Coins in a fountain

In combinatorial mathematics, coins in a fountain is an interesting problem with an even more interesting generating function. The problem is described below:

In how many different number of ways can you arrange n coins with k coins at the base such that all the coins above the base layer touch exactly two coins of the lower layer.

Solution:

The above sequence show the number of ways in which n coins can be stacked. So, for example for 9 coins we have 45 different ways of stacking them in a fountain. The number f ( n , k ) which is the solution for the above stated problem is then given by the coefficients of the polynomial of the following generating function:

Such generating function are extensively studied in

Specifically, the number of such fountains that can be created using n coins is given by the coefficients of:

This is easily seen by substituting the value of y to be 1. This is because, suppose the generating function for (1) is of the form:

∑ n ∑ k C n , k x n y k

then, if we want to get the total number of fountains we need to do summation over k. So, the number of fountains with n total coins can be given by:

∑ k C n , k x n y k

which can be obtained by substituting the value of y to be 1 and observing the coefficient of xⁿ.

Proof of generating function (1). Consider the number of ways of forming a fountain of n coins with k coins at base to be given by f ( n , k ) . Now, consider the number of ways of forming the same but with the restriction that the second most bottom layer (above the base layer) contains no gaps, i.e. it contains exactly k − 1 coins. Let this be called primitive fountain and denote it by g ( n , k ) . The two functions are related by the following equation:

This is because, we can view the primitive fountain as a normal fountain of n − k' coins with k − 1 coins in the base layer staked on top of a single layer of k coins without any gaps. Also, consider a normal fountain with a supposed gap in the second last layer (w.r.t. the base layer) in the r position. So, the normal fountain can be viewed as a set of two fountains:

1. A primitive fountain with n' coins in it and base layer having r coins.
2. A normal fountain with n − n' coins in it and the base layer having k − r coins.

So, we get the following relation:

Now, we can easily observe the generating function relation for (4) to be:

and for (3) to be:

Now, simply substituting (6) in (5) we get the relation:

F ( x , y ) = 1 1 − x y F ( x , x y ) = 1 1 − x y x y 1 − x y F ( x , x 2 y ) = ⋯ = 1 1 − x y 1 − x 2 y 1 − x 3 y ⋯