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I stumbled across the following curious empirical properties of the Bernoulli polynomials $B_n(x)$. Can anyone provide a reference or proof?

Let $k\in\mathbb{Z}$, $k\geq 2$. Then (empirically):

  1. The coefficients of $x^{n-3}$, $x^{n-5}$, $x^{n-7}, \dots$ in the polynomial $\frac{1}{n}B_n(x+k)$ are integers.

  2. Define a sequence $a_2,a_3,\dots$ by $a_2=20$, and if $b_i=a_{i+1}-a_i$ then the sequence $b_2,b_3,\dots$ is periodic of period four, with terms $16,16,16,20,16,16,16,20,\dots$. Let $n\equiv j\,(\mathrm{mod}\,4)$, $0\leq j\leq 3$. Then the coefficient of $x^i$ in $B_n(x+k)$ is negative if and only if $n\geq a_k$ and $i=j,j+4,j+8,\dots,j+4\lfloor \frac{n-a_k}{4}\rfloor$.

I have checked this for $n\leq 324$ and $2\leq k\leq 8$.

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  • $\begingroup$ I think you mean $\frac{1}{n}B_n(x+k)$ and not $\frac{1}{n-1}B_n(x+k)$. For instance the $x^2$ coefficient of $B_5(x+2)$ is $30$, which is a multiple of $5 = n$ but not of $4 = n-1$. $\endgroup$
    – Vesselin Dimitrov
    Commented May 2, 2018 at 20:15
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    $\begingroup$ @VesselinDimitrov Thanks, this has been corrected. $\endgroup$
    – Richard Stanley
    Commented May 2, 2018 at 23:43
  • $\begingroup$ A g.f. proof of (1): as $\frac{t e^{(x+k)t}}{e^t-1}=\sum_{n\geq 0} \frac{B_n(x+k)}{n!}\,t^n$ we have that \begin{align*} \frac{(n-1-2j)!}{(n-1)!}\,[x^{n-1-2j}]\, \frac{B_n(x+k)}{n}&=\,[t^{2j}]\frac{e^{kt}}{e^t-1} =\,\frac{1}{2}\,[t^{2j}]\bigg(\frac{e^{kt}}{e^t-1}+\frac{e^{-kt}}{e^{-t}-1}\bigg)\\ &=\,\frac{1}{2}\,[t^{2j}]\,e^{-(k-1)t}\bigg(\frac{e^{(2k-1)t}-1}{e^t-1}\bigg) =\,\frac{1}{2}\,\frac{1}{(2j)!}\sum_{i=0}^{2k-2}(i-k+1)^{2j} \end{align*} Hence (for $2\leq 2j\leq n-1$) $$[x^{n-1-2j}]\, \frac{B_n(x+k)}{n}={n-1 \choose 2j}\sum_{i=1}^{k-1}i^{2j}\;\;.$$ $\endgroup$
    – esg
    Commented May 4, 2018 at 16:25

2 Answers 2

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Let me address (1). First, you need a correction: I suppose you intended $\frac{1}{n}B_n(x+k)$ in the empirical calculations, and not $\frac{1}{n-1}B_n(x+k)$. For instance, the $x^2$ coefficient of $B_5(x+2)$ is $30$, which is a multiple of $5 = n$ but not of $4 = n-1$.

As to the proof of the integrality of the $x^j$ coefficient of $\frac{1}{n}B_n(x+k)$ for $j \equiv n+1 \mod{2}$, it is reduced by the "umbral" relation $$ \frac{B_n(x+1) - B_n(x)}{n} = x^{n-1} \in \mathbb{Z}[x] $$ to the vanishing of the odd Bernoulli numbers (apart from the explicit value of $B_1 = 1/2$). Just sum this relation $k-1$ times with integer-shifted arguments.

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I think I see how to prove (2), or whatever version of it is true, but I haven't checked the details. Namely, use the identity $$ B_n(x+d) = \sum_{k=0}^n {n\choose k}B_k(d)x^{n-k}, $$ together with known results on the real zeros of Bernoulli polynomials such as https://ac.els-cdn.com/0021904589900166/1-s2.0-0021904589900166-main.pdf?_tid=147e6852-c89e-44e9-a805-69dd4bc54a5b&acdnat=1525395947_25f4432d47ebd7c61d3d59b2cff877a9.

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