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?

  • $\begingroup$ Source of this question? $\endgroup$ – Gerry Myerson Feb 23 '14 at 9:10
  • $\begingroup$ From George E.Andrews's book <The Theory of Partitions> ,you can find the example in Chapter one on page 13 $\endgroup$ – Lincoln Feb 23 '14 at 9:14
  • $\begingroup$ 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? $\endgroup$ – Gerry Myerson Feb 23 '14 at 9:16
  • $\begingroup$ Unfortunately, i can't find any similar problems related to partitions with congruence. $\endgroup$ – 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.

So, how can we convert between repeated odd numbers and even numbers twice their size? Well, we can think of the even numbers as pairs of odd numbers.

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.

  • $\begingroup$ Thank you, you really help me a lot. you give another way to solve this type of problems. We can find the partitions that they both do not contain,then construct a bijection map the excluded partition, this is a wonderful idea! $\endgroup$ – Lincoln Feb 23 '14 at 12:58

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