If x₀=x₁=1, and for n≥1,

x_n+1=x_n²/(x_n-1+2x_n)

find a formula for x_n as a function of n.

Final answers:
f(n) = 1/n!!  where !! is the double factorial (odd numbers)
Also expressible as a Functional equation:  
f(n+1) = f(n)/(2n-1)

journey to the solution:
a,  b,  b^2/(a+2b), ...
Spreadsheet results
1
1
0.333333333333333
0.0666666666666667
0.00952380952380952
0.00105820105820106
9.62000962000962E-05

Try a formula for the difference :  
b-a;
b^2/(a+2b) - b = b^2/(a+2b) - b(a+2b)/(a+2b)
= (b^2 - ab - 2b^2)/(a+2b) = -(ab + b^2)/(a+2b)
= -b(a + b)/(a+2b)
Not much help.

Look at ratio from one term to the next:
n          n-th term     Ratio f(n-1)/f(n)
0    1    
1    1            1
2    0.333333333333333    3
3    0.0666666666666667    5
4    0.00952380952380952    7
5    0.00105820105820106    9
6    9.62000962000962E-05    11
7    7.4000074000074E-06    13
8    4.9333382666716E-07    15
9    2.90196368627741E-08    17
10    1.52734930856706E-09    19
11    7.27309194555742E-11    21
12    3.16221388937279E-12    23
13    1.26488555574912E-13    25
14    4.68476131758932E-15    27

The ratio of the n-th term to the (n+1)st term is: 2n+1
The ratio of the n-th term to the (n-1)st term is: 1/(2n-1)

Define eoF(n) = every other Factorial(n).
eoF(n) = (2n-1)*(2n-3)*(2n-5)* ... * 1
f(n) = 1/eoF(n)

It turns out, this is the definition of the double factorial (for odd numbers), so the formula becomes:
f(n) = 1/n!!
a short program confirms
----------------
def eoF(n):
    """ the every other factorial  """
    ans = 1
    for i in range(1,2*n + 1,2):
        ans = ans * i
    return ans

for i in range(10):
    print(i, 1/eoF(i))

Posted by Larry on 2023-07-24 10:04:12