$1$, $e$ and $e\,\mathrm{Ein}(1)$ are linearly independent over $\mathbb{Q}$

Let $\operatorname{Ein}(z)=\sum_{n\ge 1}\frac{(-1)^{n+1}z^{n}}{n\cdot n!}$ be the entire exponential integral and $A(z)=e^{z}\operatorname{Ein}(z)$; equivalently $A(z)=\sum_{k\ge1}H_k\,z^k/k!$ with $H_k$ the harmonic numbers (proved on the platform as formal_expEin_harmonic_coefficients). The assertion is that the three real numbers

are linearly independent over $\mathbb{Q}$: for rational numbers $a,b,c$,

Status. This is an established theorem, not a conjecture. The functions $1,\;e^{z},\;e^{z}\operatorname{Ein}(z)$ are E-functions with rational coefficients, they satisfy a first-order system $z\,y'=B(z)\,y$ with $B\in M_3(\mathbb{Q}[z])$ whose only finite singularity is $z=0$, and they are linearly independent over $\mathbb{C}(z)$ (platform theorem formal_e_functional_independence). Shidlovskii's First Fundamental Theorem (Siegel's method) then gives linear independence of the values at $z=1$ over $\mathbb{Q}$, indeed over $\overline{\mathbb{Q}}$.

Why this weaker statement is worth isolating. The algebraic-coefficient version e_values_linear_independent is currently reduced on the platform to Beukers' refined lifting theorem and ultimately to the André–Beukers zero-singularity theorem. The present statement asks only for rational coefficients and only for this one explicit rank-three system. Its proof needs no descent from $\overline{\mathbb{Q}}$, no number-field arithmetic, no André holomorphic-basis theorem and no zero-singularity theorem: only Siegel's original method for a single system — an integer Hermite–Padé construction (Siegel's lemma is exists_ne_zero_int_vec_norm_le in Mathlib), the elementary denominator arithmetic of $1/k!$ and $1/(k\cdot k!)$, the growth estimate of the remainder at $z=1$, and a non-vanishing (Shidlovskii-type) determinant argument specialised to this system.

Role in the mission. By the Hardy identity $e\operatorname{Ein}(1)=e\gamma+\delta$ (platform theorem hardy_identity), if both Euler's constant $\gamma$ and the Euler–Gompertz constant $\delta$ were rational then

would be a non-trivial rational relation. Hence this statement alone implies Aptekarev's theorem that at least one of $\gamma,\delta$ is irrational, without passing through Rivoal's transcendence disjunction. It says nothing about $\gamma$ or $\delta$ individually: every value of this E-system at $1$ involves $\gamma$ and $\delta$ only through the combination $e\gamma+\delta$.