Sequence of $1, 2, 2, 3, 3, 3, 4, 4, 4, 4, 5, 5, 5, 5, 5$

Question

‘A sequence is formed by writing the integers the corresponding number of times as follows : 1, 2, 2, 3, 3, 3, 4, 4, 4, 4, 5, 5, 5, 5, 5, … What is the 800 th term in this sequence?’

Answer 1

a(n(n-1)/2 + 1 ) , ... , a(n(n+1)/2) are equal to n

put n=40 : a(781) , ... , a(820) are equal to 40

so a(800) = 40

Answer 2

This is the same as asking for (the index of) the smallest triangular number that is no less than 800 (a triangular number is a positive integer of the form $\frac{n(n+1)}{2}$).

Solving $\frac{x(x+1)}{2} = 800$ gives $x=39.50312, x=-40.50312$.

Since $\frac{39 \times 40}{2} = 780$ and $\frac{40 \times 41}{2} = 820$ we see that the 800th term of the sequence must be 40.

Answer 3

It is the OEIS sequence A002024 and the formula is $a_n=[\sqrt{2n} + 1/2].$ So $a_{800}=40.$