So I am looking at ordered pairs: (1,0), (2,3), (3,9), (4,18) , all of the form (x,y)
the second differences of the y values is constant, suggesting a quadratic relationship.
let y = ax^2 + bx + c
at (1,0) ---> a + b + c = 0 ,#1
at (2,3) ---> 4a + 2b + c = 3 , #2
at (3,9) ---> 9a + 3b + c = 9 , #3
subtract #1 from #2 ----> 3a + b = 3
subtract #2 from #3 ----> 5a + b = 6
now subtract those two:
2a = 3
a = 3/2
sub into 3a + b = 3 to find b = -3/2
subbing those back into a+b+c=0 , we get c = 0
so y = (3/2)x^2 - (3/2)x = 3x(x - 1)/2
testing for the data value we did not use,
if x = 4, y = 12(3)/2 = 18 , ok then!!!
when x = 9 , (nine children)
y = 27(8)/2 = 108 <------ number of interactions for 9 children