Positivity of the coefficients of the Ehrhart polynomial of a cross-polytope Question 35996 asks about the Ehrhart polynomial $i_d(n)$ of the
standard regular cross-polytope. It can be defined equivalently by
  $$ \sum_{n\geq 0}i_d(n)x^n = \frac{(1+x)^d}{(1-x)^{d+1}}. $$
It can be shown that the coefficients of $i_d(n)$ are positive, using
Theorem 3.2 of http://math.mit.edu/~rstan/papers/cycles.pdf to show
that all zeros of $i_d(n)$ have real part $-1/2$. Is there some
"positive" formula for $i_d(n)$ that makes it transparent that the
coefficients are positive? Or at least, is there another proof that
doesn't involve the zeros of $i_d(n)$?
 A: Here is a very simple way to show the positivity.
Define $$f(d,x) = (1+x)^d/(1-x)^{d+1}.$$
Then, by induction
$$ \frac{\partial^t f(d,x)}{\partial\, d^t} = f(d,x)
 \, \ln\biggl(\frac{1+x}{1-x}\biggr)^t.$$
Putting in $d=0$ we have that the Taylor series of $f(d,x)$ with
respect to $d$ is
$$ f(d,x) = \sum_{t=0}^\infty \frac{1}{t!} (1-x)^{-1}
   \ln\biggl(\frac{1+x}{1-x}\biggr)^t\,d^t. $$
Both $(1-x)^{-1}$ and $\ln\Bigl(\frac{1+x}{1-x}\Bigr)$ have
non-negative Taylor coefficients, which completes the proof.
In summary, the coefficient of $x^nd^t$ is $2^t/t!$ times
the coefficient of $x^n$ in
$$\biggl(\sum_{k\ge 0} x^k\biggr)
  \biggl(\sum_{k\ge 0} \frac{x^{2k+1}}{2k+1}\biggr)^t.
$$
A: From the binomial theorem, $(1+x)^d=\sum_{j\geq0}\binom{d}jx^j$ while
$$\frac1{(1-x)^{d+1}}=\sum_{k\geq0}\binom{d+k}kx^k.$$
Therefore, by Cauchy Product formula and as a polynomial in $n$, $i_d(n)$ takes the form
$$i_d(n)=\sum_{k\geq0}\binom{d}{n-k}\binom{d+k}k \qquad \text{or} \qquad
i_d(n)=\sum_{k\geq0}2^k\binom{d}k\binom{n}k.$$
If we employ Zeilberger's algorithm, we find the recurrence
$$
(d+2)\cdot i_{d+2}(n)=(2n+1)\cdot i_{d+1}(n)+(d+1)\cdot i_d(n). \tag1$$
One may now proceed by induction on $d$, with 
$$i_1(n)=2n+1 \qquad \text{and} \qquad  i_2(n)=2n^2+2n+1.$$
An aside: Equation (1) reveals $n=-\frac12$ is a root of $i_d(n)$ for all $d$ odd.
