What is the nth term that goes 1,2,2,3,3,3,4,4,4,4, etc

as n occurs n times

the last position of n = n(n+1)/2

last position of (n-1) = n(n-1)/2

so the position n(n-1)/2 + 1 to n(n+1)/2 contains n

kth position contains n and

n(n-1)/2 < k <= n(n+1)/2

or n(n-1) < 2k <= n(n+1)

n^2 - n < 2k <= n^2 + n

multiply by 4

4n^2 - 4n < 8k <= 4n^2 + 4n
(2n-1) ^2 - 1 < 8k <= (2n +1)^2 - 1
(2n-1) ^2 < 8k +1 <= (2n +1)^2

so n = (ceil(sqrt(8k+1) -1)/2