Possibly solved? Be the judge yourselves.
Everyone has possibly found the rule that generates the first couple of numbers and the last couple of numbers. The rule is :
$(a,b,c)\mapsto (d,e,f) \Rightarrow d=a(b+1),\ f=c(b+1)$. Thus $(3,9,4)\mapsto (30,e,40)$. But then, what is the rule that generates $e$ ?
Be warned though, the argument that I am going to provide will not work for all triples, but hinges on a very crucial observation provided in the $4$ values given to us, and will only hold for triples following that pattern.
Always works now...
We don't need this observation now.
The observation is that:
$c$ is always even! Once we take the assumption that $c$ will always have to be even, we can proceed.
So far, we have established one assumption
Now we look for patterns.
We observe that since $c$ is always even, $c/2$ is an integer. Since $e$ has to be a two digit number, call it $XY$ where $X$ and $Y$ represent the digits. There are two observations now. One is that $X+Y=\lfloor(c/2)\rfloor+1$ in all the values that is given. The second is that either $X$ and $Y$ are consecutive with $Y>X$ or $X=Y$. $\lfloor \cdot\rfloor$ is the floor function, which sends any number to the integer just less than it.
How does that help us? Here's how.
Take $c$, divide it by $2$ and add $1$ to it. Now we have $\lfloor(c/2)\rfloor+1$. Call it $x$. If $x$ is even, divide $x$ by $2$ and set $X=Y=x/2$. Else, set $X=(x-1)/2,\ Y=(x+1)/2$. We check that in the former case $X=Y$ and in the latter case $X<Y$ with $X$ and $Y$ consecutive. Also $X+Y=x$ in both cases, so both our observations hold. And there's a nice symmetry to it!
What does this tell us?
$(a,b,c)\mapsto (d,X,Y,f)$, a $4$-tuple rather than a $3$-tuple.
Is this true?
Doing a quick check that our rule works:
$$(2,5,8)\to \lfloor8/2\rfloor=4\to 4+1=5\to X=(5-1)/2=2,\ Y=(5+1)/2=3\to XY=23$$
$$(3,5,4)\to \lfloor4/2\rfloor=2\to 2+1=3\to X=(3-1)/2=1,\ Y=(3+1)/2=2\to XY=12$$
$$(7,4,6)\to \lfloor 6/2\rfloor=3\to 3+1=4\to X=Y=4/2=2\to XY=22$$
$$(5,3,10)\to \lfloor10/2\rfloor=5\to 5+1=6\to X=Y=6/2=3\to XY=33$$
Finally, the moment has come :
$$(3,9,4)\to \lfloor4/2\rfloor=2\to 2+1=3\to X=(3-1)/2=1,\ Y=(3+1)/2=2\to XY=12$$
Putting everything together:
$$(3,9,4)\mapsto (30,12,40) \Rightarrow 3,9,4=301240$$
And that concludes the problem.
If I am right, I do owe an apology to Galen. He has done such a wonderful job and I would hate to prove him wrong. I wholeheartedly believed he was right, but in the end, I came up with something that does not agree with him. Please accept my humble apologies.