Given the recurrence formula F (n) = (n-1) (n-2) [f (n-2) + F (n-3) + (n-3) * f (n-4)] (n > 4), find the general term formula f(n)=(n-1)(n-2)[f(n-2)+f(n-3)+(n-3)*f(n-4)] (n>4) f(1)=f(2)=2 f(3)=2 f(4)=6 f(1)=f(2)=0 There's a wrong number on it This f (n) has a multiple relation with / E when n approaches infinity Give several f (n) to facilitate the test results f(5)=24 f(6)=160 f(7)=1140 f(8)=8988 The recursion above is equivalent to the recursion below f[n]=(n-1)(f[n-1]+(n-2)*f[n-3])

Given the recurrence formula F (n) = (n-1) (n-2) [f (n-2) + F (n-3) + (n-3) * f (n-4)] (n > 4), find the general term formula f(n)=(n-1)(n-2)[f(n-2)+f(n-3)+(n-3)*f(n-4)] (n>4) f(1)=f(2)=2 f(3)=2 f(4)=6 f(1)=f(2)=0 There's a wrong number on it This f (n) has a multiple relation with / E when n approaches infinity Give several f (n) to facilitate the test results f(5)=24 f(6)=160 f(7)=1140 f(8)=8988 The recursion above is equivalent to the recursion below f[n]=(n-1)(f[n-1]+(n-2)*f[n-3])

Let g (n) = f (n) / (n-1)!, H (n) = g (n) / N = f (n) / N!
Then G (n) = g (n-2) + H (n-3) + H (n-4)
The sum of N can be obtained
g(n)=1+h(1)+h(2)+...+h(n-3)
therefore
g(n+1)-g(n)=h(n-2)
perhaps
(n+1)h(n+1)-nh(n)=h(n-2)
On power series
y(x)=sum h(n)x^n,
Where the sum starts from n = 1, of course, it can also be supplemented by H (0) = 0
From the above recurrence relation, it can be concluded that
(1-x)y'(x)=x^2(y+1)
The solution is y (x) = exp (- x (x + 2) / 2) / (1-x) - 1
So f (n) is the Nth derivative of Y (x) at x = 0
As for whether there is a more elementary general term, I don't know

The recurrence formula of sequence {f (n)} is: F (n + 1) f (n-1) = f (n) ^ 2 + 1, the first two terms are: F (1) = 1, f (2) = 2. Find the general term formula

FN + 1fn-1 = FN ^ 2 + 1fn + 2fn = FN + 1 ^ 2 + 1: FN + 2fn-fn + 1fn-1 = FN + 1 ^ 2-fn ^ 2: FN + 2fn + FN ^ 2 = FN + 1fn-1 + FN + 1 ^ 2, that is, FN (FN + 2 + FN) = FN + 1 (FN + 1 + fn-1), then (FN + 2 + FN) / FN + 1 = (FN + 1 + fn-1) / FN, that is, the sequence (FN + 2 + FN) / FN + 1 is a constant column (FN +

Is it necessary to verify n = 1 at the end to find the general term by superposition method
When using the superposition method to find the general term, for example, a (n) - A (n-1) = 2N-1 lists all the items. After the superposition, is it necessary to verify whether A1 conforms to the general term formula?
Because according to my experience, A1 is consistent, but after all, a (n-1) is used in the above superposition under the premise of n ≥ 2

Definitely. Who says A1 is consistent?
Your sequence is a little special, so A1 is consistent. Not all of them are consistent
You can find a (n) = a1 + n ^ 2-1
A1 is consistent because when n = 1, n ^ 2-1 = 0
If your formula is a (n) - A (n-1) = 3n-1, is A1 still true or does it need to be verified?
It's a coincidence that you're writing about special circumstances