, 2022

If the first number in a sequence is 3 and the sequence is
generated by the function $x^2-3x+1$,
then the second number in the sequence is $$3^2-3(3)+1=1,$$ and the third number in
the sequence is $$1^2-3(1)+1=-1,$$ and
the fourth number in the sequence is $$(-1)^2-3(-1)+1=5.$$ The first four numbers
in the sequence are $3, 1, -1,
5$.
Let the first and second numbers in the sequence generated by the
function $x^2-4x+7$ be $f$ and $s$, respectively.

Then, the first three numbers in the sequence are $f,s,7$.

Since the third number in the sequence is 7, then $s^2-4s+7=7$.

Solving this equation, we get $s^2-4s=0$ or $s(s-4)=0$, which has solutions $s=0$ and $s=4$, and so the first three numbers in
the sequence could be $f,0,7$ or
$f,4,7$.

If the second number in the sequence is 0, then $f^2-4f+7=0$.

The discriminant of this equation is $(-4)^2-4(1)(7)=-12$ (less than zero) and
so there are no real solutions.

Thus, there is no first number in this sequence for which the second
number is 0.

If the second number in the sequence is 4, then $f^2-4f+7=4$, or $f^2-4f+3=0$ and so $(f-1)(f-3)=0$, which has solutions $f=1$ and $f=3$.

Therefore, if 7 is the third number in a sequence generated by the
function $x^2-4x+7$, then the first
three numbers in the sequence could be $1,4,7$ or $3,4,7$, and so the possible first numbers
in the sequence are 1 and 3.
The first two numbers in the sequence are $c,c$, and so $c^2-7c-48=c$.

Solving this equation, we get $c^2-8c-48=0$ or $(c+4)(c-12)=0$, which has solutions $c=-4$ and $c=12$.
The first number in the sequence is $a$ and the second number is $b$, and so $$a^2-12a+39=b.$$ The second number in the
sequence is $b$ and the third number
is $a$, and so $$b^2-12b+39=a.$$ Subtracting the second
equation from the first and simplifying, we get $$\begin{align*} (a^2-12a+39)-(b^2-12b+39)&=b-a\\ a^2-b^2-12a+12b&=b-a\\ a^2-b^2-11a+11b&=0\\ (a-b)(a+b)-11(a-b)&=0\\ (a-b)(a+b-11)&=0\end{align*}$$ Since $a\neq b$, then $a-b\neq 0$ and so $a+b-11=0$ or $b=11-a$.

Substituting into the first equation, we get $a^2-12a+39=11-a$ or $a^2-11a+28=0$. Factoring gives $(a-4)(a-7)=0$ and so the possible values
of $a$ are 4 and 7.

(Note that the two possible sequences are $4,7,4, \dots$ and $7,4,7,\dots$.)