
Responsive

Work involves Nato Phonetic Code

Named Charlie
Bad Combination
WMS 2009 cont
Recall the bijection from a path of exceedence n>0 to exceedence n1
Now I think an alternate way is as follows
To those that want to figure this out themselves(the hints are hidden you have to highlight them):
1. Find the intuitive bijection between excess 1 to excess 0
2. Does this still work for 1 to 2 if not which ones of 1 are not in the image of 2?
highlight to reveal hidden contentTake the first portion of the path that goes above the diagonal to the point where it touches the diagonal again remove the first vertical and last diagonal and stick them onto the first part of the path that doesn’t cross lastly swap those
Binary Trees find a bijection from the binary trees to the catalan paths
highlight to reveal hidden content Interpret left as horizontal right as vertical whenever you can go left if stuck go right so for the above picture go left , left ,right, right
Partitions
Problem 7 count the number of ones in each partition of n and sum do the same for the number of distinct parts prove you get the same answer.
The generating function for the first is simple (0+x+2x^2+…..)(1+x^2+…)…
This is equal to (x+x^2+…)(1+x+x^2+..)(1+x^2+..)…
The interpretation of this is simple as well for a partition use x^k in the first bracket to replace k in a partition then a partition with m distinct parts is counted m times.
Problem 10. prove p_e(n)p_o(n)=(1)^{n1}p_eo(n)
I someone knows a combinatorial proof that would be great. For generating functions the LHS is (1x+x^2x^3+..)(1x^2+x^4..)… which simplifies into
multiplying by and same for all odd powers of x simplifies into (1x)(1x^3)….
If anyone has a solution to a different problem in the set please post it here.
update May 22 added borders