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A sequence of natural numbers is constructed by listing the first $4$, then skipping one, listing the next $5$, skipping $2$, listing $6$, skipping $3$, and, on the $n$th iteration, listing $n+3$ and skipping $n$. The sequence begins $1,2,3,4,6,7,8,9,10,13$. What is the $500,\!000$th number in the sequence ?

Source of the problem
American Mathematics Competition
Topic

Number Theory, Sequences

7/10

Suggested Book
Challenges and Thrills of Pre-College Mathematics

You could give it a thought first…are you sure you really need a hint ?

Stuck…? Well, don’t worry. The key to solving this problem is not thinking too much about the skips. We can start with natural numbers, from 1 to 500,000. So, a useful strategy could be to find how many numbers we have actually skipped, n and then add them back accordingly.  So, now could you take things on from here ?

If you’re a tad bit doubtful of where we’re heading even now, well no problem. Clearly, we can say 999.(1000) / 2   < 500,000 < 1000.(1001) / 2 So, now can you find out how many blocks of gaps we have in the sequence ?

Now see, finding the blocks of gaps easy ! There’s just one small thing you would have to recall. We began the count from 4…so now, the number of skipped blocks in the sequence = 999 – 3 = 996.  Now to find n, from the number of blocks , we have =  (996.997) / 2 = 496,506 This stands for the number of numbers we skipped. Now concluding this is fairly easy…could you try that out yourself ?

What remains for us to do is to add back those skipped numbers to the count, 500,000 to obtain the final answer. That gives us = 500000 +496506 = 996506

And we are done !

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