Looking for a closed-form expression of a function

[JTHM]

Suppose I have a natural number x greater than 0, and I want to find some n such that x is greater than or equal to the sum of the natural numbers from 1 to n, and less than the sum of the natural numbers from 1 to n+1. Does anyone know of a closed-form expression for this function? If not, does anyone know of an efficient algorithm for calculating it?

[ConMan]

Well, the sum of natural numbers from 1 to n is the nth triangular number, given by T(n) = n(n+1)/2, so you're basically trying to find the inverse function T^-1(x), then taking the floor of this value. A quick bit of algebra gives:

x = T(y) = y(y+1)/2
y² + y - 2x = 0
y = (-1 + sqrt(1 + 8x))/2 <- taking the positive solution only since you want a natural number
n = floor(y) = floor((sqrt(8x + 1) - 1)/2)

[JTHM]

Thanks!