Let $a_1,$ $a_2,$ $a_3,$ $\dots$ be an arithmetic sequence, and let $b_1,$ $b_2,$ $b_3,$ $\dots$ be a geometric sequence.  The sequence $c_1,$ $c_2,$ $c_3,$ $\dots$ has $c_n = a_n + b_n$ for each positive integer $n.$  If $c_1 = 1,$ $c_2 = 4,$ $c_3 = 15,$ and $c_4 = 2,$ compute $c_5.$
Let the arithmetic sequence be $a_n = a + (n - 1)d,$ and let the geometric sequence be $b_n = br^{n-1}.$  Then
\begin{align*}
a + b &= 1, \\
a + d + br &= 4, \\
a + 2d + br^2 &= 15, \\
a + 3d + br^3 &= 2.
\end{align*}Subtracting pairs of equations, we get
\begin{align*}
d + br - b &= 3, \\
d + br^2 - br &= 11, \\
d + br^3 - br^2 &= -13.
\end{align*}Again subtracting pairs of equations, we get
\begin{align*}
br^2 - 2br + b &= 8, \\
br^3 - 2br^2 + br &= -24.
\end{align*}We can write these as
\begin{align*}
b(r - 1)^2 &= 8, \\
br(r - 1)^2 &= -24.
\end{align*}Dividing these equations, we get $r = -3.$  Then $16b = 8,$ so $b = \frac{1}{2}.$  Then
\begin{align*}
a + \frac{1}{2} &= 1, \\
a + d - \frac{3}{2} &= 4.
\end{align*}Solving for $a$ and $d,$ we find $a = \frac{1}{2}$ and $d = 5.$

Hence,
\begin{align*}
c_5 &= a_5 + b_5 \\
&= a + 4d + br^4 \\
&= \frac{1}{2} + 4 \cdot 5 + \frac{1}{2} \cdot (-3)^4 \\
&= \boxed{61}.
\end{align*}