2024 Gauss 8 Problem 11

In the sequence 1, 3, 4, 7, \ldots, each number beginning with the 4 is the sum of the two numbers before it. This means that the next number in the sequence is 4 + 7 = 11. The smallest number greater than 100 that appears in the sequence is

\textbf{(A)}\ 133\quad \textbf{(B)}\ 199\quad \textbf{(C)}\ 113\quad \textbf{(D)}\ 101\quad \textbf{(E)}\ 123

If there are no answer choices shown, enter a numerical answer.

Full credit to this problem is given to the CEMC, you may view all Gauss contests here.