Title: On the Uniqueness of Ein(1) among Linear Combinations of the Euler-Mascheroni and Euler-Gompertz Constants Show Chinese title

Title: 关于Ein(1)在欧拉-马歇罗尼常数和欧拉-贡普茨常数线性组合中的唯一性

Abstract: From a well-known equation of Hardy, one can derive a simple linear combination of the Euler-Mascheroni constant ($\gamma=0.577215\ldots$) and Euler-Gompertz constant ($\delta=0.596347\ldots$): $\gamma+\delta/e=\textrm{Ein}\left(1\right)$. Although neither $\gamma$ nor $\delta$ is currently known to be irrational, this linear combination has been shown to be transcendental (by virtue of the fact that it appears as an algebraic point value of a particular E-function). Moreover, both pairs ($\gamma$,$\delta$) and ($\gamma$,$\delta/e$) are known to be disjunctively transcendental. In light of these observations, we investigate the impact of the coefficient $\alpha$ in combinations of the form $\gamma+\alpha\delta$, and find that $\alpha=1/e$ is the unique coefficient value such that canonical Borel-summable divergent series for $\gamma$ and $\delta$ can be linearly combined to force conventional convergence of the resulting series. We further indicate how this uniqueness property extends to a sequence of generalized linear combinations, $\gamma^{\left(n\right)}+\alpha\delta^{\left(n\right)}$, with $\gamma^{\left(n\right)}$ and $\delta^{\left(n\right)}$ given by (ordinary and conditional) moments of the Gumbel(0,1) probability distribution. Show Chinese abstract

Abstract: 从哈代的一个已知方程出发，可以推导出欧拉-马斯刻若尼常数（$\gamma=0.577215\ldots$）和欧拉-贡珀茨常数（$\delta=0.596347\ldots$）的一个简单线性组合：$\gamma+\delta/e=\textrm{Ein}\left(1\right)$。尽管目前尚不知晓$\gamma$或$\delta$是否为无理数，但这一线性组合已被证明是超越数（由于它作为特定E函数的代数点值而出现）。 此外，两对($\gamma$,$\delta$)和($\gamma$,$\delta/e$)已被知为可析超越的。 鉴于这些观察，我们研究系数$\alpha$在形式为$\gamma+\alpha\delta$的组合中的影响，并发现$\alpha=1/e$是唯一的系数值，使得对于$\gamma$和$\delta$的规范 Borel-可求和发散级数可以线性组合以强制结果级数的常规收敛。 我们进一步说明这个唯一性性质如何扩展到一个广义线性组合序列，$\gamma^{\left(n\right)}+\alpha\delta^{\left(n\right)}$，其中$\gamma^{\left(n\right)}$和$\delta^{\left(n\right)}$由 Gumbel(0,1) 概率分布的（普通和条件）矩给出。