How are these prime numbers occuring here?

Question

Consider the series: $$T_k = \sum_{i=0}^k\dfrac{1}{i!(k-i)!}$$

where $k \in \mathbb{Z}$ and $k \geq 0$.

Thus this series yields: $$1, 2, 2, \frac{4}{3}, \frac{2}{3}, \frac{4}{15}, \frac{4}{45}, \frac{8}{315}, \frac{2}{315}, \frac{4}{2835}, \frac{4}{14175}, \frac{8}{155925}, \frac{4}{467775}, \frac{8}{6081075}, \frac{8}{42567525}, \frac{16}{638512875}, ...$$

Now consider the series of the denominators of the above series: $$1, 1, 1, 3, 3, 15, 45, 315, 315, 2835, 14175, 155925, 467775, 6081075, 42567525, 638512875, ...$$

Divide a term by its preceding term in the series: $$1, 1, 3, 1, 5, 3, 7, 1, 9, 5, 11, 3, 13, 7, 15, ...$$

Why does this happen?

Answer 1

Hint:

Insert powers of $2$

$$1\cdot2^0, 1\cdot2^1, 3\cdot2^0, 1\cdot2^2, 5\cdot2^0, 3\cdot2^1, 7\cdot2^0, 1\cdot2^3, 9\cdot2^0, 5\cdot2^1, 11\cdot2^0, 3\cdot2^2, 13\cdot2^0, 7\cdot2^1, 15\cdot2^0, ...$$ to get a familiar sequence:

$$1,2,3,4,5,6,7,8,9,10,11,12,13,14,15,\cdots$$

Answer 2

As @robjohn correctly pointed out, $T_n=\frac{2^n}{n!}$.

Then note that $n!\equiv0\pmod5\forall n\ge5$, and so can be written as $n!=5p\cdot2^q$ for some $p,q\in\Bbb Z^+$

Hence we have $$T_n=\frac{2^n}{5p\cdot2^q}=\frac{2^{n-q}}{5p}$$

If you can show that $n\ge q$, then the multiplication of odd numbers explains perfectly why the result over your division is odd.