The sequence a_n is determined by a_n = n + {sqrt(n)}, where {x} is the nearest integer to x and halves round up. Find the smallest k for which a_k .. a_k+2000 is a sequence of 2001 consecutive integers.

David

PS. Of course, the comment that halves round up is completely vacuous; it's just there to confuse you. The square root of an integer is either an integer or irrational (easily proved by contradiction).

David

After more effort than I'm sure should have been necessary, I get 999001 for question 4.

Does this look reasonable David?

James.

Another way of saying the question is that:

{sqrt(k)}, {sqrt(k+1)}, ..., {sqrt(k + 2000)}

are all the same. This means that the difference between any two of sqrt(k), sqrt(k+1), ..., sqrt(k + 2000) must be less than 1. (Though this condition may not be sufficient.)

So certainly:

sqrt(k + 2000) < 1 + sqrt(k)

And:

k + 2000 < 1 + k + 2sqrt(k)
1999 < 2sqrt(k)

So:

k > (1999)² /4 = (1998 + 1)² /4 = 999² + 1998/2 + 1/4

So k is greater than 999² + 999 = 1000 * 999 = 999000. So the lowest possible value it could be is 999001 (which is what James got as his final answer). We now just have to show that this value works, i.e. that:

{sqrt(999001)}, {sqrt(999002)}, ..., {sqrt(1001001)} are all the same.

To do this it suffices to demonstrate that:

i) 999.5 < sqrt(999001)
ii) sqrt(1001001) < 1000.5

The first one is equalivalent to:

(1000 - 0.5)² < 999001

i.e.

1000000 - 1000 + 0.25 < 999001 which is true.

But ii) doesn't appear to be true. It seems from my calculations that {sqrt(1001001)} is actually 1001 whereas the first few terms are only 1000. Have I made a silly mistake somewhere?

The next case to try would be the one in which each {sqrt(a_i )} is 1001. Let the minimal integer for which {sqrt(x)} = 1001. You quickly find x = 1001001 so the next thing to try is:

{sqrt(1001001)}, {sqrt(1001002)}, ..., {sqrt(1003001)}

This will be the solution if and only if {sqrt(1003001)} = 1001. So we need to investigate whether:

1003001 < 1001.5²

i.e.

1003001 < 1000000 + 3000 + 1.5²

or

1 < 1.5²

This last equation is true and so (by working backwards) it follows {sqrt(1003001)} = 1001. Therefore I think the answer is 1001001.

Oh. It seems that I made the stupid mistake of thinking that we were looking for 2000 consecutive values, when we actually want 2001.

The way I did it (or at least, the way I should have done it) is as follows.

We need to find the smallest integer r such that there are 2001 integer square roots between r - 1/2 and r + 1/2. That is equivalent to saying (r + 1/2)² - (r - 1/2)² > = 2001, which simplifies to r > = 1000.5. So we take r = 1001, and then the corresponding value of k is the smallest integer greater than (1001 - 1/2)² = (1000 + 1/2)² = 1000000 + 1000 + 1/4 which is k = 1001001.

James.