2014 AMC 10A Problem 24

After $n$ full iterations, the number of listed terms is $4+5+\cdots+(n+3)=\frac{n(n+7)}2$. We need the largest $n$ with $\frac{n(n+7)}2<500000$. Since $996\cdot1003=998988$, after $996$ iterations there are $499494$ listed numbers. The first number listed in iteration $997$ is one more than the total of all listed and skipped numbers so far, namely $996^2+4\cdot996+1=996001$. The $500000$th listed number is the $500000-499494=506$th number of this next block, so it is $996001+505=996506$. Thus, A is the correct answer.