Finding general terms of a series

Questions
,
1

I want to find general terms of this series I already used command ‘‘FindSequenceFunction’’. Help me

FindSequenceFunction[{E^-m, E^-m (2 - m), 
  FullSimplify[ (1 - 2 m + m^2/2)], FullSimplify[(-m + m^2 - m^3/6)], 
  FullSimplify[ (m^2/2 - m^3/6 + m^4/24)], 
  FullSimplify[ (-(m^3/6) + m^4/24 - m^5/120)]}, m]
2

I will venture to guess that the middle terms of the first and second elements are not quite what was intended. This would make the task more difficult.

3

Just to be clear, what is sought is to find the general term a_n of the sequence

\eqalign{ a_0&=e^{-m}, \cr a_1&=e^{-m} (2-m), \cr a_2&=\frac{1}{2} (m-4) m+1, \cr a_3&=-\frac{1}{6} m ((m-6) m+6), \cr a_4&=\frac{1}{24} m^2 ((m-4) m+12), \cr a_5&=-\frac{1}{120} m^3 ((m-5) m+20), \cr }

Right?

There’s not much of a pattern there. A factor of (-1)^n/n! Another factor of m^{n-2} for n\ge2, which mysteriously becomes e^{-m} for n=0,1, perhaps. Finally, again for n\ge2, a factor of the form (m-b_n)m+n(n-1), where b_n is a seemingly random sequence, b_2=4, b_3=6, b_4=4, b_5=5.

Except for the factor e^{-m}, the term a_1 could be seen to fit the pattern of a_n for n\ge2. Then b_n=2n for n=1,2,3 and b_n=n for n=4,5 — who knows what happens for n>5?

Here’s the sequence based on the pattern for a_2, a_3, if two terms constitute a pattern:

Table[
   Subscript[a, n] == (-1)^n/n! m^(n - 2) ((m - 2 n) m + n (n - 1)),
   {n, 0, 5}] // Column // TeXForm
\eqalign{ a_0&=1, \cr a_1&=2-m, \cr a_2&=\frac{1}{2} ((m-4) m+2), \cr a_3&=-\frac{1}{6} m ((m-6) m+6), \cr a_4&=\frac{1}{24} m^2 ((m-8) m+12), \cr a_5&=-\frac{1}{120} m^3 ((m-10) m+20), \cr }

In fact, we see that both a_0 and a_1 in the OP’s sequence would fit this formula if the mysterious factor e^{-m} were a mistake and could be dropped from each term.

Where this sequence differs from the OP’s suggests that the OP’s sequence is much more complicated, that is, assuming it contains no errors. In my experience, I have to give FindSequenceFunction[] a couple dozen or more terms for it to discover a formula for a complicated sequence.

4

Ignoring the exponential terms, the closest I arrived is this:

Table[(-m)^(n - 1)/(n + 1)!
  (n + n^2 + (m - a[n]) m),
 {n, -1, 4}]

where a[n] should come from

FindSequenceFunction[{0, 2, 4, 6, 4, 5}, n]

if only it gave an answer. I was disappointed that this too gave no answer:

FindSequenceFunction[{2, 6, 24, 120}, n]