# Let $P_1(r;n)$ denote the number of partitions of n into parts that are either even and not congruent to 4r-2(mod 4r)

Let $P_1(r;n)$ denote the number of partitions of n into parts that are either even and not congruent to 4r-2(mod 4r) or odd and congruent to 2r-1 or 4r-1(mod 4r). Let $P_2(r;n)$ denote the number of partitions of n in which only even parts may be repeated and all odd parts are congruent to 2r-1 modulo 2r. Then $P_1(r;n)=P_2(r;n)$.

The problem is very complicated, first i want to use the generation function to prove it ,but r is also a variation ,it's unknown. Can we find a bijection between the two sets?

• Source of this question? – Gerry Myerson Feb 23 '14 at 9:10
• From George E.Andrews's book <The Theory of Partitions> ,you can find the example in Chapter one on page 13 – Lincoln Feb 23 '14 at 9:14
• Then I guess you're expected to use the methods of Chapter 1 to solve it. Are there any similar problems where a proof is given? – Gerry Myerson Feb 23 '14 at 9:16
• Unfortunately, i can't find any similar problems related to partitions with congruence. – Lincoln Feb 23 '14 at 9:24

The difference between a $P_1$ partition and a $P_2$ partition is that a $P_1$ partition cannot contain even numbers congruent to $-2$ (mod $4r$), while a $P_2$ partition cannot repeat odd numbers. So to find a bijection, our first hope would be that these "excluded partitions" somehow map onto each other.
The additional fact we have about both $P_1$ and $P_2$ partitions is that odd parts are always congruent to -1 (mod $2r$). $~$ Interestingly, these values are exactly 1/2 of the values which are excluded from $P_1$ partitions.
This bijection in fact maps $P_1$ and $P_2$ onto each other quite cleanly.
For $r=25$, we have that odd parts (in both $P_1$ and $P_2$) must always end with "49" or "99", and in $P_2$ the odd parts cannot be repeated, while in $P_1$ there are no parts ending in "98". So, where $P_1$ has repetitions of odd numbers, we pair them up (leaving at most one unpaired), yielding numbers ending in "98" for the $P_2$ partition. In the other direction, we split every $P_2$ number ending in "98" into two parts of half the size to get a $P_1$ partition.