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it's my first answer here.

I do have an algorithm to calculate the factorial of a given number $n$ recursively, but without using any multiplication like the usual one, only sums are available.

I've ended up with an algorithm with a complexity function $T(n) = 1 + nT(n - 1)$.

Opening the recurrence relation ends up in:

$T(n) = 1 + n T(n - 1)$

$T(n) = 1 + n[1 + (n - 1)T(n - 2)]$

$T(n) = 1 + n + n^2 T(n - 2) - n T(n - 2)$

$T(n) = 1 + n + n^2[1 + (n - 2) * T(n - 3)] - n T(n - 2)$

$T(n) = 1 + n + n^2 + n^3T(n - 3) - n T(n - 2) - n^2T(n - 3)$

Because of the subtractions, I'm not able to reach good upper-bound for $T(n)$. Any tips ?

Before asking the question, I've managed to state that $T(n) \in \mathcal{O}(n^n)$. Which is 'loose' upper bound, but I'm neither able to figure out a tighter upper bound.

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Solving your recurrence, assuming the base case $T(0) = 1$, we get $$ \begin{align*} T(n) &= 1 + nT(n-1) \\ &= 1 + n + n(n-1)T(n-2) \\ &= 1 + n + n(n-1) + n(n-1)(n-2)T(n-3) \\ &= \cdots \\ &= 1 + n + n(n-1) + \cdots + n(n-1)\cdots2 + n! \\ &= n! \left(\frac{1}{0!} + \frac{1}{1!} + \frac{1}{2!} + \frac{1}{3!} + \cdots + \frac{1}{n!}\right) \\ &= n! \left(e - O\left(\frac{1}{(n+1)!}\right)\right) \\ &= e \cdot n! - O\left(\frac{1}{n}\right). \end{align*} $$ In particular, $T(n) = \Theta(n!) = \Theta(\sqrt{n}(n/e)^n)$.

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  • $\begingroup$ thank you, avoid unfolding the multiplications lead to a way simpler solution $\endgroup$ – aajjbb Apr 15 '17 at 22:25

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