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In my research in theoretical physics, I have arrived at some coefficients $a_{n,m}$ depending on two integers, $n\geq 1$ and $0\leq m\leq n$:

$$ a_{n,m}=\sum_{j=0}^{n-1} {2j \choose j+m} \left(\frac{n}{4}\right)^j s(n,j+1) $$

where $s(n,j+1)$ are Stirling numbers of the first kind. Although this expression does not make it manifest (to me), these coefficients are zero when $n+m$ is even. For physical reasons, I am convinced that when $m+n$ is odd, $a_{n,m}>0$, but I haven't been able to prove it.

I would like to find a proof that $a_{n,m}\geq 0$.

I have found various ways to rewrite these coefficients. For instance, in terms of power series involving modified Bessel functions of the first kind $I_m$:

$$ a_{n,m}= \left. \frac{d^{n-1} \,\left( (1-z)^{\frac{n-2}{2}} \, I_m\left(-\frac{n}{2} \log (1-z)\right) \right)}{d z^{n-1}} \right|_{z=0} $$

but I am not able to conclude that $a_{n,m}\geq 0$ from this expression either.

Alternatively, when $n$ is a multiple of 4, these coefficients are integers, so I was hoping that there might be a combinatorial argument for this particular case. However, I have been unable to produce it.

Any help would be greatly appreciated.

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    $\begingroup$ Have you tried using the representation at mathoverflow.net/questions/34151 for the Stirling numbers and then applying the techniques at math.upenn.edu/~wilf/AeqB.html? $\endgroup$
    – Steve Huntsman
    Commented Sep 16, 2016 at 20:11
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    $\begingroup$ If we define $b_{n,m} := 4^{n-1}a_{n,m}$ then $b_{n,m}$ is always an integer, so it might be easier to find a combinatorial interpretation. For example, empirically it seems that $b_{n,n-1} = n^{n-1}$, and $b_{n,n-3}$ appears to factor into small prime factors. $\endgroup$
    – Timothy Chow
    Commented Sep 16, 2016 at 20:30
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    $\begingroup$ I am convinced that there is no closed formula in general, unlike some special cases $(n,n−1),(n,n−3)$. Not even a finite recurrence. Perhaps the best we can say is this: $b_{n,n-2k-1}=\alpha_k\times n^{2k+1}\binom{n}{k+1}\times P_k(n)$ for some polynomial $P_k\in\mathbb{Z}[n]$ of degree $2k-1$ and for some positive $\alpha_k\in\mathbb{Q}$. More importantly, each $P_k(y)$ has positive coefficients! $\endgroup$
    – T. Amdeberhan
    Commented Sep 17, 2016 at 1:07
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    $\begingroup$ The fact that $n^m\vert b_{n,m}$ can be proved. In $a_{n,m}$ replace $x=\frac{n}4$. Notice that although $\sum_{j=0}^{n-1}$, in fact $\sum_{j=m}^{n-1}$ since $\binom{2j}{j+m}=0$ when $j<m$. Consequently, $a_{n,m}$ as a polynomial in $x$ has an obvious factor $x^m$. That means $n^m$ factors out up on retreating to $\frac{n}4$ in place of $x$. $\endgroup$
    – T. Amdeberhan
    Commented Sep 17, 2016 at 1:51
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    $\begingroup$ I think the $a_{n,m}$ are Fourier coefficient of the polynomials $P_n(x)=\big( nx/2 +n/2-1\big)^{\underline n}$ (falling factorial), with respect to a certain symmetric measure on $[-1,1]$ quite concentrated at $\pm 1$. I'll make a computation to reconstruct the measure. Since the $P_n$ are odd/even according to the parity of $n$, and since they are also large at $\pm 1$, this hopefully would explain both positivity and vanishing properties for the $a_{n,m}$. $\endgroup$
    – Pietro Majer
    Commented Sep 17, 2016 at 16:20

1 Answer 1

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The numbers $a_{n,m}$ are in fact the Fourier coefficients of the polynomial $$P_n(x)=\prod_{j=1}^{n-1} \Big( \frac{nx}{2} + \frac{n}{2}-j\Big) $$ with respect to the Chebyshev measure $d\sigma:=(1-x^2)^{-1/2}dx$ on $[-1,1]$, and its orthogonal bases of the Chebyshev polynomials of the first kind. Precisely, for $0\le m\le n$ $$a_{n,m}=\frac{1}{\pi} \int_{-1}^1 P_n(x)T_m(x)d\sigma \ .$$

Changing variable, we have a trigonometric version: $$a_{n,m}=\frac{1}{\pi}\int_0^\pi P_n(\cos \theta)\cos (m\theta)d\theta \ .$$

Note that the polynomials $P_n$ and $T_m$ are odd resp. even, according to the parity of $n-1$, respectively $m$, so the integrand $P_n(x)T_m(x)$ has the same parity of $n+m-1$. On the other hand, the Chebyshev measure is symmetric, which explains the vanishing property $a_{n,m}=0$ whenever $n+m$ is even. Moreover, for odd $n+m$ the integrand is positive and concentrated about $\pm 1$; this should hopefully yield to the desired estimate $a_{n,m}>0$. I'll try some computation and in case add details later.

To compute the integral we may also use the Chebyshev-Gauss Quadrature formula on $N$ nodes, which is exact on polynomials of degree less than $2N$. Therefore, for $2N\ge n+m$ we have $$a_{n,m}=\frac{1}{N}\sum_{k=1}^{N} P_n\Big(\cos\big( \frac{2k-1}{2N}\pi \big) \Big)\ \cos\big( m\frac{2k-1}{2N}\pi\big) \ .$$

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  • $\begingroup$ Thanks for your detailed discussion. Indeed, this is precisely the problem that led me to these coefficients. However, I don't understand what you mean by the claim "for $n+m$ odd the integrand is positive". The polynomials that you call $P_n(x)$ have $n-1$ real roots in $(-1,1)$, so for instance, for $T_0(x)=1$, the function $P_n(x)T_0(x)=P_n(x)$ changes sign $n-1$ times in $(-1,1)$. $\endgroup$
    – Tomeu Fiol
    Commented Sep 17, 2016 at 20:10
  • $\begingroup$ Sorry, maybe it is not very clear. The complete sentence is "for n+m odd the integrand is positive about $\pm 1$ . $P_nT_m$ of course changes sign several times in (-1,1), but it is positive and large close to $\pm 1$. $\endgroup$
    – Pietro Majer
    Commented Sep 17, 2016 at 21:28
  • $\begingroup$ Ok, it is clear now. Thanks also for mentioning the discrete Chebyshev transform. In the past, I also tried to prove $a_{n,m} \geq 0$ from the last formula in your answer, but I didn't manage to. $\endgroup$
    – Tomeu Fiol
    Commented Sep 17, 2016 at 22:55

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