Good asymptotic for this recursion? Consider some initial integer values and let the integer sequence continue like 
$$\begin{align}f(n) &= f(n-1) \\
&+ n(n+1)(n+2)\dots(n+5) f(n-2) \\
&- \frac{ n(n+1)(n+2)\dots(n+5)}{2} f(n-3) \\
&- \frac{ n(n+1)(n+2)\dots(n+5)}{3} f(n-4) \\
&- \frac{ n(n+1)(n+2)\dots (n+5)}{6} f(n-5) \\
&+ K.
\end{align}$$
For some positive integer $K$.
Consider the cases where eventually the sequence remains strictly increasing.
What are possible closed forms if any ? 
What are the best asymptotics !?
I assume the best(?) asymptotics are something like 
$ A ( n ! )^6 $ or $ A ( n!)^5 $ ? 
( $ A $ some constant )
Big those estimates seem poor and even kinda ignore the variable $K$.
So i doubt my own guesses.
Not sure how to handle it.
Reminder me of the factorial of hypergeo.
Clearly this is just part of a bigger question ; the similar recursion equations
$$ \sum_i p_i(n) f(n-i) = 0 $$
And perhaps it is useful to think of the related equations 
$$ \sum_i p_i(x) f(x/i) = 0 $$
$$ \sum_i p_i(x) f(x/2^i) = 0 $$
And look at their Taylor series solutions ?
Those solutions might be found and understood by systems of equations ...
 A: Under the assumptions that $f$ is positive and strictly increasing we have:
$$
\begin{align}f(n) &= f(n-1) \\
&+ n(n+1)(n+2)\dots(n+5) f(n-2) \\
&- \frac{ n(n+1)(n+2)\dots(n+5)}{2} f(n-3) \\
&- \frac{ n(n+1)(n+2)\dots(n+5)}{3} f(n-4) \\
&- \frac{ n(n+1)(n+2)\dots (n+5)}{6} f(n-5) \\
&+ K.
\\&\Leftrightarrow\\
f(n)&=f(n-2) \\
&+ (n+1)(n+2)\dots(n+6) f(n-3) \\
&- \frac{ (n+1)(n+2)\dots(n+6)}{2} f(n-4) \\
&- \frac{ (n+1)(n+2)\dots(n+6)}{3} f(n-5) \\
&- \frac{ (n+1)(n+2)\dots (n+6)}{6} f(n-6) \\
&+ K\\
&+ n(n+1)(n+2)\dots(n+5) f(n-2) \\
&- \frac{ n(n+1)(n+2)\dots(n+5)}{2} f(n-3) \\
&- \frac{ n(n+1)(n+2)\dots(n+5)}{3} f(n-4) \\
&- \frac{ n(n+1)(n+2)\dots (n+5)}{6} f(n-5) \\
&+ K\\
&\le f(n-2) \\
&+ (n+1)(n+2)\dots(n+6) f(n-3) \\
&+ n(n+1)(n+2)\dots(n+5) f(n-2) \\
&+ 2K\\
&=  \\
&+ (n+1)(n+2)\dots(n+6) f(n-3) \\
&+ (1+n(n+1)(n+2)\dots(n+5)) f(n-2) \\
&+ 2K
\end{align}$$
Further we have then
$$
\begin{align}
f(n) &\le\\
&+ (n+1)(n+2)\dots(n+6) f(n-3) \\
&+ (1+n(n+1)(n+2)\dots(n+5)) f(n-2) \\
&+ 2K\\
&\le \\
&+ 2(1+n(n+1)(n+2)\dots(n+5)) f(n-2) \\
&+ 2K
\end{align}
$$
For all $n\ge N$ for some $N$ we further have:
$$
f(n)\le 
 c\cdot 2(1+n(n+1)(n+2)\dots(n+5)) f(n-2) 
$$
You then can solve this recurrence equation, e.g. per WolframAlpha and obtain some bounds using it (you can get an even better than I initially guessed).
