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Both questions are answered in the 2011 Ph.D. thesis of G.A. Kalugin:

$$\alpha_{n,k}=\sum_{m=0}^k\frac{1}{m!}{{2n-1}\choose{k-m}}\sum_{q=0}^m(-1)^q (q+n)^{m+n-1}.$$ There are equivalent representations in terms of Stirling numbers or Bernoulli polynomials.

In section 4.2.3 Kalugin proves that the coefficients are positive, unimodal, and log-concave.

Both questions are answered in the 2011 Ph.D. thesis of G.A. Kalugin, see also arXiv:1011.5940

$$\alpha_{n,k}=\sum_{m=0}^k\frac{1}{m!}{{2n-1}\choose{k-m}}\sum_{q=0}^m(-1)^q (q+n)^{m+n-1}.$$ There are equivalent representations in terms of Stirling numbers or Bernoulli polynomials.

In section 4.2.3 Kalugin proves that the coefficients are positive, unimodal, and log-concave.

Both questions are answered in the 2011 Ph.D. thesis of G.A. Kalugin:

$$\alpha_{n,k}=\sum_{m=0}^k\frac{1}{m!}{{2n-1}\choose{k-m}}\sum_{q=0}^m(-1)^q (q+n)^{m+n-1}.$$ There are equivalent representations in terms of Stirling numbers or Bernoulli polynomials. Kalugin also proves that the coefficients are positive, unimodal, and log-concave.

Both questions are answered in the 2011 Ph.D. thesis of G.A. Kalugin:

$$\alpha_{n,k}=\sum_{m=0}^k\frac{1}{m!}{{2n-1}\choose{k-m}}\sum_{q=0}^m(-1)^q (q+n)^{m+n-1}.$$ There are equivalent representations in terms of Stirling numbers or Bernoulli polynomials.

In section 4.2.3 Kalugin proves that the coefficients are positive, unimodal, and log-concave.

Both questions are answered in the Ph.D. thesis of Kalugin:

$$\alpha_{n,k}=\sum_{m=0}^k\frac{1}{m!}{{2n-1}\choose{k-m}}\sum_{q=0}^m(-1)^q (q+n)^{m+n-1}.$$

Both questions are answered in the 2011 Ph.D. thesis of G.A. Kalugin:

$$\alpha_{n,k}=\sum_{m=0}^k\frac{1}{m!}{{2n-1}\choose{k-m}}\sum_{q=0}^m(-1)^q (q+n)^{m+n-1}.$$ There are equivalent representations in terms of Stirling numbers or Bernoulli polynomials. Kalugin also proves that the coefficients are positive, unimodal, and log-concave.