B +树中的最大和最小元素

Question

I am trying to calculate how much would be the maximum and minimum number of elements in a B+ Tree with 4 levels of depth. (Root + 2 middle and leafs). when n=75 (it's the order, which means each node has 75 pointers and 74 elements).

The root max amount of elements are n-1 = 74 . And the max amount of the root children are n which means 74 + 74*n . (Depth 2). each one of children have also a max of n children which is 74 + 74*n + (n^2)*74 . (Depth 3).

Which means depth 4 with the leafs are (max amount of children): 74 + 74*n + (n^2)*74 + (n^3)*74 ?

Answer 1

Okay so I have the answer, if anyone come across this:

Max - because the leafs should contain all of the values, the MAXIMUM value is: (n-1)n^3 = 74*(75^3)

Min: the root minimum is 2, and each of the two middle levels pointers minimum is floor() of n/2 and the minimum of elements is (n/2-1). which means the formula to calculate is 2*(n/2)^2*(n/2-1) = 2*(38^2)*37