On Radical Mersenne Prime Containing Series, and why the number 5 doesn't appear here? As part of a project on operators I was studying the function: 
$$ L =\left( \sqrt{x^3} - x \right) + \left( \sqrt{\sqrt{x^3}} - \sqrt{x} \right) + \left( \sqrt{\sqrt{\sqrt{x^3}}} - \sqrt{\sqrt{x}}\right) + ... $$ 
It can be written explicitly as $$L(x) = \sum_{k=1}^{\infty} \left( x^{\frac{3}{2^n}} - x^{\frac{1}{2^{n-1}}  }\right) $$
In order to get a better understanding of the function I decided to taylor expand it around the point $x=1$. This leads to the series:
$$ L(x) = (x-1) - \frac{2}{7}(x-1)^2 + \frac{8}{21}(x-1)^3 - \frac{80}{93}(x-1)^4 + \frac{5360}{1953} (x-1)^5 -\frac{133760}{11811}(x-1)^6... $$
The coefficients in the series were bizarre looking and the numerators didn't appear to be part of any known sequences in OEIS so I decided to study the denominators in depth, specifically I prime factorized each denominator (and numerator too for those curious).
$$ L(x) = (x-1) - \frac{2}{7}(x-1)^2 + \frac{2^3}{3\times 7}(x-1)^3 - \frac{2^5 \times 5}{3\times 31}(x-1)^4 + \frac{2^5 \times 5 \times 67}{3^2 \times 7 \times 31} (x-1)^5 -\frac{2^7 × 5 × 11 × 19}{3 \times 31 \times 127}(x-1)^6... $$
This was continued for about 50 terms surprisingly the terms had a LOT of common structure. The mersenne primes appear in all the terms (and after skipping ahead to the 100th term, it appears the first 10 mersenne primes definitely are present although the 9th mersenne prime shows as early as term 60).
The Question:
I wanted to start with something very simple before going after crazier conjectures involving this sequence of denominators: how can I quickly prove that the number 5 will never be a divisor of any of the denominators?
An approach:
If we let the first coefficent be 1. Then the generally we have
$$ a(n) = \sum_{i=1}^{\infty} \left[ \prod_{k=0}^{n-1} \left[ \frac{3}{2^i} - k \right] \right] - \sum_{i=1}^{\infty} \left[ \prod_{k=0}^{n-1} \left[ \frac{2}{2^i} - k \right] \right] $$
Example:
$$ a(2) = \frac{3}{2} \frac{1}{2}  + \frac{3}{4}(-\frac{1}{4}) + \frac{3}{8}(-\frac{5}{8}) + \frac{3}{16} (-\frac{13}{16}) ...  - \frac{1}{2}(-\frac{1}{2}) - \frac{1}{4} (-\frac{3}{4}) - \frac{1}{8}(-\frac{7}{8}) ... $$
[Intentionally left unsimplified so the pattern is easier to deduce] 
Somehow we need to get a closed from to these product series and then show that after simplification the denominator term will never contain a 5. Getting a closed form feels out of reach for me at the moment.

$$\prod_{k=0}^{n-1} \left(\frac{3}{2^i}-k\right) = \sum_{j=1}^n \frac{3^j}{2^{ij}} \left[{n \atop j}\right]$$
where $\left[{n \atop j}\right]$ is a Stirling number of the first kind.
Then
$$\sum_{i=1}^\infty \prod_{k=0}^{n-1} \left(\frac{3}{2^i}-k\right) = \sum_{j=1}^n \frac{3^j}{2^j-1} \left[{n \atop j}\right] $$
Similarly 
$$ \prod_{k=0}^{n-1} \left(\frac{2}{2^i}-k\right) = \sum_{j=1}^n \frac{2^j}{2^j-1} \left[{n \atop j}\right] $$
so your 
$$a(n) = \sum_{j=1}^n \frac{3^j-2^j}{2^j-1} \left[{n \atop j}\right] $$
Now the $5$-adic order of $2^j - 1$ is
$i > 0$ iff $j$ is divisible by $4 \cdot 5^{i-1}$ but not by $4 \cdot 5^i$, and the same is true for $3^j-2^j$. Therefore the 
$5$-adic order of $a(n)$ is nonnegative, i.e. $5$ does not appear in the denominator of $a(n)$. 
