Solving $U_{n+1}=(U_{n})^{2} (n+2)$

Question

I need help solving the recurrence relation:

$U_{n+1}=(U_{n})^{2} (n+2)$, with $U(1)=2$.

I've tried wolfram alpha, but something really horrible came up. The methods I've tried have just failed so I need some ideas please.

Answer 1

$U_{n+1} = (U_n)^2(n + 2)$

$U_{n+1} = (U_{n-1})^4(n + 1)^2(n + 2)$

$U_{n+1} = (U_{n-2})^8(n + 0)^4(n + 1)^2(n + 2)$

Consider the first factor of our partial expansion of $U_{n+1}$: $(U_{n-2})^8$. This factor grows at a rate of $\times 2$ for each substitution. We can replace it by $2^{n}$ because we know there are $n$ substitutions before we reach a base case for $2$.

Next, look at the generated factors each time we perform the substitution. The smallest factor we can generate is $2$ because there are $n$ substitutions before reaching the base case. Further, we see that the powers are in descending powers of $2$. So, what we have is that these factors form a product $\prod_{i=2}^{n} (i+2)^{(2^{(n - i)})}$

Therefore, $$U_n = 2^n\prod_{i=2}^{n} (i+2)^{\left(2^{(n - i)}\right)}$$

Answer 2

$$U_2=(U_1)^2 \times (2+1) = 2^2 \times 3$$ $$U_3=(U_2)^2 \times (3+1) = (2^2 \times 3)^2 \times 4 =2^{2^2} \times 3^{2^1} \times 4^{2^0}$$ So now we have found a final product - $$U_{n+1}= 2^{2^n} \times ..... \times n ^{2^2} \times (n+1)^{2^1} \times (n+2)^{2^0}$$

so You have to solve this:- The General Term $$ U_n=\prod_{i=0}^n(i+2)^{2^{(n-i)}} $$

Its general solution can not be found easily in terms of n but can be bounded by some big-O function , i hope so.