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.
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.