Calculate \(\sqrt{n}\) and \(t_n\) for the first \(20\) or so terms in the sequence.

So

• \(t_n=1\) for \(2\) values of \(n\),
• \(t_n=2\) for \(4\) values of \(n\),
• \(t_n=3\) for \(6\) values of \(n\),
• \(t_n=4\) for \(8\) values of \(n\).

Conjecture: \(t_n=k\) for \(2k\) values of \(n\).

Note that \(\sqrt{2020} \approx 44.94\) and so \(t_{2020}=45\).

This means that before this point in the sequence, we have included all terms with \(t_{n} \leq 44\).

According to our conjecture, the number of terms with \(t_n \leq 44\) should be

Note that \(\sqrt{1980}\approx 44.497\) giving \(t_{1980}=44\), and \(\sqrt{1981} \approx 44.508\) giving \(t_{1981} = 45\).

Since \(t_{1981}=t_{2020}=45\), then each of the terms from \(t_{1981}\) to \(t_{2020}\) equals \(45\).

Therefore, there are \(40\) terms that equal \(45\).

Thus, the required sum equals

where there are \(44\) copies of \(2\).

Therefore, the sum equals \(88\frac{8}{9}\).

Lastly, we prove that for each positive integer \(k\), there are \(2k\) terms \(t_n\) that equal \(k\).

In order to have \(t_n=k\), we need \(k - \frac{1}{2} \leq \sqrt{n} < k+\frac{1}{2}\) (in other words, \(\sqrt{n}\) needs to round to \(k\)).

Since \(n\) and \(k\) are positive,

• \(k-\frac{1}{2} \leq \sqrt{n}\) is equivalent to \((k-\frac{1}{2})^2 \leq n\), and
• \(\sqrt{n} < k+\frac{1}{2}\) is equivalent to \(n < (k+\frac{1}{2})^2\).

Therefore, we need

Since \(n\) is an integer,

There are \((k^2+k)-(k^2-k+1)+1=2k\) such values of \(n\), as required. \(\square\)