how to find the nth number in the sequence? consider the sequence of numbers below,  


2 5 10 18 31 52 . . .  


the sequence goes on like this.
My Question is,
How to find the nth term in the sequence?
thanks.
 A: One could try to stare at it for long enough and notice, that if one "demands" the first and the second term to be $2$ and $5$ respectively then the third term is $(2 + 5 + 3) = 10$ and similarly for the other terms. 
After you have noticed that $a_{n+2} = a_n + a_{n+1} + 3$ you could try to solve it using generating functions.
To do so, multiply both sides with $x^n$ then sum both sides $\sum_{n \geq 0}$. Set $G(x) = \sum_{n \geq 0} a_n x^n$. After some manipulation and using that $\sum_{n \geq 0}x^n = \frac{1}{1-x}$ I arrived at 
$$ \frac{1}{x^2}( G(x) -a_0 - a_1x) = G(x)  + \frac{1}{x}(G(x) - a_0) + \frac{3}{1-x}$$
And after some more manipulations and using $a_0 = 2$ and $a_1 = 5$ I arrived at
$$ G(x) = - \frac{x + 2}{(1-x) (x^2 + x - 1)}$$
Now one would look up the corresponding power series to read off the $n$-th coefficient (which corresponds to  the $n$-th term of your sequence) but for this one it seems that there isn't an easy power series which is somewhat disappointing. (For a list of potential power series check for example late H. S. Wilf's book on page 52/53). 
Nonethelss, the mechanics I have outlined above work in a wide variety of cases so you might as well keep this approach in the back of your mind and use it when you need it to solve something else in the future.
A: $$a_{n+2}=a_{n}+a_{n+1}+3, \; (n\in \mathbb{N})$$
A: This is listed at OEIS as A006327. Title of the sequence says:

a(n) = Fibonacci(n) - 3. Number of total preorders.
(Formerly M1371)

A: It's usually a good start to consider any famous sequences you know and apply operations to them until a pattern pops out. In this case, consider the fibonacci sequence minus a constant.
A: Let the given sequence $S_n :\{2, 5 ,10, 18, 31, 52\}$
Taking differences, $S_n':\{3, 5, 8, 13,21\}$
Again taking differences, $S_n'':\{2, 3 ,5, 8\}$
Again taking differences, $S_n''':\{1,2,3\}$
If $T_r,T_r',T_r'',T_{r}'''$ be the $r$th terms of series $S_r,S_n',S_n'',S_{n}'''$ respectively.
Clearly, $T_r'''=T_{r+1}''-T_r''$ and $T_r'''=r$
So, $T_r''$ is of order $2,=Ar^2+Br+C$(say)
then $r=T_r'''=T_{r+1}''-T_r''=A(2r+1)+B=2Ar+A+B$
Comparing the coefficients of $r,2A=1\implies A=\frac12$
Comparing the constant terms, $A+B=0\implies B=-A=-\frac12$
$\implies T_r''=\frac{r^2-r}2+C, 2=T_1''=C\implies C=2\implies  T_r''=\frac{r^2-r}2+2=\frac{r^2}2-\frac r2+2$
Again, $ T_r''=T_{r+1}'-T_r'$
So, $T_r'$ is of order $3,=Dr^3+Er^2+Fr+G$(say)
$\frac{r^2}2-\frac r2+2=T_r''=T_{r+1}'-T_r'=D\{(r+1)^3-r^3\}+E\{(r+1)^2-r^2\}+F\{(r+1)-r\}=3Dr^2+r(3D+2E)+D+E+F$
Comparing the coefficients of the different powers of $r,$ we get $D=\frac16,E=-\frac12,F=\frac73$ so that $T_r'=\frac16r^3-\frac12r^2+\frac73r+G$
Now, $3=T_1'=\frac161^3-\frac121^2+\frac73+G\implies G=1$
$\implies T_r'=\frac16r^3-\frac12r^2+\frac73r+1$
Again, $ T_r'=T_{r+1}-T_r$
So, $T_r'$ is of order $4,=Pr^4+Qr^3+RFr^2+Sr+T$(say)
$\frac16r^3-\frac12r^2+\frac73r+1= T_r'=T_{r+1}-T_r=P\{(r+1)^4-r^4\}+Q\{(r+1)^3-r^3\}+S\{(r+1)-r\}=4Pr^3+r^2(6P+3Q)+r(4P+3Q+2R)+P+Q+R+S$
Comparing the coefficients of the different powers of $r,$ we shall get  the values of $P,Q,R,S$.
$T$ can be determined using the value of any term $T_r$ like the earlier case.
A: There is no single answer, using Lagrange extrapolation one can make a polynomial were the next number is $0,1,-1,e^{i\pi},etc.$ of course some one once mentioned here the next number with the least entropy is the most natural answer but they didn't mention how to get it.
A: The sequence can be expressed in many ways.
As Matt N. and M. Strochyk mentioned:
$$ a_{n+2}= a_{n}+a_{n+1}+3,$$
$$ a_1 = 2 \quad (n\in \mathbb{N})$$
Or as this one for example:
$$ a_{n+1}= a_{n}+\frac{(n-1)n(2n-1)}{12}-\frac{(n-1)n}{4}+2n+1,$$
$$ a_1 = 2 \quad (n\in \mathbb{N})$$
It's interesting that the term:
$$ b_n = \frac{(n-1)n(2n-1)}{12}-\frac{(n-1)n}{4}+2n+1$$ gives five Fibonacci numbers $(3, 5, 8, 13, 21)$ for $1 \leq n \leq 5$.
