I came across the following expressions:

$$\begin{align} \widehat{S}_{2n} &:= \sum_{1 \leq k_1<\cdots<k_n \leq 2n} \prod_{\ell=1}^n (k_\ell-2\ell)^2, \\ \widehat{T}_{2n+1}&:=\sum_{1 \leq k_1 <\cdots <k_n \leq 2n} \prod_{\ell=1}^n (k_\ell-(2\ell+1))(k_\ell-2\ell) \end{align} $$

and I suspect that they equal the secant $S_{2n}$ and tangent $T_{2n+1}$ numbers , respectively. The secant and tangent numbers may be defined using Taylor series: $$\begin{align} \sec x &= \sum_{n=0}^\infty \frac{S_{2n}}{(2n)!} x^{2n}, \\ \tan x &= \sum_{n=0}^\infty \frac{T_{2n+1}}{(2n+1)!} x^{2n+1} .\end{align} $$

I have verified that $\widehat{S}_{2n}= S_{2n}$ and that $\widehat{T}_{2n+1}=T_{2n+1}$ for $n \leq 10$, but I couldn't find a proof for the general case. I should also mention that the transformation $k_i \mapsto m_i+i$, which maps strictly increasing sequences to non-decreasing sequences produces the nicer-looking formulas:

$$\begin{align} \widehat{S}_{2n} &:= \sum_{0 \leq m_1 \leq \cdots \leq m_n \leq n} \prod_{\ell=1}^n (m_\ell-\ell)^2, \\ \widehat{T}_{2n+1}&:=\sum_{0 \leq m_1 \leq \cdots \leq m_n \leq n} \prod_{\ell=1}^n (m_\ell-(\ell+1))(m_\ell-\ell). \end{align} $$

In search of a proof, I have tried generalizing this pattern. For example, the numbers $$\widehat{S}^{(N)}_{2n}:= \sum_{0 \leq m_1 \leq \cdots \leq m_n \leq n} \prod_{\ell=1}^n (\ell-m_\ell)^N, $$ appear to be the Taylor coefficients of the function $$f_N(x) = \frac{1}{1-\frac{1^N x}{1-\frac{2^N x}{1-\frac{3^N x}{1-\dots}}}}, $$

for any natural number $N$. That is, it seems that $$f_N(x) = \sum_{n=0}^\infty \widehat{S}^{(N)}_{2n} x^n. $$

That made me think that a proof could be obtained using a "continued-fraction-to-power-series" formula. Unfortunately, I do not know of such a formula.

I would appreciate help in confirming or denying the equalities $\widehat{S}_{2n}= S_{2n}$ and $\widehat{T}_{2n+1}=T_{2n+1}$ for all $n$. Also, a proof (or disproof) that $\widehat{S}^{(N)}_{2n}$ are indeed related to continued fractions as above would be great. Thanks!


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