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Faà di Bruno's formula implies that $$d_k(n) = \sum_{m_1,\dots,m_k\geq 0\atop 1\cdot m_1+\cdots+k\cdot m_k=k} \frac{(2k)!}{m_1!\,2!^{m_1}\,m_2!\,4!^{m_2}\,\cdots\,m_k!\,(2k)!^{m_k}}\cdot (n)_{m_1+\cdots+m_k}\cdot \prod_{j=1}^k d_j(1)^{m_j},$$ where $(n)_t=n(n-1)\cdots(n-t+1)$ is the falling factorial.

Correspondingly, the coefficient of $n^\ell$ in $d_k(n)$ is given by $$\sum_{m_1,\dots,m_k\geq 0\atop 1\cdot m_1+\cdots+k\cdot m_k=k} \frac{(2k)!}{m_1!\,2!^{m_1}\,m_2!\,4!^{m_2}\,\cdots\,m_k!\,(2k)!^{m_k}}\cdot s(m_1+\cdots+m_k,\ell)\cdot \prod_{j=1}^k d_j(1)^{m_j},$$ where $s(z,t)$ are the (signed) Stirling numbers of first kind.

It remains to notice that and $d_k(1)=(-1)^{k-1}(2^{2k}-2)B_{2k}$.

ADDED. Equivalently, in terms of Bell polynomials $\mathcal{B}_{n,k}()$, we have $$[n^\ell]\ d_k(n) = \sum_{m=1}^{2k} s(m,\ell)\cdot \mathcal{B}_{2k,m}(0,d_1(1),0,d_2(1),\dots).$$

Faà di Bruno's formula implies that $$d_k(n) = \sum_{m_1,\dots,m_k\geq 0\atop 1\cdot m_1+\cdots+k\cdot m_k=k} \frac{(2k)!}{m_1!\,2!^{m_1}\,m_2!\,4!^{m_2}\,\cdots\,m_k!\,(2k)!^{m_k}}\cdot (n)_{m_1+\cdots+m_k}\cdot \prod_{j=1}^k d_j(1)^{m_j},$$ where $(n)_t=n(n-1)\cdots(n-t+1)$ is the falling factorial.

Correspondingly, the coefficient of $n^\ell$ in $d_k(n)$ is given by $$\sum_{m_1,\dots,m_k\geq 0\atop 1\cdot m_1+\cdots+k\cdot m_k=k} \frac{(2k)!}{m_1!\,2!^{m_1}\,m_2!\,4!^{m_2}\,\cdots\,m_k!\,(2k)!^{m_k}}\cdot s(m_1+\cdots+m_k,\ell)\cdot \prod_{j=1}^k d_j(1)^{m_j},$$ where $s(z,t)$ are the (signed) Stirling numbers of first kind.

It remains to notice that and $d_k(1)=(-1)^{k-1}(2^{2k}-2)B_{2k}$.

ADDED. Equivalently, in terms of Bell polynomials $\mathcal{B}_{n,k}()$, we have $$[n^\ell]\ d_k(n) = \sum_{m=1}^k s(m,\ell)\cdot \mathcal{B}_{2k,m}(0,d_1(1),0,d_2(1),\dots).$$

Faà di Bruno's formula implies that $$d_k(n) = \sum_{m_1,\dots,m_k\geq 0\atop 1\cdot m_1+\cdots+k\cdot m_k=k} \frac{(2k)!}{m_1!\,2!^{m_1}\,m_2!\,4!^{m_2}\,\cdots\,m_k!\,(2k)!^{m_k}}\cdot (n)_{m_1+\cdots+m_k}\cdot \prod_{j=1}^k d_j(1)^{m_j},$$ where $(n)_t$ is the falling factorial.

Correspondingly, the coefficient of $n^\ell$ in $d_k(n)$ is given by $$\sum_{m_1,\dots,m_k\geq 0\atop 1\cdot m_1+\cdots+k\cdot m_k=k} \frac{(2k)!}{m_1!\,2!^{m_1}\,m_2!\,4!^{m_2}\,\cdots\,m_k!\,(2k)!^{m_k}}\cdot s(m_1+\cdots+m_k,\ell)\cdot \prod_{j=1}^k d_j(1)^{m_j},$$ where $s(z,t)$ are the (signed) Stirling numbers of first kind.

It remains to notice that and $d_k(1)=(-1)^{k-1}(2^{2k}-2)B_{2k}$.

Faà di Bruno's formula implies that $$d_k(n) = \sum_{m_1,\dots,m_k\geq 0\atop 1\cdot m_1+\cdots+k\cdot m_k=k} \frac{(2k)!}{m_1!\,2!^{m_1}\,m_2!\,4!^{m_2}\,\cdots\,m_k!\,(2k)!^{m_k}}\cdot (n)_{m_1+\cdots+m_k}\cdot \prod_{j=1}^k d_j(1)^{m_j},$$ where $(n)_t=n(n-1)\cdots(n-t+1)$ is the falling factorial.

Correspondingly, the coefficient of $n^\ell$ in $d_k(n)$ is given by $$\sum_{m_1,\dots,m_k\geq 0\atop 1\cdot m_1+\cdots+k\cdot m_k=k} \frac{(2k)!}{m_1!\,2!^{m_1}\,m_2!\,4!^{m_2}\,\cdots\,m_k!\,(2k)!^{m_k}}\cdot s(m_1+\cdots+m_k,\ell)\cdot \prod_{j=1}^k d_j(1)^{m_j},$$ where $s(z,t)$ are the (signed) Stirling numbers of first kind.

It remains to notice that and $d_k(1)=(-1)^{k-1}(2^{2k}-2)B_{2k}$.

ADDED. Equivalently, in terms of Bell polynomials $\mathcal{B}_{n,k}()$, we have $$[n^\ell]\ d_k(n) = \sum_{m=1}^{2k} s(m,\ell)\cdot \mathcal{B}_{2k,m}(0,d_1(1),0,d_2(1),\dots).$$

Faà di Bruno's formula implies that $$d_k(n) = \sum_{m_1,\dots,m_k\geq 0\atop 1\cdot m_1+\cdots+k\cdot m_k=k} \frac{(2k)!}{m_1!\,2!^{m_1}\,m_2!\,4!^{m_2}\,\cdots\,m_k!\,(2k)!^{m_k}}\cdot (n)_{m_1+\cdots+m_k}\cdot \prod_{j=1}^k d_j(1)^{m_j},$$ where $(n)_t$ is the falling factorial.

It remains to notice that $d_k(1)=(-1)^{k-1}(2^{2k}-2)B_{2k}$.

Faà di Bruno's formula implies that $$d_k(n) = \sum_{m_1,\dots,m_k\geq 0\atop 1\cdot m_1+\cdots+k\cdot m_k=k} \frac{(2k)!}{m_1!\,2!^{m_1}\,m_2!\,4!^{m_2}\,\cdots\,m_k!\,(2k)!^{m_k}}\cdot (n)_{m_1+\cdots+m_k}\cdot \prod_{j=1}^k d_j(1)^{m_j},$$ where $(n)_t$ is the falling factorial.

Correspondingly, the coefficient of $n^\ell$ in $d_k(n)$ is given by $$\sum_{m_1,\dots,m_k\geq 0\atop 1\cdot m_1+\cdots+k\cdot m_k=k} \frac{(2k)!}{m_1!\,2!^{m_1}\,m_2!\,4!^{m_2}\,\cdots\,m_k!\,(2k)!^{m_k}}\cdot s(m_1+\cdots+m_k,\ell)\cdot \prod_{j=1}^k d_j(1)^{m_j},$$ where $s(z,t)$ are the (signed) Stirling numbers of first kind.

It remains to notice that and $d_k(1)=(-1)^{k-1}(2^{2k}-2)B_{2k}$.