In the final section, we show several identities for hypergeometric Bernoulli numbers which the convergents of the continued fraction expansion of the generating function of hypergeometric Bernoulli numbers entail.
In [1, 19] by studying the convergents of the continued fraction of
$$ \frac{x/2}{\tanh x/2}=\sum_{n=0}^{\infty }B_{2 n} \frac{x^{2 n}}{(2 n)!}, $$
some identities of Bernoulli numbers are obtained. In this section, the nth convergent of the generating function of hypergeometric Bernoulli numbers is explicitly given. As an application, we give some identities of hypergeometric Bernoulli numbers in terms of binomial coefficients.
The generating function on the left-hand side of (1) can be expanded as a continued fraction
$$ \frac{1}{{}_{1} F_{1}(1;N+1;x)} =1-\cfrac{x}{N+1+\cfrac{x}{N+2-\cfrac{(N+1)x}{N+3+\cfrac{2 x}{N+4-\cfrac{(N+2)x}{N+5+ \ddots }}}}} $$
(14)
(cf. [29, (91.2)]). Its nth convergent \(P_{n}(x)/Q _{n}(x)\) (\(n\ge 0\)) is given by the recurrence relation
$$\begin{aligned} &P_{n}(x)=a_{n}(x)P_{n-1}(x)+b_{n}(x)P_{n-2}(x)\quad(n \ge 2), \end{aligned}$$
(15)
$$\begin{aligned} &Q_{n}(x)=a_{n}(x)Q_{n-1}(x)+b_{n}(x)Q_{n-2}(x)\quad(n \ge 2), \end{aligned}$$
(16)
with initial values
$$\begin{aligned} &P_{0}(x)=1, \qquad P_{1}(x)=(N+1)-x; \\ &Q_{0}(x)=1, \qquad Q_{1}(x)=N+1, \end{aligned}$$
where, for \(n\ge 1\), \(a_{n}(x)=N+n\), \(b_{2n}(x)=n x\) and \(b_{2 n+1}(x)=-(N+n)x\).
We have explicit expressions of both the numerator and the denominator of the nth convergent of (14).
Theorem 4
For
\(n\ge 1\), we have
$$\begin{aligned} &P_{2 n-1}(x)=\sum_{j=0}^{n}(-1)^{j} \binom{n}{j}\prod_{l=1}^{2 n-j-1}(N+l) \cdot x^{j}, \\ &P_{2 n}(x)=\sum_{j=0}^{n}(-1)^{j} \binom{n}{j}\prod_{l=1}^{2 n-j}(N+l) \cdot x^{j}, \end{aligned}$$
and
$$\begin{aligned} &Q_{2 n-1}(x)=\sum_{j=0}^{n-1}\sum _{k=0}^{j}(-1)^{j-k}(2 n-j-1)_{k} \binom{n-k-1}{j-k}\prod_{l=k+1}^{2 n-j-1}(N+l) \cdot x^{j}, \\ &Q_{2 n}(x)=\sum_{j=0}^{n}\sum _{k=0}^{j}(-1)^{j-k}(2 n-j)_{k} \binom{n-k-1}{j-k}\prod_{l=k+1}^{2 n-j}(N+l) \cdot x^{j}. \end{aligned}$$
Remark
Here we use the convenient values
$$ \binom{n}{k}=0\quad (0\le n< k), \qquad\binom{-1}{0}=1, $$
and recognize the empty product as 1. Otherwise, we should write \(Q_{2 n}(x)\) as
$$ Q_{2 n}(x)=\sum_{j=0}^{n-1}\sum _{k=0}^{j}(-1)^{j-k}(2 n-j)_{k} \binom{n-k-1}{j-k}\prod_{l=k+1}^{2 n-j}(N+l) \cdot x^{j}+n! x^{n}. $$
If we use the unsigned Stirling numbers of the first kind, \([{n\atop k} ]\), of which the generating function is given by
$$ \sum_{n=k}^{\infty }(-1)^{n-k} \begin{bmatrix}n\cr k \end{bmatrix}\frac{z^{n}}{n!}=\frac{ (\log (1+z) )^{k}}{k!} $$
[9, (7.50)], we can express the products as
$$ \prod_{l=1}^{2 n-j-1}(N+l)=\sum _{i=1}^{2 n-j} \begin{bmatrix} 2 n-j\cr i\end{bmatrix} N ^{i-1} $$
or
$$ \prod_{l=k+1}^{2 n-j-1}(N+l)=\sum _{i=1}^{2 n-j-k} \begin{bmatrix} 2 n-j-k\cr i\end{bmatrix}(N+k)^{i-1} . $$
Proof of Theorem 4
The proof is done by induction on n. It is easy to see that for \(n=0\), we have \(P_{0}(x)=Q_{0}(x)=1\), and for \(n=1\), we have \(P_{1}(x)=(N+1)-x\) and \(Q_{1}(x)=N+1\). Assume that the results hold up to \(n-1(\ge 2)\). Then by using the recurrence relation in (16)
$$\begin{aligned} &(N+2 n)P_{2 n-1}(x)+n P_{2 n-2}(x)\cdot x \\ &\quad =(N+2 n)\sum_{j=0}^{n}(-1)^{j} \binom{n}{j}\prod_{l=1}^{2 n-j-1}(N+l) \cdot x^{j} \\ &\qquad{} +n\sum_{j=0}^{n-1}(-1)^{j} \binom{n-1}{j}\prod_{l=1}^{2 n-j-2}(N+l) \cdot x^{j+1} \\ &\quad=(N+2 n)\prod_{l=1}^{2 n-1}(N+l) \\ &\qquad{} +(N+2 n)\sum_{j=1}^{n}(-1)^{j} \binom{n}{j}\prod_{l=1}^{2 n-j-1}(N+l) \cdot x^{j} \\ &\qquad{} -n\sum_{j=1}^{n}(-1)^{j} \binom{n-1}{j-1}\prod_{l=1}^{2 n-j-1}(N+l) \cdot x^{j}. \end{aligned}$$
Since
$$ (N+2 n)\binom{n}{j}-n\binom{n-1}{j-1}=(N+2 n-j)\binom{n}{j}, $$
we get
$$\begin{aligned} &(N+2 n)P_{2 n-1}(x)+n P_{2 n-2}(x)\cdot x \\ &\quad =\sum_{j=0}^{n}(-1)^{j} \binom{n}{j}\prod_{l=1}^{2 n-j}(N+l)\cdot x ^{j} \\ &\quad =P_{2 n}. \end{aligned}$$
Next,
$$\begin{aligned} &(N+2 n+1)P_{2 n}(x)-(N+n)P_{2 n-1}(x)\cdot x \\ &\quad =(N+2 n+1)\sum_{j=0}^{n}(-1)^{j} \binom{n}{j}\prod_{l=1}^{2 n-j}(N+l) \cdot x^{j} \\ &\qquad{} -(N+n)\sum_{j=0}^{n}(-1)^{j} \binom{n}{j}\prod_{l=1}^{2 n-j-1}(N+l) \cdot x^{j+1} \\ &\quad=(N+2 n+1)\prod_{l=1}^{2 n}(N+l) \\ &\qquad{} +(N+2 n+1)\sum_{j=1}^{n}(-1)^{j} \binom{n}{j}\prod_{l=1}^{2 n-j}(N+l) \cdot x^{j} \\ &\qquad{} +(N+n)\sum_{j=1}^{n}(-1)^{j} \binom{n}{j-1}\prod_{l=1}^{2 n-j}(N+l) \cdot x^{j} \\ &\qquad{} -(N+n) (-1)^{n}\prod_{l=1}^{n-1}(N+l) \cdot x^{n+1}. \end{aligned}$$
Since
$$ (N+2 n+1)\binom{n}{j}+(N+n)\binom{n}{j-1}=(N+2 n-j+1) \binom{n+1}{j}, $$
we get
$$\begin{aligned} &(N+2 n+1)P_{2 n}(x)-(N+n)P_{2 n-1}(x)\cdot x \\ &\quad =\sum_{j=0}^{n+1}(-1)^{j} \binom{n+1}{j}\prod_{l=1}^{2 n-j+1}(N+l) \cdot x^{j} \\ &\quad =P_{2 n+1}. \end{aligned}$$
Concerning \(Q_{n}(x)\),
$$\begin{aligned} &(N+2 n)Q_{2 n-1}(x)+n Q_{2 n-2}(x)\cdot x \\ &\quad =(N+2 n)\sum_{j=0}^{n-1}\sum _{k=0}^{j}(-1)^{j-k}(2 n-j-1)_{k} \binom{n-k-1}{j-k}\prod_{l=k+1}^{2 n-j-1}(N+l) \cdot x^{j} \\ &\qquad{} +n\sum_{j=0}^{n-1}\sum _{k=0}^{j}(-1)^{j-k}(2 n-j-2)_{k} \binom{n-k-2}{j-k}\prod_{l=k+1}^{2 n-j-2}(N+l) \cdot x^{j+1} \\ &\quad=\prod_{l=1}^{2 n}(N+l)-n\sum _{k=0}^{n-1}(-1)^{n-k}(n)_{k} \binom{n-k-2}{n-k-1}\prod_{l=k+1}^{n-1}(N+l)\cdot x^{n} \\ & \qquad{}+\sum_{j=1}^{n-1}\sum _{k=0}^{j}(-1)^{j-k}(2 n-j-1)_{k} \\ &\qquad{} \times \binom{n-k-1}{j-k}(N+2 n)\prod_{l=k+1}^{2 n-j-1}(N+l) \cdot x ^{j} \\ & \qquad {}-n\sum_{j=1}^{n-1}\sum _{k=0}^{j-1}(-1)^{j-k}n(2 n-j-1)_{k} \binom{n-k-2}{j-k-1}\prod_{l=k+1}^{2 n-j-1}(N+l) \cdot x^{j}. \end{aligned}$$
Since \(N+2 n=(N+2 n-j)+j\),
$$\begin{aligned} &\prod_{l=k+1}^{2 n-j-1}(N+l)=\prod _{l=k+2}^{2 n-j}(N+l)-(2 n-j-k-1) \prod _{l=k+2}^{2 n-j-1}(N+l), \\ &(j-k)\binom{n-k-1}{j-k}-n\binom{n-k-2}{j-k-1}=-(k+1) \binom{n-k-2}{j-k-1} \end{aligned}$$
and \((2 n-j-1)_{k}+k(2 n-j-1)_{k-1}=(2 n-j)_{k}\), we obtain
$$\begin{aligned} &(N+2 n)Q_{2 n-1}(x)+n Q_{2 n-2}(x)\cdot x \\ &\quad=\prod_{l=1}^{2 n}(N+l)+n! x^{n} \\ &\qquad{} +\sum_{j=1}^{n-1}\sum _{k=0}^{j}(-1)^{j-k}(2 n-j-1)_{k} \binom{n-k-1}{j-k}\prod_{l=k+1}^{2 n-j}(N+l) \cdot x^{j} \\ &\qquad{} +\sum_{j=1}^{n-1}\sum _{k=0}^{j}(-1)^{j-k}j(2 n-j-1)_{k} \\ &\qquad{} \times \binom{n-k-1}{j-k}(N+2 n)\prod_{l=k+1}^{2 n-j-1}(N+l) \cdot x ^{j} \\ &\qquad{} -\sum_{j=1}^{n-1}\sum _{k=0}^{j-1}(-1)^{j-k}n(2 n-j-1)_{k} \\ &\qquad{} \times \binom{n-k-2}{j-k-1}(N+2 n)\prod_{l=k+1}^{2 n-j-1}(N+l) \cdot x ^{j} \\ &\quad=\prod_{l=1}^{2 n}(N+l)+n! x^{n} \\ & \qquad{}+\sum_{j=1}^{n-1}\sum _{k=0}^{j}(-1)^{j-k}(2 n-j)_{k} \binom{n-k-1}{j-k}\prod_{l=k+1}^{2 n-j}(N+l)\cdot x^{j} \\ &\quad =Q_{2 n}. \end{aligned}$$
Similarly,
$$\begin{aligned} &(N+2 n+1)Q_{2 n}(x)-(N+n)x Q_{2 n-1}(x) \\ &\quad =(N+2 n+1)\sum_{j=0}^{n}\sum _{k=0}^{j}(-1)^{j-k}(2 n-j)_{k} \binom{n-k-1}{j-k}\prod_{l=k+1}^{2 n-j}(N+l)\cdot x^{j} \\ &\qquad{} -(N+n)\sum_{j=0}^{n-1}\sum _{k=0}^{j}(-1)^{j-k}(2 n-j-1)_{k} \binom{n-k-1}{j-k}\prod_{l=k+1}^{2 n-j-1}(N+l) \cdot x^{j+1} \\ &\quad =(N+2 n+1)\prod_{l=k+1}^{2 n}(N+l) \\ &\qquad{} +\sum_{j=1}^{n}\sum _{k=0}^{j}(-1)^{j-k}(2 n-j)_{k} \binom{n-k-1}{j-k}\prod_{l=k+1}^{2 n-j+1}(N+l)\cdot x^{j} \\ &\qquad{}\times \sum_{j=1}^{n}\sum _{k=0}^{j}(-1)^{j-k}(2 n-j)_{k} j \binom{n-k-1}{j-k}\prod_{l=k+1}^{2 n-j}(N+l) \cdot x^{j} \\ &\qquad{} +\sum_{j=1}^{n}\sum _{k=0}^{j-1}(-1)^{j-k}(2 n-j)_{k} \binom{n-k-1}{j-k-1}\prod_{l=k+1}^{2 n-j+1}(N+l)\cdot x^{j} \\ & \qquad{}-\sum_{j=1}^{n}\sum _{k=0}^{j-1}(-1)^{j-k}(2 n-j)_{k}(n-j+1) \binom{n-k-1}{j-k-1}\prod_{l=k+1}^{2 n-j}(N+l) \cdot x^{j}. \end{aligned}$$
Since
$$ \binom{n-k-1}{j-k}+\binom{n-k-1}{j-k-1}=\binom{n-k}{j-k} $$
and
$$\begin{aligned} &j\binom{n-k-1}{j-k}-(n-j+1)\binom{n-k-1}{j-k-1} \\ &\quad =k\binom{n-k}{j-k}-(k+1)\binom{n-k-1}{j-k-1}, \end{aligned}$$
we get
$$\begin{aligned} &(N+2 n+1)Q_{2 n}(x)-(N+n)x Q_{2 n-1}(x) \\ &\quad =(N+2 n+1)\prod_{l=k+1}^{2 n}(N+l) \\ &\qquad{} +\sum_{j=1}^{n}\sum _{k=0}^{j}(-1)^{j-k}(2 n-j)_{k} \binom{n-k}{j-k}\prod_{l=k+1}^{2 n-j+1}(N+l)\cdot x^{j} \\ &\qquad{} +\sum_{j=1}^{n}\sum _{k=0}^{j}(-1)^{j-k}(2 n-j)_{k} \biggl(k \binom{n-k}{j-k}-(k+1)\binom{n-k-1}{j-k-1} \biggr) \\ &\qquad{} \times \prod_{l=k+1}^{2 n-j}(N+l)\cdot x^{j}. \end{aligned}$$
Since
$$\begin{aligned} &(-1)^{j-k-1}(2 n-j)_{k+1}\binom{n-k-1}{j-k-1}\prod _{l=k+2}^{2 n-j+1}(N+l) \\ &\qquad{} +(-1)^{j-k}(2 n-j)_{k} \biggl(k\binom{n-k}{j-k}-(k+1) \binom{n-k-1}{j-k-1} \biggr)\prod_{l=k+1}^{2 n-j}(N+l) \\ &\qquad{} +(-1)^{j-k+1}(2 n-j)_{k}\binom{n-k}{j-k}\prod _{l=k+1}^{2 n-j}(N+l) \\ &\quad =(-1)^{j-k-1}(2 n-j+1)_{k+1}\binom{n-k-1}{j-k-1}\prod _{l=k+2}^{2 n-j+1}(N+l) \\ &\qquad{} +(-1)^{j-k-1}(k+1) (2 n-j)_{k+1}\binom{n-k-1}{j-k-1}\prod _{l=k+2} ^{2 n-j}(N+l), \end{aligned}$$
we have
$$\begin{aligned} &(N+2 n+1)Q_{2 n}(x)-(N+n)x Q_{2 n-1}(x) \\ &\qquad{} +\sum_{j=1}^{n}\sum _{k=0}^{j}(-1)^{j-k}(2 n-j+1)_{k} \binom{n-k}{j-k}\prod_{l=k+1}^{2 n-j+1}(N+l) \cdot x^{j} \\ &\quad =Q_{2 n+1}(x). \end{aligned}$$
□
Some more identities of hypergeometric Bernoulli numbers
Since \(P_{2 n-1}(x)\), \(P_{2 n}(x)\) and \(Q_{2 n}(x)\) are polynomials with degree n and \(Q_{2 n-1}(x)\) is a polynomial with degree \(n-1\), by the approximation property of the continued fraction, we have the following.
Lemma 8
Let
\(P_{n}(x)/Q_{n}(x)\)
denote the
nth convergent of the continued fraction expansion of (14). Then for
\(n\ge 0\),
$$ Q_{n}(x)\sum_{\kappa =0}^{\infty }B_{N,\kappa } \frac{x^{\kappa }}{ \kappa !}\equiv P_{n}(x)\quad {\bigl(\operatorname{mod} x^{n+1}\bigr)}. $$
By this approximation property, the coefficients \(x^{j}\) (\(0\le j \le n\)) of
$$ Q_{n}(x)\sum_{\kappa =0}^{\infty }B_{N,\kappa } \frac{x^{\kappa }}{ \kappa !}-P_{n}(x) $$
are nullified. By Theorem 4,
$$\begin{aligned} &Q_{2 n}(x)\sum_{\kappa =0}^{\infty }B_{N,\kappa } \frac{x^{\kappa }}{ \kappa !} \\ &\quad =\sum_{h=0}^{\infty }\sum _{j=0}^{\min \{h,n\}}\sum_{k=0}^{j}(-1)^{j-k}(2 n-j)_{k} \binom{n-k-1}{j-k}\prod_{l=k+1}^{2 n-j}(N+l) \cdot \frac{B_{N,h-j}}{(h-j)!}x^{h} \end{aligned}$$
and
$$ P_{2 n}(x)=\sum_{h=0}^{n}(-1)^{h} \binom{n}{h}\prod_{l=1}^{2 n-h}(N+l) \cdot x^{h}. $$
Therefore,
$$\begin{aligned} &\sum_{j=0}^{\min \{h,n\}}\sum _{k=0}^{j}(-1)^{j-k}(2 n-j)_{k} \binom{n-k-1}{j-k}\prod_{l=k+1}^{2 n-j}(N+l)\cdot \frac{B_{N,h-j}}{(h-j)!} \\ &\quad = \textstyle\begin{cases} (-1)^{h}\binom{n}{h}\prod_{l=1}^{2 n-h}(N+l) &(0\le h\le n); \\ 0 &(h>n). \end{cases}\displaystyle \end{aligned}$$
Similarly, since
$$\begin{aligned} &Q_{2 n-1}(x)\sum_{\kappa =0}^{\infty }B_{N,\kappa } \frac{x^{\kappa }}{ \kappa !} \\ &\quad =\sum_{h=0}^{\infty }\sum _{j=0}^{\min \{h,n-1\}}\sum_{k=0}^{j}(-1)^{j-k}(2 n-j-1)_{k} \binom{n-k-1}{j-k} \\ &\qquad{} \times \prod_{l=k+1}^{2 n-j-1}(N+l)\cdot \frac{B_{N,h-j}}{(h-j)!}x ^{h} \end{aligned}$$
and
$$ P_{2 n-1}(x)=\sum_{h=0}^{n}(-1)^{h} \binom{n}{h}\prod_{l=1}^{2 n-h-1}(N+l) \cdot x^{h}, $$
we have
$$\begin{aligned} &\sum_{j=0}^{\min \{h,n\}}\sum _{k=0}^{j}(-1)^{j-k}(2 n-j-1)_{k} \binom{n-k-1}{j-k}\prod_{l=k+1}^{2 n-j-1}(N+l) \cdot \frac{B_{N,h-j}}{(h-j)!} \\ &\quad= \textstyle\begin{cases} (-1)^{h}\binom{n}{h}\prod_{l=1}^{2 n-h-1}(N+l) &(0\le h\le n); \\ 0 &(h>n). \end{cases}\displaystyle \end{aligned}$$
Theorem 5
We have
$$\begin{aligned} &\sum_{j=0}^{\min \{h,n\}}\sum _{k=0}^{j}(-1)^{j-k}(2 n-j)_{k} \binom{n-k-1}{j-k}\prod_{l=k+1}^{2 n-j}(N+l)\cdot \frac{B_{N,h-j}}{(h-j)!} \\ &\quad = \textstyle\begin{cases} (-1)^{h}\binom{n}{h}\prod_{l=1}^{2 n-h}(N+l) &(0\le h\le n); \\ 0 &(h>n), \end{cases}\displaystyle \end{aligned}$$
(17)
and
$$\begin{aligned} &\sum_{j=0}^{\min \{h,n\}}\sum _{k=0}^{j}(-1)^{j-k}(2 n-j-1)_{k} \binom{n-k-1}{j-k}\prod_{l=k+1}^{2 n-j-1}(N+l) \cdot \frac{B_{N,h-j}}{(h-j)!} \\ &\quad = \textstyle\begin{cases} (-1)^{h}\binom{n}{h}\prod_{l=1}^{2 n-h-1}(N+l) &(0\le h\le n); \\ 0 &(h>n). \end{cases}\displaystyle \end{aligned}$$
(18)
In particular, when \(N=1\), we have the relations for the classical Bernoulli numbers.
Corollary 4
We have
$$\begin{aligned} &\sum_{j=0}^{\min \{h,n\}}\sum _{k=0}^{j}(-1)^{j-k}(2 n-j)_{k} \binom{n-k-1}{j-k}\frac{(2 n-j+1)!}{(k+1)!(2 n-h+1)!}\cdot \frac{B _{h-j}}{(h-j)!} \\ &\quad = \textstyle\begin{cases} (-1)^{h}\binom{n}{h} &(0\le h\le n); \\ 0 &(h>n), \end{cases}\displaystyle \end{aligned}$$
(19)
and
$$\begin{aligned} &\sum_{j=0}^{\min \{h,n\}}\sum _{k=0}^{j}(-1)^{j-k}(2 n-j-1)_{k} \binom{n-k-1}{j-k}\frac{(2 n-j)!}{(k+1)!(2 n-h)!}\cdot \frac{B_{h-j}}{(h-j)!} \\ &\quad = \textstyle\begin{cases} (-1)^{h}\binom{n}{h} &(0\le h\le n); \\ 0 &(h>n). \end{cases}\displaystyle \end{aligned}$$
(20)
Remark
Since
$$\begin{aligned} &\sum_{k=0}^{j}(-1)^{j-k} \frac{(2 n-j)_{k}}{(k+1)!}\binom{n-k-1}{j-k} \\ &\quad = \textstyle\begin{cases} \frac{1}{j+1}\binom{n}{j} &\text{if $j$ is even}; \\ \frac{1}{4 j}\binom{j-2}{(j-1)/2}^{-1}\binom{n-(j+1)/2}{(j-1)/2} \binom{n}{(j-1)/2} &\text{if $j$ is odd $\ge 3$}; \\ \frac{1}{2} &\text{if $j=1$}, \end{cases}\displaystyle \end{aligned}$$
we can write (19) as
$$\begin{aligned} &\sum_{j=0}^{ \lfloor \frac{h}{2} \rfloor }\frac{(2 n-2 j+1)!}{2 j+1} \binom{n}{2 j}\frac{B_{h-2 j}}{(h-2 j)!} + \frac{(2 n)!}{2}\frac{B_{h-1}}{(h-1)!} \\ &\qquad{} +\sum_{j=1}^{ \lfloor \frac{h-1}{2} \rfloor }\frac{(2 n-2 j)!}{4(2 j+1)} \binom{2 j-1}{j}^{-1}\binom{n-j-1}{j}\binom{n}{j} \frac{B _{h-2 j-1}}{(h-2 j-1)!} \\ &\quad = \textstyle\begin{cases} (-1)^{h}\binom{n}{h}(2 n-h+1)! &\text{if $1\le h\le n$}; \\ 0 &\text{if $n< h\le 2 n+1$}. \end{cases}\displaystyle \end{aligned}$$
Since
$$\begin{aligned} &\sum_{k=0}^{j}(-1)^{j-k} \frac{(2 n-j-1)_{k}}{(k+1)!} \binom{n-k-1}{j-k} \\ &= \textstyle\begin{cases} \frac{ ((j/2)! )^{2}}{(j+1)!}\binom{n}{j/2} \binom{n-j/2-1}{j/2} &\text{if $j$ is even}; \\ 0 &\text{if $j$ is odd}, \end{cases}\displaystyle \end{aligned}$$
we can write (20) as
$$\begin{aligned} &\sum_{j=0}^{ \lfloor \frac{h}{2} \rfloor }\frac{(j!)^{2}(2 n-2 j)!}{(2 j+1)!} \binom{n}{j}\binom{n-j-1}{j} \frac{B_{h-2 j}}{(h-2 j)!} \\ &\quad = \textstyle\begin{cases} (-1)^{h}\binom{n}{h}(2 n-h)! &\text{if $1\le h\le n$}; \\ 0 &\text{if $n< h\le 2 n$}. \end{cases}\displaystyle \end{aligned}$$
Here the empty summation is recognized as 0, as usual.