Question 35996 asks about the Ehrhart polynomial $i_d(n)$ of the standard regular crosspolytope. It can be defined equivalently by $$ \sum_{n\geq 0}i_d(n)x^n = \frac{(1+x)^d}{(1x)^{d+1}}. $$ 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 $i_d(n)$?

1$\begingroup$ Possibly related?: mathoverflow.net/questions/308178/… $\endgroup$ – Sam Hopkins Feb 8 at 20:25

2$\begingroup$ @SamHopkins: It is related but not so useful in answering my question, since positivity is proved by the same Theorem 3.2. $\endgroup$ – Richard Stanley Feb 8 at 20:32

1$\begingroup$ I don't understand the difficulty: 1/(1x) has Maclaurin series with positive coefficients, so that so do all its powers (which are known anyway), and multiplying by $(1+x)^d$ preserves this. Since the radius of convergence of the thing on the right is $1$, uniqueness yields the result. Moreover, it also follows that the coefficients form a log concave sequence ... (since those of $1/(1x)$ and of $1+x$ do). $\endgroup$ – David Handelman Feb 8 at 22:19

$\begingroup$ Now I see: $i_d (n)$ is supposed to be a polynomial itself in $x$. So ignore my remarks ... $\endgroup$ – David Handelman Feb 9 at 0:37

1$\begingroup$ It would be nice to include hyperlink to mentioned MO 35996 question mathoverflow.net/questions/35996/ehrhartpolynomial $\endgroup$ – Alexander Chervov Feb 9 at 19:26
Here is a very simple way to show the positivity. Define $$f(d,x) = (1+x)^d/(1x)^{d+1}.$$ Then, by induction $$ \frac{\partial^t f(d,x)}{\partial\, d^t} = f(d,x) \, \ln\biggl(\frac{1+x}{1x}\biggr)^t.$$ Putting in $d=0$ we have that the Taylor series of $f(d,x)$ with respect to $d$ is $$ f(d,x) = \sum_{t=0}^\infty \frac{1}{t!} (1x)^{1} \ln\biggl(\frac{1+x}{1x}\biggr)^t\,d^t. $$ Both $(1x)^{1}$ and $\ln\Bigl(\frac{1+x}{1x}\Bigr)$ have nonnegative 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 $$\biggl(\sum_{k\ge 0} x^k\biggr) \biggl(\sum_{k\ge 0} \frac{x^{2k+1}}{2k+1}\biggr)^t. $$