This week, we have a guest post written by Alexander B. on generating functions, which are useful tools for solving advanced counting problems.

How would you use generating functions to solve the following?

>

How many non-negative integer solutions are there to \( a + b + 2c = n \) for any positive integer \(n\)?

Share a question which can be approached using Generating Functions.

If you are interested in contributing a post, please contact me at Calvin@Brilliant.org.

## Comments

Sort by:

TopNewestAfter you finish number 1, try this:

How many non-negative integer solutions are there to \(a+b+2c=n\) for any positive integer \(n\)

for distinct \(a, b, c\)? – Logan Dymond · 3 years, 9 months agoLog in to reply

– Mark Hennings · 3 years, 9 months ago

If \(d_n\) is the number of solutions with \(a=b\), then \[ d_n \; = \; \left\{ \begin{array}{lcl} \tfrac12n+1 & \quad & n \mbox{ even} \\ 0 & \quad & n \mbox{ odd} \end{array}\right. \] If \(e_n\) is the number of solutions with \(a=c\) or \(b=c\) then \[ e_n \; = \; 1 + \big\lfloor \tfrac13n\big\rfloor \] If \(f_n\) is the number of solutions with \(a=b=c\) then \[ f_n \; = \; \left\{ \begin{array}{lcl} 1 & \quad & 4\,|\,n \\ 0 & \quad & 4 \not| n \end{array} \right. \] and the number of solutions with \(a,b,c\) distinct is \[ c_n - d_n - 2e_n + 2f_n \] where \(c_n = \big\lfloor \tfrac14(n+2)^2\big\rfloor\) is as in my other post.Log in to reply

The number of solutions, \(c_n\), is the coefficient of \(x^n\) in the expansion of \[ \frac{1}{(1-x)^2(1-x^2)} \; = \; \frac{1}{(1-x)^3(1+x)} \; = \; \left(\sum_{j=0}^\infty {j+2 \choose 2}x^j\right)\left(\sum_{k=0}^\infty (-1)^kx^k\right) \] and so \[ \begin{array}{rcl} c_n & = & \sum_{m=0}^n (-1)^{n-m}{m+2 \choose 2} \; = \; \tfrac12(-1)^n\sum_{m=0}^n(-1)^m(m+1)(m+2) \\ & = & \tfrac14(-1)^n\sum_{m=0}^n (-1)^m\big[(m+1)^2+(m+2)^2-1\big] \\ & = & \tfrac14(-1)^n\left[\sum_{m=0}^n (-1)^m(m+1)^2 - \sum_{m=1}^{n+1}(-1)^m(m+1)^2 - \sum_{m=0}^n(-1)^m\right] \\ & = &\tfrac14(-1)^n\left[1 + (-1)^n(n+2)^2 - \sum_{m=0}^n(-1)^m\right] \; = \; \tfrac14\big[(n+2)^2 - \varepsilon_n\big] \end{array} \] where \[ \varepsilon_n \; = \; \left\{ \begin{array}{lcl} 0 & \qquad & n \mbox{ even} \\ 1 & \qquad & n \mbox{ odd} \end{array} \right. \] and hence \[ c_n \; = \; \left \lfloor \tfrac14(n+2)^2 \right\rfloor \] – Mark Hennings · 3 years, 9 months ago

Log in to reply

Awesome..Till now, I was completely unknown to this concept,(I had seen it in problems but never understood )but now I am very interested in it...I am very happy now..thnx Alexander and Calvin. – Kishan K · 3 years, 9 months ago

Log in to reply

Can you use generating functions when you are picking 2 or more

distinctelements from one set? – Michael Tong · 3 years, 9 months agoLog in to reply

What would it look like for picking 3 distinct objects? – Calvin Lin Staff · 3 years, 9 months ago

Log in to reply

– Mike Kong · 3 years, 9 months ago

If \(f(x) = x^4 + x^5 + x^6 + x^7\) then it would be \((f(x))^3 - x^4 f(x^2) - x^5 f(x^2) - x^6f(x^2) - x^7f(x^2)\)?Log in to reply

Your formula looks like \( f(x)^3 - f(x) f(x^2) \).

If it works, why does it work?

If it doesn't work, why doesn't it work? – Calvin Lin Staff · 3 years, 9 months ago

Log in to reply

To your edit-- It would be \(f(x)^3 - f(x^3) - \) some other things to get rid of the products in which only a pair are the same. I'll think about it. – Michael Tong · 3 years, 9 months ago

Log in to reply

I don't think your example works.

Hint:Just because you learnt an advanced technique, doesn't mean that you should forget your basics. – Calvin Lin Staff · 3 years, 9 months agoLog in to reply

– Michael Tong · 3 years, 9 months ago

I know mine doesn't work, my sentence comes out weird in latex but what I meant was "this minus some other things." Your first one works because you take the product and then subtract the case when two of the same element is being multiplied by each other, but when you scale up to three you have to worry about all three of them being the same as well as them being not pairwise distinct (e.g. \(x^4 \times x^4 \times x^7\)). However, while a term being multiplied three times occurs only once, the situation where two terms plus another distinct term occurs thrice. This makes me come to the formula of \(f(x)^3 - 3f(x)f(x^2) + 2f(x^3)\). It's multiplying by three, then taking out three times the pairwise non-distinct as well as the triple non-distinct, and then adding two of the triple non-distincts back in since that only occurs once in the original expansion of \(f(x)^3\). Though I'm a bit doubtful that this one works either..Log in to reply

Principle of Inclusion and Exclusion.

To elaborate on my hint, use theThis tells you why the formula looks the way it does (and also why it gets ugly quickly). Interestingly, it only involves terms of the form \( f(x^n) \), so if you have a simple description of \( f(x) \), then you might be able to get a simple description for the generating function of distinct elements. – Calvin Lin Staff · 3 years, 9 months ago

Log in to reply