easy

what are the next two numbers? 0,1,4,5,8,?,?

without mistakes:

$\begin{array}{r|rcll} n &\\ \hline 1& a_1 &=& 0\\ & && & +1 \\ 2& a_2 &=& 1\\ & && & +3 \\ 3& a_3 &=& 4 \\ & && & +1 \\ 4& a_4 &=& 5 \\ & && & +3 \\ 5& a_5 &=& 8 \\ & && & +1 \\ 6& a_6 &=& 9 \\ & && & +3 \\ 7& a_7 &=& 12 \\ & && & +1 \\ 8& a_8 &=& 13 \\ & && & +3 \\ 9& a_9 &=& 16 \\ & && & +1 \\ 10& a_{10} &=& 17 \\ \cdots \end{array}$

$\begin{array}{rcll} a_{n}= \left\{ \begin{array}{rl} 2\cdot n-2 \qquad & (\text{n odd})\\ 2\cdot n-3 \qquad & (\text{n even}) \end{array} \right. \end{array}$

$\color{red} \boxed{~ \color{black} \begin{array}{rcll} a_n &=& (2\cdot n - 2) -\frac12 [~ 1+(-1)^n ~] \end{array} ~}$

Thanks for that answer, heureka.........

could you explain how you derived your final formula?

...so far:

$\begin{array}{rcll} a_{n}= \left\{ \begin{array}{ll} 2\cdot n-2 = (2\cdot n - 2) + 0\qquad & (\text{n odd})\\ 2\cdot n-3 = (2\cdot n - 2) -1 \qquad & (\text{n even}) \end{array} \right. \end{array}$

...composite:

$\begin{array}{rcll} a_n = (2\cdot n - 2) -1 \cdot s \end{array}$

s is a switch. We need a function for a switch.

if n is even, s is on or equal 1,

and if n is odd, s is off or equal 0.

As the next we see we can switch with $(-1)^n$

This function does switch beween -1 and 1 if n is odd or if n is even.

$\begin{array}{lcr} (-1)^{\text{odd number}} &=& -1 \\ (-1)^{\text{even number}} &=& 1 \end{array}$

We must modify this "switch" function to get our function.

But it is easy to do this.

First step we add 1.

$\begin{array}{lcr} 1 + (-1)^{\text{odd number}} &=& -1+1 = 0\\ 1 + (-1)^{\text{even number}}&=& 1+1 = 2\\ \end{array}$

Second step, the number 2 must be number 1, so we divide by 2
$\begin{array}{lcr} \dfrac{1 + (-1)^{\text{odd number}} } {2} &=& \frac{0}{2} = 0\\\\ \dfrac{1 + (-1)^{\text{even number}} } {2}&=& \frac{2}{2} = 1\\ \end{array}$

So our switch s is ready:

$\begin{array}{lrcr} & s &=& \dfrac{1 + (-1)^n } {2} \\ & s &=& \frac12 [~ 1+(-1)^n ~]\\ n & \text{ is odd} & s = 0 \\ n & \text{ is even} & s = 1 \end{array}$