Sequence $a * \sum_{k=0}^{n} \frac{x^k}{k!}F[n-k]$ For computing a spiral I found the sequence
$$ F[0]=1, \\ F[n+1] = a * \sum_{k=0}^{n} \frac{x^k}{k!}F[n-k].$$
Does anyone have an idea what a closed formula would look like or what I should google to find out what it is?
Edit:
What would already help a lot is to know when $F[n+1]$ becomes negative;
Also so far I got this: For $F^0[n+1] = F[n+1]$ and $F^i[n+1] = F^{i-1}[n+1]-F^{i-1}[n+1]$ as well as $\ \overline{n} := a \frac{x^n}{n!} $ I think it holds:
$$
F^{k+1}[n+1] = \sum_{j=0}^{n-k-1} \ \overline{j} \ F^{k+1} \ [n-j] + \sum_{j=0}^{k} \ \sum_{i=0}^{k-j} {{k-j}\choose{i}} (-1)^i \ \overline{n-j-i} \ F^j \ [j]
$$
But still I am not able to find a closed formula.
 A: Use the lag operator $\hat{L}$, which maps a sequence $\{F_n\}$ to $\{\hat{L}F_n\}$ given by
$$\hat{L}F_n=\left\{\begin{array}{ll}
F_{n-1}, & n\geq 1,\\
0, & n=0.
\end{array}\right.$$
The sequence $\{F_n\}$ gets shifted to the right by one step and the new first term is $0$. Then we have
$$F_{n+1}=a\sum_{k=0}^\infty\frac{x^k}{k!}\hat{L}^k F_n=a\exp(x\hat{L})\,F_n.$$
We can obtain the $m$-step formula
$$F_{n+m}=a^m\exp(mx\hat{L})\,F_n.$$
Then set $n=0$ to obtain
$$F_m=a^m\sum_{k=0}^\infty\frac{(mx)^k}{k!}\hat{L}^kF_0=a^m$$
because only the $k=0$ term survives.
A: So we are to solve
$$ \bbox[lightyellow] {  
\left\{ \matrix{
  F(0) = 1 \hfill \cr 
  F(n + 1) = a\sum\limits_{k = 0}^n {{{x^{\,k} } \over {k!}}F(n - k)}  \hfill \cr}  \right.
} \tag{0}$$
Let's call $H(z)$ the z-Tranform of $F(n)$
$$ \bbox[lightyellow] {  
\eqalign{
  & H(z) = \sum\limits_{n = 0}^\infty  {F(n)z^{\,n} }  = F(0) + \sum\limits_{n = 1}^\infty  {F(n)z^{\,n} }  =   \cr 
  &  = F(0) + z\sum\limits_{n = 0}^\infty  {F(n + 1)z^{\,n} }  \cr} 
}$$
and let's define
$$ \bbox[lightyellow] {  
E(z,x) = \sum\limits_{k = 0}^\infty  {{{x^{\,k} } \over {k!}}z^{\,k} }  = e^{\,x\,z} 
}$$
We also recall that  the Iverson bracket $[P]$ is defined to be:
$$ \bbox[lightyellow] {  
\left[ P \right] = \left\{ {\begin{array}{*{20}c}
   1 & {P = TRUE}  \\
   0 & {P = FALSE}  \\
 \end{array} } \right.
}$$
Now the sum that appears in (0) is the convolution of $f(n)=x^n/n!$ and $F(n)$
and the relevant z-Transform is therefore the product of the single ones
$$ \bbox[lightyellow] {  
\sum\limits_{n = 0}^\infty  {\left( {\sum\limits_{k = 0}^n {{{x^{\,k} } \over {k!}}F(n - k)} } \right)z^{\,n} }  = e^{\,x\,z} H(z)
}$$
That premised, if we take the z-Transform of both sides of the recurrence we get
$$ \bbox[lightyellow] {  
\eqalign{
  & F(n + 1) = a\sum\limits_{k = 0}^n {{{x^{\,k} } \over {k!}}F(n - k)}   \cr 
  & \quad \quad  \Downarrow   \cr 
  & \sum\limits_{n = 0}^\infty  {F(n + 1)z^{\,n} }  = a\sum\limits_{n = 0}^\infty  {\left( {\sum\limits_{k = 0}^n {{{x^{\,k} } \over {k!}}F(n - k)} } \right)z^{\,n} }  = a\,e^{\,x\,z} H(z)  \cr 
  & \quad \quad  \Downarrow   \cr 
  & H(z) = 1 + z\sum\limits_{n = 0}^\infty  {F(n + 1)z^{\,n} }  = a\sum\limits_{n = 0}^\infty  {\left( {\sum\limits_{k = 0}^n {{{x^{\,k} } \over {k!}}F(n - k)} } \right)z^{\,n} }  = 1 + z\,a\,e^{\,x\,z} H(z)  \cr 
  & \quad \quad  \Downarrow   \cr 
  & H(z) = {1 \over {1 - z\,a\,e^{\,x\,z} }} \cr} 
} \tag{1}$$
Thus
$$ \bbox[lightyellow] {  
\eqalign{
  & H(z) = {1 \over {1 - z\,a\,e^{\,x\,z} }} = \sum\limits_{j = 0}^\infty  {\left( {z\,a\,e^{\,x\,z} } \right)^{\,j} }  = \sum\limits_{j = 0}^\infty  {z^{\,j} \,a^{\,j} \,e^{\,j\;x\,z} }  =   \cr 
  &  = \sum\limits_{j = 0}^\infty  {\sum\limits_{k = 0}^\infty  {{{j^{\,k} x^{\,k} } \over {k!}}\,a^{\,j} z^{\,j + k} } \,}  = \sum\limits_{n = 0}^\infty  {\left( {\sum\limits_{k = 0}^n {{{\left( {n - k} \right)^{\,k} x^{\,k} } \over {k!}}\,a^{\,n - k} } } \right)z^{\,n} \,}  =   \cr 
  &  = \sum\limits_{n = 0}^\infty  {\left( {\sum\limits_{k = 0}^n {{{\left( {n - k} \right)^{\,k} \left( {x/a} \right)^{\,k} } \over {k!}}\,} } \right)\left( {a\,z} \right)^{\,n} \,}  \cr} 
} \tag{2}$$
that is:
$$ \bbox[lightyellow] {  
\eqalign{
  & F(x,a,n) = a^{\,n} \sum\limits_{k = 0}^n {{{\left( {n - k} \right)^{\,k} \left( {x/a} \right)^{\,k} } \over {k!}}\,}  =   \cr 
  &  = a^{\,n} \sum\limits_{k = 0}^{n - 1 + \left[ {0 = n} \right]} {{{\left( {n - k} \right)^{\,k} \left( {x/a} \right)^{\,k} } \over {k!}}\,}  \cr} 
} \tag{3}$$
For $0<n$ the above polynomial is of degree $n-1$, and a calculation 
of the roots (in $x/a$, and excluding the coefficient in $a^n$) for the first values of $n$ gives that all roots are real and negative.
