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Is there any way to get Mathematica to find the coefficients of the product of sums? As an example (the problem I am trying to solve): Coefficients for a Taylor expansion of $e^{z^2}$ centered around $z=1$. We can write this as: $e^{z^2} = e^{(z-1)^2+2(z-1)+1} = \left(\sum_{n=0}^\infty \frac{(z-1)^{2n}}{n!}\right)\left(\sum_{n=0}^\infty \frac{2^n(z-1)^{n}}{n!}\right)e$. I would like to find the coefficients of the above sum but written as just one sum over n.

Some analysis has led to find that we can write the above as: $\sum_n^\infty \sum_{0 \leq 2m \leq n} \frac{e2^{n-2m}}{m!(n-2m)!}(z-1)^n$, which is of the form I wanted. Is there any way for Mathematica to have found this?

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Here are two ways to find the $n^{th}$ term of the series expansion:

c1[n_] = SeriesCoefficient[Exp[z^2],{z,1,n}, Assumptions->n>=0]
DifferenceRoot[Function[{\[FormalY], \[FormalN]}, {-2 \[FormalY][\[FormalN]] - 
  2 \[FormalY][1 + \[FormalN]] + (2 + \[FormalN]) \[FormalY][
    2 + \[FormalN]] == 0, \[FormalY][0] == E, \[FormalY][1] == 2 E}]][n]

and:

c2[n_] = D[Exp[z^2], {z, n}]/n! /. z -> 1

(2^n Sqrt[π] HypergeometricPFQRegularized[{1/2, 1}, {1 + 1/2 (-1 - n), 1 - n/2}, 1])/n!

Check:

s = Series[Exp[z^2], {z, 1, 5}];
s //TeXForm

$e+2 e (z-1)+3 e (z-1)^2+\frac{10}{3} e (z-1)^3+\frac{19}{6} e (z-1)^4+\frac{13}{5} e (z-1)^5+O\left((z-1)^6\right)$

as compared to:

Table[c1[n], {n, 0, 5}] //TeXForm

$\left\{e,2 e,3 e,\frac{10 e}{3},\frac{19 e}{6},\frac{13 e}{5}\right\}$

and:

Table[c2[n], {n, 0, 5}] //TeXForm

$\left\{e,2 e,3 e,\frac{10 e}{3},\frac{19 e}{6},\frac{13 e}{5}\right\}$

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