I like to think of helices on cylinders as images of lines under planar curling, so here's an approach incorporating that idea.
Imagine the circle rolling up a line drawn in the plane, with $P$ the point of tangency along the line and $Q$ a distinguished point on the circumference of the circle. (So, $Q$ traces out a cycloid in the plane.) Now, imagine the plane is made of thin paper, but the circle is made of stiff cardboard. If we curl the plane into a cylinder but the circle remains flat, the plane of the circle will be tangent to the cylinder along the (vertical) line passing through the point $P$. The point $P$ will follow a helix, and the point $Q$ --which lies in the plane of the circle-- traces out the helical cycloid in space. While the path that $P$ (and $Q$) takes is decidedly different after curling than before, the displacement vector between $P$ and $Q$ in the plane of the circle is the same throughout.
Before curling, if the circle of radius $r$ rolls along a horizontal track in the $uv$-plane, then we have $P=(r t, 0)$, and the point $Q$ traces out the standard cycloid:
$$\begin{align}
u &= r ( t - \sin t ) \\
v &= r ( 1 - \cos t )
\end{align}$$
The displacement vector from $P$ to $Q$ at time $t$ is given by
$$d := PQ = [m, n] = r[-\sin t, 1 - \cos t ] = - 2 r \sin\frac{t}{2} \; [\cos\frac{t}{2},-\sin\frac{t}{2}]$$
Rotating the plane through an angle, say, $\theta := \rm{atan\frac{c}{a}}$, and writing $b$ for $\sqrt{a^2+c^2}$, the circle's point of tangency along the tilted track is given by ...
$$P_2 = (r t \cos\theta,r t \sin\theta) = \left( \frac{r t a}{b}, \frac{r t c}{b}\right)$$
... and the displacement vector, $d_2$, to the point $Q_2$ tracing out the tilted cycloid is given by ...
$$\begin{align}
d_2 :&= [m \cos\theta - n \sin \theta, m \sin\theta + n \cos \theta ] \\
&= \frac{1}{b} [a m - c n, c m + a n] \\
&= -\frac{2r}{b}\sin\frac{t}{2}\left[\cos\left(\theta-\frac{t}{2}\right), \sin\left(\theta - \frac{t}{2} \right) \right] \\
\end{align}$$
Now the fun part: Curl the $(u,v)$-plane around a cylinder of radius $a$, such that the $uv$-origin aligns with the $xyz$-point $(a,0,0)$ and the $v$-axis runs parallel to the $z$ axis. The curled, tilted track will coincide with a helix. The horizontal distance travelled by $P_2$ in the plane becomes the length of horizontal circular arc travelled by $P_3$ around the cylinder; upon dividing by the radius, $a$, of the cylinder, this length becomes the angular "distance" --$s := \frac{rt}{b}$-- travelled by $P_3$; the vertical distances match. Therefore:
$$P_3 = \left(a\cos s, a\sin s, \frac{rtc}{b} \right)=(a\cos s,a\sin s,c s)$$
As for the displacement vector: The $u$ direction of the tangent plane coincides with the horizontal vector tangent to the cylinder at $P_3$; the $v$ direction coincides with the $z$ direction. Thus, the transformation from $uv$-coordinates to $xyz$-coordinates is given by
$$[1,0]\to[-\sin s, \cos s, 0]\hspace{0.5in}[0,1]\to[0,0,1]$$
The image, $d_3$, of the displacement vector $d_2$, then, is
$$d_3 = -\frac{2r}{b}\sin\frac{t}{2}\left[-\sin s \cos\left(\theta-\frac{t}{2}\right), \cos s \cos\left(\theta-\frac{t}{2}\right), \sin\left(\theta - \frac{t}{2} \right) \right] $$
and the path of $Q_3$, which traces the helical cycloid, is given by
$$
Q_3 = P_3 + d_3 = \left\{
\begin{align}
x &= a \cos s &+ \frac{2 r}{b} \sin s \sin\frac{bs}{2r} \cos\left(\theta-\frac{bs}{2r}\right)\\
y &= a \sin s &- \frac{2 r}{b} \cos s \sin\frac{bs}{2r} \cos\left(\theta-\frac{bs}{2r}\right)\\
z &= c s &- \frac{2 r}{b} \sin\frac{b s}{2r}\sin\left(\theta-\frac{bs}{2r}\right)
\end{align}
\right.
$$
Here's a picture with $a=c=1$ and $r=1/2$:
Note: The above does not simply curl the planar cycloid around the cylinder. Since
$$x^2 + y^2 = a^2 + \frac{4r^2}{b^2} \sin^2\frac{bs}{2r} \cos^2\left(\theta-\frac{bs}{2r}\right) \ge a^2$$
we see that most of the helical cycloid lies outside the surface of the cylinder.
Best Answer
Piecewise vs Parametric
As discussed in the comments this can be done piecewise or parametrically. The difference is minor. Any pieces of a piecewise function can be joined into a single function by the following:
$$f(t)=\begin{cases} f_1(t) & a_0<t\le a_1 \\ f_2(t) & a_1 < t \le a_2 \\ f_3(t) & a_2<t\le a_3 \\ & etc\end{cases}$$
$$f(t)=p[a_0,a_1,t]f_1(t)+p[a_1,a_2,t]f_2(t)+p[a_2,a_3,t]f_3(t)+\cdots$$
where $p[n,m,x]=\frac{1}{2}\left(1-\frac{|x-n|}{x-n}\cdot\frac{|x-m|}{x-m}\right)$. The function $p$ returns $1$ when $n<x<m$ and $0$ otherwise.
Piecewise Solution
$a$ is the horizontal "radius" of the pipe. I.e. the horizontal distance between the two sides is $2a$.
$b$ is the vertical "radius" of the pipe. I.e. the vertical distance between the two sides is $2b$.
$r$ is the radius of the corner pieces.
$$f(t)=\begin{cases} \{a,-(b-r)(2t-1)\} & 0\le t\le 1 \\ \left\{a-r+r \cos \left(\frac{1}{2} \pi (t-1)\right),-b+r-r \sin \left(\frac{1}{2} \pi (t-1)\right)\right\} & 1<t\le 2 \\ \{-(a-r) (2t-5),-b\} & 2<t\le 3 \\ \left\{-a+r-r \sin \left(\frac{1}{2} \pi (t-3)\right),-b+r-r \cos \left(\frac{1}{2} \pi (t-3)\right)\right\} & 3<t\le 4 \\ \{-a,(b-r) (2t-9)\} & 4<t\le 5 \\ \left\{-a+r-r \cos \left(\frac{1}{2} \pi (t-5)\right),b-r+r \sin \left(\frac{1}{2} \pi (t-5)\right)\right\} & 5<t\le 6 \\ \{(a-r) (2t-13),b\} & 6<t\le 7 \\ \left\{a-r+r \sin \left(\frac{1}{2} \pi (t-7)\right),b-r+r \cos \left(\frac{1}{2} \pi (t-7)\right)\right\} & 7<t\le 8 \end{cases}$$
This varies the parameter $t$ over the range $[0,8]$. If a range of $[0,2\pi]$ is required then just apply a scaling factor: $g(t)=f\left(\frac{4t}{\pi}\right)$
Image $(a=1,b=1,r=0.5)$
Image $(a=8,b=5,r=2)$