# Generating function of Young Diagram from a given semiperimeter

so my question is:

What is the generating function for the number of Young diagrams of a given semiperimeter?

My approach: knowing that there exists a diagram with zero boxes, $$a_0=1$$$$a_1=2$$ $$...$$ so I got the sequence $$A(q)=2^0,2^1,...,2^q$$ and that corresponds to a sequence of semiperimeters, $$1,2,...,n$$ which gives a straightforward polynomial $$f(q) = 1/(1-2q).$$ Is it right? Thank you.

• What Young diagram has perimeter 2? – Alien May 14 '20 at 4:20
• None, but if it has semiperimeter $2$ the diagram can be written with two boxes. – Luís Felipe May 14 '20 at 4:34
• Then why do you have $$a_1=2$$? – Alien May 14 '20 at 4:35
• Because the young diagram whose semiperimeter is $1$, is represented by $2$ boxes and so on. – Luís Felipe May 14 '20 at 4:46
• You can't have a young diagram whose semiperimeter is 1. I challenge you to draw one. I'm assuming by semiperimeter you mean half the perimeter. – Alien May 14 '20 at 4:55

I believe you meant to write $$A(q)=1,0,1,2,4,8,16,...$$ and this is true. We will demonstrate on Young Tableaus with semiperimeter 5 and the same process can be repeated for all $$n\geq 2$$.
The first edge must be horizontal from left to right and the last edge must be vertical for a non-degenerate Tableau, so the number of possible Tableaus in the nxm Rectangular hull is the number of ways to order n-1 vertical edges and m-1 horizontal edges or $${m + n - 2 \choose m-1}$$ It is now easy to see that the number of Young Tableaus of semiperimeter 5 becomes the combinatorial sum $${3 \choose 0}+{3\choose 1}+{3\choose 2}+{3\choose 3}=2^3$$ since $$m+n=5$$.
The same process can be generalized for all $$n\geq 2$$ with $$a_n=2^{n-2}$$, and the special cases $$a_0=1$$ and $$a_1=0$$.
This gives the generating function $$A(x)=\frac{(1-x)^2}{1-2x}$$ or $$\frac{1-2x+x^2}{1-2x}$$ or $$1+\frac{x^2}{1-2x}$$