Here is my question that I tried to solve:
Let $(a_n)_{n \in \mathbb{N}}$ and
$\displaystyle \frac{1}{1-x-x^2-x^3}=\sum_{n=1}^\infty a_n x^n$ $ \quad $ holds.
According to above find all positive integers that satisfy the equaility
$a_n=(n+1)^2$
My attempt: In the first place, i sensed that there is no n positive integer that holds the desired. With that starting point, I remembered that with using tribonacci numbers which are denoted by $T_n$ that satisfy
$T_0=T_1=0, T_2=1$ and for $\, \,n \geq 3 $ $\quad$ $ T_n=T_{n-1}+T_{n-2}+T_{n-3}$
the power series converges the function which is desired ;
$\displaystyle \sum_{n=1}^\infty \,T_n \, \, x^n=\frac{1}{1-x-x^2-x^3}$
So I started to check is there any numbers that satisfy the equatily $\,$$T_n = a_n$ $\, $ for $ n \geq 1$
$T_1= 0 \neq 4=(1+1)^2$
$T_2= 1 \neq 9=(2+1)^2$
$T_3= 1 \neq 16=(3+1)^2$
$T_4= 2 \neq 25=(4+1)^2$
$T_5= 4 \neq 36=(5+1)^2$
$T_6= 7 \neq 49=(6+1)^2$
$T_7= 13 \neq 64=(7+1)^2$
$T_8= 24 \neq 81=(8+1)^2$
$T_9= 44 \neq 100=(9+1)^2$
$T_{10}= 81 \neq 121=(10+1)^2$
$T_{11}= 149 \neq 144=(11+1)^2$
$T_{12}= 274 \neq 169=(12+1)^2$
$T_{13}= 504 \neq 196=(13+1)^2$
$T_{14}= 927 \neq 225=(14+1)^2$
$T_{15}= 3136 \neq 256=(15+1)^2$
and so on.
I sensed that for every $n >10 $
$T_n > (n+1)^2$.
If i could show that inequaility my problem is solved. But here is where is stuck. Thought that maybe I could use Binet's formulas for tribonacci numbers to show
$ \displaystyle \lim _{n \to \infty} \frac{a_n}{T_n}=0$
but it is not the answer that I seek for(Maybe there are positive integers that holds)
It is not necessary to continue solving the problem the way i started - other way of solutions is helpful also- but i want to solve in my way. Thanks in advance for your help and tips.
Induction works after the initial segment: If $n>4$ and $T_k>(k+1)^2$ for $k=n-1$ and for $k=n-2$, then $$ T_n-(n+1)^2\ge n^2+(n-1)^2-(n+1)^2=n^2-4n>0$$