The $n$th term of the sequence is For the bonus (what numbers never appear in the sequence),


They appear to be: Following this rule, the next two numbers would be: because


f(n) = (n^2) % 100 so next is f(16) = 256 % 100 = 56


Here's an argument to show that Reasoning Not sure about the primes bit yet.


Before I start, consider function $f$ that satisfies $A_x = f(x)$. That is, to find the $x$-th number in the sequence, we can plug in $x$ to $f$. So we want to find the values of $f(21), f(22), f(23), f(24), f(25)$. Inspired from hint 2, instead of mapping x to $f(x)$, what if we try to map the relation of $x$ and $f(f(x))$ instead? We get this following ...


Next sequence defined as: no. of characters x from previous one following an x. Ordered from 1 to 9. Example: aabbb, 2a3b, 12131a1b ... So next one: 212223 and then 114213, 31121314, 41122314, 31221324, 21322314, 21322314, ...


Next one is 0.

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