Count all the values

Relevant wiki: Floor Function

$\begin{array} { l r l } & \frac{n^2}{2017} - \frac{(n-1)^2}{2017} &\geq 1 \\ \Leftrightarrow & \frac{n^2-(n^2+1-2n)}{2017} &\geq 1 \\ \Leftrightarrow & \frac{2n-1}{2017} &\geq 1 \\ \Leftrightarrow & 2n-1 &\geq 2017 \\ \Leftrightarrow & n &\geq 1009\\ \end{array}$

From above, when $n \geq 1009$ its difference of a value from its previous value is greater then 1. Therefore the floor function will have a unique value when $n \geq 1009 .$ The total number of values for this case is $(2017-1009)+1=1009 .$

Conversely, when $n \leq 1008$ , then the difference between consecutive values is less than 1, hence we will obtain every integer in that range. Since $\left \lfloor \frac{1008^2}{2017} \right \rfloor = 503 ,$ each value from 0 to 503 will be obtained so there are 504 unique values.

This means the total number of unique values is $1009+504=1513.$

I've written the following program in Pyth :

K2017l{m/^d2KSK

You can try it online here . Note that Pyth is a golfing language, and its aim is to make programs as short as possible.

How does it work?

K2017l{m/^d2KSK  - Full program.

K2017            - Assign 2017 to a variable K.
       m     SK  - Map over the range [1, K], with a preassigned variable d.
         ^d2     - Square d.
        /   K    - Integer division with K.
      {          - Remove duplicate elements.
     l           - Length.
                 - Print the result implicitly.

The lazy solution (python3): len(set(x*x//2017 for x in range(1, 2018)))

Explanation (also valid code):

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# The i'th element of the sequence is implemented by f(i). "//" is integer division which is equivalent to floor(a/b)
f = lambda i: (i*i)//2017

# The full sequence is i in the range of [1, 2018)
seq = (f(i) for i in range(1, 2018))

# Finally, we need to get the number of unique numbers in the sequence. We do this by initializing a "set"
# which ignores duplicate entries.
len(set(seq))

$\sqrt{2017}=44.9$ which means that $\lfloor x^2/2017 \rfloor=0$ if $0\leq x\leq 44$

$\sqrt{2017*2}=63.5$ which means that $\lfloor x^2/2017 \rfloor\leq 1$ if $0\leq x\leq 63$ , which also means that there is 2 distinct numbersin the sequence up to x=63 (which are 0 and 1).

Now the fun part:

$\sqrt{2017*500}=1004.24$ and $\sqrt{2017*501}=1005.24$ , this means that for x>=1005 every number in the sequence is unique.

Since $\sqrt{2017*500}=1004.24$ we know there exist 500 unique numbers(including 0) for $0\leq x\leq 1004$

The entire sequence contains 500+2017-1004=1513 unique numbers.

One-liner in Python2.7:

print len({i**2/2017 for i in xrange(1,2018)})

Php code solution

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<?php  
$a = array();  
for($i=1;$i<=2017;$i++){  
    $a[] = floor(pow($i,2)/2017);  
}  
$a = array_unique($a);  
echo count($a);  //prints 1513
?>