How many attempts are needed?

Let $f(x)= \tan x$ , then the coefficients of its Maclaurin series are $\dfrac {f^{(n)}(0)}{n!}$ , where $n = 0, 1,2,...$ non-negative integers. The Maclaurin series up to $n=9$ is as follows (Eqn. 29) (note that $f^{(n)}(0) = 0$ for even $n$ ):

$\begin{aligned} \tan x & = x + \frac 13 x^3 + \frac 2{15}x^5 + \frac {17}{315}x^7 + \frac {62}{2835} + ... \\ & = \frac {f'(0)}{1!}x + \frac {f'''(0)}{3!}x^3 + \frac {f^{(5)}(0)}{5!}x^5 + \frac {f^{(7)}(0)}{7!}x^7 + \frac {f^{(9)}(0)}{9!}x^9 + ... \end{aligned}$

Therefore, we have:

\(\begin{array} {} f'(0) = 1 \cdot 1! = 1 & = \color{red}{2^0} \\ f'''(0) = \frac 13 \cdot 3! = 2 & = \color{red}{2^1} \\ f^{(5)}(0) = \frac 2{15} \cdot 5! = 16 & = \color{red}{2^4} \\ f^{(7)}(0) = \frac {17}{315} \cdot 7! = 272 & = \color{blue}{\text{Not a power of 3 } - \text{ accepted.}} \end{array} \)

$\implies n = \boxed{7}$ .

$f^{(n)}(0)$ for odd $n$ is also given by (Eqn. 50) :

$\frac {d^{2n+1}}{dx^{2n+1}} \tan x \ \bigg|_{x=0} = \frac {(-1)^n2^{2n+2}(2^{2n+2}-1)B_{2n+2}}{2n+2}$

where $B_n$ is a Bernoulli number .

The first few $f^{(n)}(0)$ are given below:

$\begin{array} {lrll} n = 0, & f'(0) & = \dfrac {(1)(4)(3)\left(\frac 16 \right)}2 & = 1 \\ n = 1, & f'''(0) & = \dfrac {(-1)(16)(15)\left(-\frac 1{30} \right)}4 & = 2 \\ n = 2, & f^{(5)}(0) & = \dfrac {(1)(64)(63)\left(\frac 1{42} \right)}6 & = 16 \\ n = 3, & f^{(7)}(0) & = \dfrac {(-1)(256)(255)\left(-\frac 1{30} \right)}8 & = \color{#3D99F6}{272} \\ n = 4, & f^{(9)}(0) & = \dfrac {(1)(1024)(1023)\left(\frac 5{66} \right)}{10} & = 7936 \end{array}$