It is equal to 10@(c1@(c1@(c1)c1)c1...c1 layers... )10 = chick number
eg 3 layers = c1@(c1@(c1@(c1)c1)c1)c1 (total 3 c1@('s）with 2 c1)s and 1c1 at end
4 layers =c1@(c1@(c1@(c1@(c1)c1)c1)c1)c1
etc.
N layers = total N c1@(‘s, excluding the innermost c1, N-1 c1)s, 1 c1 at end
where c1 =
10@(10@(10)10)10 = mini chick number = c1
the notation @ is defined as follows
a@b = a^(ab)b = a^...(a*b arrows)...^b
eg.
3@3 = 3^(3*3)3 = 3^(9)3 = 3^^^^^^^^^3
method: solve ab to find amount of arrows and put ab arrows between a and b
a@@b = a@b^(a@b)a@b
in order to solve a@...@b (where the amount of @s are more than 1) deconstruct it into a similar equation of a@b, except replace a and b with a@...(one less than original)...@b.
equation: x^(xy)y =x^…xy arrows…^y
in the case of a@b, x = a and y = b
in the case of a@@b = x=a@b and y = a@b（also works with a@(2)b)
in the case of a@(n)b, x=a@(n-1)b y= a@(n-1)b where a@(n)b = a@...(n @s)...b and n > 1
For the case of a@(n)b where n >1, repeat the n-1 step until you are left with only one @ between all as and bs. Then, use the first case (a@b) to simplify it to have no @s. Clearer example below
Simplification example
3@@3 = 3@3^(3@3*3@3)3@3 = 3^(9)3^(3^(9)3*3^(9)3 )^3 = (3^^^^^^^^^3) ^...(3^^^^^^^^^3*3^^^^^^^^^3 arrows)...^(3^^^^^^^^^3)
in general, if n>1,
a@(n)b = a@...(n @'s)...@b = a@(n-1)b^(a@(n-1)b*a@(n-1)b)a@b
Simplification example 2
3@(3)3 = 3@@@3 = 3@@3^(3@@3*3@@3)3@@3 = (3@3^(3@3)3@3)^(3@3^(3@3)3@3*3@3^(3@3)3@3b) (3@3^(3@3)3@3)