From 46d23f5d8b8a55956db561dea5ad05afb47dc3ed Mon Sep 17 00:00:00 2001
From: Youssef <>
Date: Fri, 21 Jun 2024 09:31:25 +0200
Subject: [PATCH] simple improvements

---
 _posts/2024-04-23-domino-and-tromino-tiling.markdown | 4 ++--
 1 file changed, 2 insertions(+), 2 deletions(-)

diff --git a/_posts/2024-04-23-domino-and-tromino-tiling.markdown b/_posts/2024-04-23-domino-and-tromino-tiling.markdown
index 3be9da9..1085f6d 100644
--- a/_posts/2024-04-23-domino-and-tromino-tiling.markdown
+++ b/_posts/2024-04-23-domino-and-tromino-tiling.markdown
@@ -33,14 +33,14 @@ For $n = 4$, there are 11 ways:
 
 ![](/images/4-board_tiling.svg)
 {: refdef}
-*Figure 3: Possible tilings of a $2 \times 4$ board - 11 in total* 
+*Figure 3: Possible tilings of a $2 \times 4$ board (11 in total)* 
 {: refdef}
 ### A bit of maths
 
 Now, if we let $T_k$ be the number of ways of tiling a $2 \times k$ board, we can think of $T_k$ in terms of smaller instances (i.e. in terms of $T_{k-i}$ where $i \in 1, \dots, k-1$).
 
 This leads us to the following recursive formula:
-$T_k = T_{k-1} + T_{k-2} + 2 \times T_{k-3} + 2 \times T_{k-4} + \dots + 2 \times T_1 :(1)$
+$T_k = T_{k-1} + T_{k-2} + 2 \times (T_{k-3} + \dots + T_1) :(1)$
 
 $\space \space \space \space = T_{k-1} + T_{k-2} + 2 \times \sum_{i=3}^{k-1}T_{k-i} :(2)$