**Question** : Letting $k,n$ be positive integers, let's define a sequence $\{a_i\}\ (i=0,1,\cdots, kn)$ as
$$(1+x+\cdots+x^k)^n=\sum_{i=0}^{kn}a_ix^i.$$
Then, is the 'special' difference-sequence $\{d^Na_i\}$ a unimodal sequence for every non-negative integer $N$? If the answer is yes, then prove that. If the answer is no, then find a counterexample.

**Definition** :

*1.* Let's call a sequence $\{a_i\}\ (i=0,1,\cdots,kn)$ which satisfies the following condition **'a unimodal sequence'**.

Condition : $\ $There exists an integer $t$ such that $$a_0\le a_1\le \cdots\le a_t\ge a_{t+1}\ge a_{t+2}\ge\cdots.$$

*2.* Let's define $\{d^Na_i\}\ (N\in\mathbb N)$ as the following:

$$d^Na_i=\max(d^{N-1}a_i-d^{N-1}a_{i-1},0)\ \ \ (i=0,1,\cdots, kn),$$ $$d^0a_i=a_i, \ a_{-1}=0.$$

Please note that this is **not** an usual difference-sequence. (we may call this **'a special difference-sequence'**)

**Motivation** :

When I saw Pascal's triangle, I found this property about $a_i=\ _nC_i$. Then, I generalized this property.

**Example** : Let's see the $(k,n)=(1,8)$ case.

$$a_i(=d^0a_i) : 1\ \ 8\ \ 28\ \ 56\ \ 70\ \ 56\ \ 28\ \ 8\ \ 1$$

$$d^1a_i : 1\ \ 7\ \ 20\ \ 28\ \ 14\ \ 0\ \ 0\ \ 0\ \ 0$$

$$d^2a_i : 1\ \ 6\ \ 13\ \ 8\ \ 0\ \ 0\ \ 0\ \ 0\ \ 0$$

$$d^3a_i : 1\ \ 5\ \ 7\ \ 0\ \ 0\ \ 0\ \ 0\ \ 0\ \ 0$$

$$d^4a_i : 1\ \ 4\ \ 2\ \ 0\ \ 0\ \ 0\ \ 0\ \ 0\ \ 0$$

$$d^5a_i : 1\ \ 3\ \ 0\ \ 0\ \ 0\ \ 0\ \ 0\ \ 0\ \ 0$$

$$d^6a_i : 1\ \ 2\ \ 0\ \ 0\ \ 0\ \ 0\ \ 0\ \ 0\ \ 0$$

$$d^7a_i : 1\ \ 1\ \ 0\ \ 0\ \ 0\ \ 0\ \ 0\ \ 0\ \ 0$$

$$d^Na_i (N\ge 8): 1\ \ 0\ \ 0\ \ 0\ \ 0\ \ 0\ \ 0\ \ 0\ \ 0$$

Hence, we can see that $\{d^N{_8C_i}\}\ (i=0,1,\cdots,8)$ is a unimodal sequence for every non-negative integer $N$.

**Remark** : This question has been asked previously on math.SE without receiving any complete answers:
https://math.stackexchange.com/questions/492651/about-the-unimodality-of-the-coefficients-sequence-of-1x-cdotsxkn

The above definition about 'difference-sequences' might be somewhat unusual, but I won't change its definition. This is because this definition would keep the unimodality. If we define $d^Na_i=d^{N-1}a_i-d^{N-1}a_{i-1} (i=0,1,\cdots)$ as usual, then we get

$$d^1a_i : 1\ \ 7\ \ 20\ \ 28\ \ 14\ \ -14\ \ -28\ \ -20\ \ -7$$

in the above example, which is not unimodal.

truncatedfinite difference operator, in particular since you use that symbol before givin a non-standard definition of it; this is bound to confuse readers. $\endgroup$ – Marc van Leeuwen Sep 19 '13 at 9:17