10, 3, 6, 4, 6, 2, 8, 4, 8, 4, 4, 2, 12,\\
0, 2, 1, 16,3, 2, 0, 2, 1, 9, 0, 2, 0, 9, 6, 1, 0, 0, -, 2, 1, 1, 0, 1, 2, 3, 2, 1, 0, 2, \\
\\
In the first $40$ terms [not that many, this is speculative] \\
Every zero in $A$ after the first is matched with a $2$ in the divisor sequence, this is the primality of a number and is built in. \\