As per suggestion, this is being posted as a separate question
Given the prime-factor concatenation of n in base-m (I'll use the notation $ PFC_m(n) $), if n is a composite number ($c_k$), does $\exists n \in c_k$ s.t $ \frac{PFC_m(n)}{n} =1 $?
To clarify the notation $\frac{PFC_m(n)}{n}$ with some examples:
- $\frac{PFC_{10}(4)}{4}=\frac{22}{4} = 5.5 $
- $\frac{PFC_{10}(6)}{6}=\frac{23}{6} = 3.8\overline{3} $
- $\frac{PFC_{2}(100)}{100}=\frac{1010}{100} = 10.1$, where 10.1 (base-2)$\rightarrow$ 2.5 (base-10)
Clearly, if n is prime the above statement is trivially true, but in the composite case ($n \in c_k$) I don't believe this is ever true, though we can get arbitrarily close to 1 for certain arbitrarily large values of n (conjecture).
Working on a proof but am still a bit of an amateur at this stuff.
Here's a few interesting examples that hint at those values of n where $ \frac{PFC_m(n)}{n} \approx 1$
- $\frac{PFC_{10}(9409)}{9409} = \frac{9797}{9409} = 1.041237113402062$, where $9409 = 97\cdot97$ (97 is prime)
- $\frac{PFC_{10}(9998200081)}{9998200081} = \frac{99991 99991}{9998200081}=1.00010000900081$ where $9998200081 = 99991\cdot99991$ (99991 is prime)
Some examples in base-2.
Note: right-hand side of $\frac{PFC_2(n)}{n}$ is reported in base-10 for simplicity
- $\frac{PFC_{2}(1111000001)}{1111000001} = \frac{1111111111}{1111000001}=1.064516129032258$ where 1111000001 (base-2) $\rightarrow 961 = 31\cdot31$(base-10) and 31 (base-10)$\rightarrow$ 11111 (base-2) (31 is prime)
- $\frac{PFC_{2}(11111100000001)}{11111100000001} =\frac{11111111111111}{11111100000001}=1.015748031496063$ where 11111100000001 (base-2) $\rightarrow 16129 = 127\cdot127$(base-10) and 127 (base-10) $\rightarrow$ 1111111 (base-2) (127 is prime)
See the pattern and the reason for the conjecture that we can get arbitrarily close to 1 for certain composite numbers! I've written a program to check for counterexamples and tested it for n = [2,10000] and base-m = [2,16] and haven't found any.
If someone wants to prove it or is inspired to share a related result I'd love to see it.
I think it's impossible.
Let $n = p_1 p_2 \cdots p_k$ for $k \ge 2$, and each $p_i$ is $l_i$ digits long in base $m$ (that is, $l_i = \lfloor\log_m p_i\rfloor + 1$)
We are trying to solve $$n = p_1 p_2 \cdots p_k = p_1 + m^{l_1}p_2 + m^{l_1 + l_2}p_3 + \cdots + m^{l_1 + \cdots + l_{k-1}}p_k$$
where the RHS represents the prime-factor concatenation.
To demonstrate this concretely, let's say $n = 91 \times 13 \times 7$. The order of the primes is arbitrary. The concatenated number in base 10 is $91137 = 91 \times 10^3 + 13 \times 10^1 + 7$. How did we obtain this? Well $7$ is $1$ digit in base 10. Thus we multiply $13$ by $10$ to shift it over one place. $137$ is $3$ digits in base 10, so we multiply $91$ by $10^3$ to shift it over $3$ places. Work out some examples by hand to convince yourself this is true.
Move last term to LHS and factor $p_k$:
$$p_k (p_1 \cdots p_{k-1} - m^{l_1 + \cdots + l_{k-1}}) = p_1 + m^{l_1}p_2 + m^{l_1 + l_2}p_3 + \cdots + m^{l_1 + \cdots + l_{k-2}}p_{k-1}$$
However since $p_1 < m^{l_1}, p_2 < m^{l_2}, \dots , p_{k-1} < m^{l_{k-1}}$, when we multiply the inequalities we find $p_1 \cdots p_{k-1} - m^{l_1 + \cdots + l_{k-1}}$ is negative, a contradiction.