Three cases: 
1. distinct parts
2. all elements are odd
3. number of parts is odd.


We now focus the bijection between case 1 and case 2.




# Mathematica code 🏗




```
distinctPartitionsOf13 = (n |-> Select[IntegerPartitions[n], DuplicateFreeQ[#] &])[13];
oddElementPartitionsOf13 = IntegerPartitions[13, Infinity, Range[1, 13, 2]];
oddPartsPartitionsOf13 = (n |-> Select[IntegerPartitions[n], OddQ@Length@# &])[13];

Length /@ {distinctPartitionsOf13, oddElementPartitionsOf13, 
 oddPartsPartitionsOf13} (* {18,18,52} *)
```


### distinctToOddElement



```
(* Helper function to split even numbers until no evens remain *)
splitEven[n_] := 
 If[EvenQ[n], Flatten[{splitEven[n/2], splitEven[n/2]}], n]

(* Function to convert a partition with distinct parts to odd parts *)
distinctToOddElement[partition_List] := Flatten[splitEven /@ partition]

generatedOddElementPartitionsOf13 = distinctToOddElement/@ distinctPartitionsOf13;

Sort@Map[Sort, generatedOddElementPartitionsOf13 ] == 
 Sort@Map[Sort, oddElementPartitionsOf13 ]

```

### oddElementToDistinct


```
TODO
```




# Reference


Use the method found on OEIS [A000009][2].



Bijection: given n = L1\*1 + L2\*3 + L3\*5 + L7\*7 + ..., a partition into odd parts, write each Li in binary, Li = 2^a1 + 2^a2 + 2^a3 + ... where the aj's are all different, then expand n = (2^a1 * 1 + ...)\*1 + ... by removing the brackets and we get a partition into distinct parts. For the reverse operation, just keep splitting any even number into halves until no evens remain.


 [1]: https://mathworld.wolfram.com/FerrersDiagram.html
 [2]: https://oeis.org/A000009