Semiprime numbers which, along with their prime factors, generate many semiprimes by concatenation There's something quite interesting about the number $1191$:

*

*this number is a semiprime ($1191= 3 \cdot 397$),

*the concatenation of its prime factors in any order are semiprimes ($3397$ and $3973$ are semiprimes), but not just that,

*if you concatenate $1191$ and its prime factors $3$ and $397$ in any order, the result is always a semiprime (i.e. $11913397$, $11913973$, $31191397$, $39711913$, $33971191$, and $39731191$ are all semiprimes).

I call such numbers semiprime-to-the-bones, or STB.
Definitions: Let $s = p \cdot q$ be a semiprime.

*

*I call $s$ an STB number if these following 8 concatenations are all semiprimes: $p \mathop{^\smallfrown} q$, $q \mathop{^\smallfrown} p$, $s \mathop{^\smallfrown} p \mathop{^\smallfrown} q$, $s \mathop{^\smallfrown} q \mathop{^\smallfrown} p$, $p \mathop{^\smallfrown} s \mathop{^\smallfrown} q$, $q \mathop{^\smallfrown} s \mathop{^\smallfrown} p$, $p \mathop{^\smallfrown} q \mathop{^\smallfrown} s$, $q \mathop{^\smallfrown} p \mathop{^\smallfrown} s$, where $a \mathop{^\smallfrown} b$ means the number whose decimal representation is the decimal representation of $a$ followed by that of $b$ (i.e. concatenation).


*I call $s$ a nice STB number if it is an STB number whose reversal is also semiprime.


*I call $s$ a super STB number if it is an STB number and the following are all also semiprimes:
$s \mathop{^\smallfrown} p$,
$s \mathop{^\smallfrown} q$,
$p \mathop{^\smallfrown} s$,
$q \mathop{^\smallfrown} s$.
I've searched numbers up to $3000$, but the only STB number I've found is $1191$.

Questions:

*

*Can you find a larger example of STB numbers (or nice STB numbers).

*Do super STB numbers exist?


 A: Here I want to prove that there are no even STB numbers, here's the proof: Let S be an even semiprime ($S= A\times2$), it is very obvious that the last digit of $A$ is either $1, 3, 7,$ or $9$ (with $10= 5\times2$ as an exceptional case), so the concatenation of $A$ and $2$ will be looked like these: $\_\_\_ 12$, or $\_\_\_32$, or $\_\_\_72$, or $\_\_\_92$, we can see that these four numbers are all divisible by $4$, so the concatenation of $A$ and $2$ will never be a semiprime!. So I've completed the proof that there are no even STB numbers. And yes I wanna add that if $p= q$ then I will call it as a 
"weak" STB numbers or  "weak" super STB numbers.
A: First I want to tell you (I've verified it) that Jerry Jan's conclusion is absolutely true, and this STB number 560523= 3* 186841 is almost/near super STB number (!!). this STB number 560523 has satisfied 10 conditions of those 12 conditions required (!). And I did further research on it, and finally I have a strong/good method to find type 1 super STB numbers, so just simply follow this 2 simple steps: (Let S= 3*N be an STB number, and  All dots here means concatenation), 
Step 1: Find an STB number S= 3*N, such that both S.3 and 3.S are semiprimes ("not impossible at all").
Step 2: Find a reversible prime ("an Emirp") of the form 10000...00003, and let 10000...00003 be X, and its reversal 30000...00001 be Y (We need both X and Y to be prime numbers)
If S.N= NY and N.S= NX, then S is a (type 1) super STB number!. ADDITIONAL INFORMATION: If we follow that method/ steps, then the next type 1 super STB number after 9 will have more than 660000 digits ! (I mean it). The STB number 28222149 is VERY NEARLY a super STB number (!!), 28222149 has satisfied 11 conditions of those 12 conditions required. And there are SO MANY cases where the number of digits of 30000...00001 is one less than the number of digits of its reversal 10000...00003, and after calculate some things together, I conclude that there's more than 90 % chances that the next (type 1) super STB number will have EXACTLY 11 digits.
A: I want give you a valuable information, I did some observations on super STB numbers, and I conclude that there are only two types of super STB numbers: 
(Let S=M*N be a super STB number),the first type is super STB numbers  that is divisible by 3 (i.e. S=3*N), which is Joriki or Bernard has found one of them (9=3*3, very trivial one and weak ?). 
The second type is super STB numbers that is not divisible by 3 but with S+M+N divisible by 3, do you know what it means? It means that we need all these six concatenations S.M.N, S.N.M, M.S.N, N.S.M, M.N.S, and N.M.S to be all semiprimes of the form 3c, yes we need six semiprimes of the form 3c, The reason is this:
S= M*N, we MUST avoid S+M or S+N to be divisible by 3, because if one of them (S+M or S+N) divisible by 3, then either S.M or S.N will have more than two prime factors ( remember that 9= 3*3 is type 1 super STB number). Dots here means concatenation. 
Okay folks, so just check those and you will find the first non trivial super STB number. :D
