COMBINATORIAL EXPANSIONS OF MACDONALD AND LLT POLYNOMIALS

\quad\quad In 1987, Ian Macdonald introduced a special family of symmetric polynomials $H_{\mu}(X;q,t)$. These are now known as Macdonald polynomials, written as $H_{\mu}(X;q,t)$ $=\sum_{\lambda\vdash n}K_{\lambda,\mu}(q,t)s_{\lambda}(X)$, a sum over Schur functions $s_{\lambda}(X)$, a basis for the ring of symmetric functions. Macondald conjectured that $K_{\lambda,\mu}(q,t)\in \mathbb{N}[q,t]$, i.e., have positive coefficients. Shortly after, a more natural form of these polynomials was introduced, $\tilde{H}{\mu}(X;q,t)$. Written in the Schur basis, $\tilde{H}{\mu}(X;q,t)=\sum_{\lambda\vdash n}\tilde{K}{\lambda,\mu}(q,t)s{\lambda}(X)$ where $\tilde{K}{\lambda,\mu}(q,t)=t^{n(\mu)}K{\lambda,\mu}(q,1/t).$ In 2001, using algebraic geometry, Mark Haiman showed $\tilde{K}{\lambda,\mu}(q,t)\in\mathbb{N}[q,t]$. Since then, it has been a major open problem to find a combinatorial interpretation for $\tilde{K}{\lambda,\mu}(q,t)$. We prove a new formula for $\tilde{K}{\lambda,\mu}(q,t)$ when $\mu=(n-k-1,2,1^{k-1})$ in terms of statistic on Standard Young Tableau. Using this formula, we then prove a special case of a conjecture due to Lynne Butler in 1994 on the change of Schur coefficients from a hook shape to an augmented hook shape. \par \quad\quad In 1997, Alain Lascoux, Bernard Leclerc, and Jean-Yves Thibon introduced a new family of symmetric polynomials, now known as LLT polynomials. In 2005, Jim Haglund, Mark Haiman, and Nick Loehr showed how to write Macdonald polynomials as a sum of LLT polynomials. Thus, a combinatorial formula for Macdonald polynomials can be derived from a combinatorial formula for LLT polynomials. In 2020, Alex Abreu and Antonio Nigro showed that if $G$ is an indifference graph, then $LLT{G}(q)=\sum_{\sigma\leq\textbf{m}}(q-1)^{n-\ell(\lambda(\sigma))}q^{wt_{G}(\sigma)}e_{\lambda(\sigma)}$. Using this expansion of the LLT polynomials into the $e$-basis, we prove a combinatorial formula for the coefficients of $s_{\lambda}$ when $\lambda=(n-k,1^{k})$ or $\lambda=(n-k-1,2,1^{k-1})$.