# On the set $\{\sum_{k=1}^n \lambda_ka_k:\ a_1,\ldots,a_k\ \text{are distinct elements of}\ A\}$

For a field $$F$$ let $$p(F)=p$$ if the characteristic of $$F$$ is a prime $$p$$, and $$p(F)=+\infty$$ if $$F$$ is of characteristic zero.

In 2007 I considered the linear extension of the Erdos-Heilbronn conjecture, and conjectured (cf. Sun and Zhao - Linear extension of the Erdos-Heilbronn conjecture) that, for any nonzero elements $$\lambda_1,\dotsc,\lambda_n$$ of a field $$F$$ with $$p(F)\ne n+1$$ and a finite subset $$A$$ of $$F$$ we have $$\begin{multline*} \lvert\{\lambda_1a_1+\ldots+\lambda_n a_n:\ \text{a_1,\dotsc,a_n are distinct elements of A}\}\rvert \\\ge\min\{p(F)-\delta,\, n(|A|-n)+1\}, \end{multline*}$$ where $$\delta=1$$ if $$n=2$$ and $$\lambda_1+\lambda_2=0$$, and $$\delta=0$$ otherwise.

Motivated by the above as well as Question 316142 of mine, here I ask the following question.

QUESTION: Is my following conjecture true?

Conjecture. Let $$\lambda_1,\ldots,\lambda_n\ (n\ge3)$$ be positive integers with $$\gcd(\lambda_1,\ldots,\lambda_n)=1$$ and $$\lambda_{k+1}-\lambda_k\in\{0,1\}$$ for all $$k=1,\ldots,n-1$$. Let $$F$$ be a field with $$p(F)>n+1$$. Then, for any finite subset $$A$$ of $$F$$ with $$|A|\ge n+\delta_{n,3}$$ we have $$\begin{multline*} \biggl\lvert\biggl\{\sum_{k=1}^n\lambda_ka_k:\ \text{a_1,\dotsc,a_n are distinct elements of A}\biggr\}\biggr\rvert \\\ge\min\biggl\{p(F),\ (\lambda_1+\ldots+\lambda_n)(|A|-n)+\sum_{k=1}^{\lfloor n/2\rfloor}(n+1-2k)(\lambda_{n+1-k}-\lambda_k)+1\biggr\}. \end{multline*}$$

Now let me explain where the lower bound comes from. Suppose that $$A$$ is just the subset $$\{1,\ldots,m\}$$ of the rational field $$\mathbb Q$$. For the set $$S=\{\lambda_1a_1+\ldots+\lambda_na_n:\ \text{a_1,\dotsc,a_n are distinct elements of A}\},$$ its minimal element should be $$\sum_{k=1}^n\lambda_k(n+1-k)$$, while its maximal element should be $$\sum_{k=1}^n\lambda_k(m-n+k)$$. Note that $$\begin{multline*} \biggl\lvert\biggl\{\sum_{k=1}^n\lambda_k(n+1-k),\ \dotsc,\ \sum_{k=1}^n\lambda_k(m-n+k)\biggr\}\biggr\rvert \\=(\lambda_1+\ldots+\lambda_n)(m-n)+\sum_{k=1}^{\lfloor n/2\rfloor}(n+1-2k)(\lambda_{n+1-k}-\lambda_k)+1. \end{multline*}$$ If $$\lambda_k=k$$ for all $$k=1,\dotsc,n$$, then $$\sum_{k=1}^{\lfloor n/2\rfloor}(n+1-2k)(\lambda_{n+1-k}-\lambda_k)=\sum_{k=1}^{\lfloor n/2\rfloor}(n+1-2k)^2=\frac{n(n^2-1)}6.$$