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The sequence OEIS A260695 can be seen as the interweaving of nonzero Hultman numbers $\mathcal{H}(n,k)$ for $k=1$ and $k=2$. Namely, $$a(n) = \mathcal{H}(n,1+(n\bmod 2)).$$

It is known that $$\mathcal{H}(n,k) = \frac{c(n+2,k)}{\binom{n+2}2},$$ where $c(\cdot,\cdot)$ are unsigned Stirling numbers of first kind.

Noticing that $c(n+2,2)=(n+1)!H_{n+1}$, we obtain the same formulae as in Mikhail's answer.

The sequence OEIS A260695 can be seen as the interweaving of nonzero Hultman numbers $\mathcal{H}(n,k)$ for $k=1$ and $k=2$. Namely, $$a(n) = \mathcal{H}(n,1+(n\bmod 2)).$$

It is known that $\mathcal{H}(n,k)$ is nonzero only when $n-k$ is odd, in which case its value is given by $$\mathcal{H}(n,k) = \frac{c(n+2,k)}{\binom{n+2}2},$$ where $c(\cdot,\cdot)$ are unsigned Stirling numbers of first kind.

Noticing that $c(n+2,2)=(n+1)!H_{n+1}$, we obtain the same formulae as in Mikhail's answer.

The sequence OEIS A260695 can be seen as the interweaving of nonzero Hultman numbers $\mathcal{H}(n,k)$ for $k=1$ and $k=2$. Namely, $$\mathrm{A260695}(n) = \mathcal{H}(n,1+(n\bmod 2)).$$

It is known that $$\mathcal{H}(n,k) = \frac{c(n+2,k)}{\binom{n+2}2},$$ where $c(\cdot,\cdot)$ are unsigned Stirling numbers of first kind.

Noticing that $c(n+2,2)=(n+1)!H_{n+1}$, we obtain the same formulae as in Mikhail's answer.

The sequence OEIS A260695 can be seen as the interweaving of nonzero Hultman numbers $\mathcal{H}(n,k)$ for $k=1$ and $k=2$. Namely, $$a(n) = \mathcal{H}(n,1+(n\bmod 2)).$$

It is known that $$\mathcal{H}(n,k) = \frac{c(n+2,k)}{\binom{n+2}2},$$ where $c(\cdot,\cdot)$ are unsigned Stirling numbers of first kind.

Noticing that $c(n+2,2)=(n+1)!H_{n+1}$, we obtain the same formulae as in Mikhail's answer.

The sequence OEIS A260695 can be seen as the interweaving of nonzero Hultman numbers $\mathcal{H}(n,k)$ for $k=1$ and $k=2$. Namely, $$\mathrm{A260695}(n) = \mathcal{H}(n,1+(n\bmod 2)).$$

It is known that $$\mathcal{H}(n,k) = \frac{c(n+2,k+1)}{\binom{n+2}2},$$ where $c(\cdot,\cdot)$ are unsigned Stirling numbers of first kind.

Noticing that $c(n+2,2)=(n+1)!H_{n+1}$, we obtain the same formulae as in Mikhail's answer.

The sequence OEIS A260695 can be seen as the interweaving of nonzero Hultman numbers $\mathcal{H}(n,k)$ for $k=1$ and $k=2$. Namely, $$\mathrm{A260695}(n) = \mathcal{H}(n,1+(n\bmod 2)).$$

It is known that $$\mathcal{H}(n,k) = \frac{c(n+2,k)}{\binom{n+2}2},$$ where $c(\cdot,\cdot)$ are unsigned Stirling numbers of first kind.

Noticing that $c(n+2,2)=(n+1)!H_{n+1}$, we obtain the same formulae as in Mikhail's answer.