Is this the expansion for any known function?

Question

The expansion is $$\sum_{n=1}^{\infty}\frac{x^n}{n!(n-1)!}\left[c(1+c)\dots((n-1)^2+c)\right]$$

So the first 3 terms are $cx$, $\dfrac{c(1+c)}{2}x^2$, $\dfrac{c(1+c)(4+c)}{12}x^3$.

Answer 1

If I am not mistaken, your problem is $$\sum_{n=1}^{\infty}\frac{x^n}{n!(n-1)!}\,\,\prod_{k=0}^{n-1}(c+k^2)$$

Using Pochhammer symbols $$\prod_{k=0}^{n-1}(c+k^2)=c \left(1-i \sqrt{c}\right)_{n-1} \left(1+i \sqrt{c}\right)_{n-1}$$

$$\prod_{k=0}^{n-1}(c+k^2)=\frac{\sqrt{c}\, \sinh \left(\pi \sqrt{c}\right) }{\pi }\,\Gamma \left(n-i \sqrt{c}\right)\, \Gamma \left(n+i \sqrt{c}\right)$$

All the ingredients are now present to define the Gaussian hypergeometric function $$\sum_{n=1}^{\infty}\frac{x^n}{n!(n-1)!}\,\,\prod_{k=0}^{n-1}(c+k^2)=c\, x \,\, _2F_1\left(1-i \sqrt{c},1+i \sqrt{c};2;x\right)$$

If $a_n$ represents the summand, then $$\frac{a_{n+1}}{a_n}=\frac{ \left(n^2+c\right)}{n (n+1)}x$$ which shows the radius of convergence; so, the function is defined for any $x<1$.