Let e ( n ) be the number of partitions of n with even number of even parts and let o ( n ) denote the number of partitions with odd number of even parts. In Enumerative Combinatorics 1, it is claimed that it is easy to see that ∑ n ≥ 0 ( e ( n ) − o ( n ) ) x n = 1 ( 1 − x ) × ( 1 + x 2 ) × ( 1 − x 3 ) × ( 1 + x 4 ) × . . . . I have been racking my head over this for the past few hours, and I can't see any light.Noticed, that e ( n ) − o ( n ) = 2 e ( n ) − p ( n ) where p ( n ) is the number of partitions of n, so the above claim is equivalent to showing ∑ n ≥ 0 e ( n ) x n = 1 2 1 ( 1 − x ) ( 1 − x 3 ) ( 1 − x 5 ) . . . ( 1 ( 1 − x 2 ) ( 1 − x 4 ) . . . . + 1 ( 1 + x 2 ) ( 1 + x 4 ) . . . . . . . . ), and similarly, it is equivalent to ∑ n ≥ 0 o ( n ) x n = 1 2 1 ( 1 − x ) ( 1 − x 3 ) ( 1 − x 5 ) . . . ( 1 ( 1 − x 2 ) ( 1 − x 4 ) . . . . − 1 ( 1 + x 2 ) ( 1 + x 4 ) . . . . . . . . ), but these identities appear more difficult than the original one.

Question

Let   e  (  n  ) be the number of partitions of   n with even number of even parts and let   o  (  n  ) denote the number of partitions with odd number of even parts. In Enumerative Combinatorics 1, it is claimed that it is easy to see that       ∑          n      ≥      0        (  e  (  n  )  −  o  (  n  )  )      x    n    =      1          (      1      −      x      )      ×      (      1      +              x        2            )      ×      (      1      −              x        3            )      ×      (      1      +              x        4            )      ×      .      .      .      . I have been racking my head over this for the past few hours, and I can&#039;t see any light.Noticed, that   e  (  n  )  −  o  (  n  )  =  2  e  (  n  )  −  p  (  n  ) where   p  (  n  ) is the number of partitions of   n, so the above claim is equivalent to showing       ∑          n      ≥      0        e  (  n  )      x    n    =      1    2        1          (      1      −      x      )      (      1      −              x        3            )      (      1      −              x        5            )      .      .      .        (      1          (      1      −              x        2            )      (      1      −              x        4            )      .      .      .      .        +      1          (      1      +              x        2            )      (      1      +              x        4            )      .      .      .      .      .      .      .      .        ), and similarly, it is equivalent to       ∑          n      ≥      0        o  (  n  )      x    n    =      1    2        1          (      1      −      x      )      (      1      −              x        3            )      (      1      −              x        5            )      .      .      .        (      1          (      1      −              x        2            )      (      1      −              x        4            )      .      .      .      .        −      1          (      1      +              x        2            )      (      1      +              x        4            )      .      .      .      .      .      .      .      .        ), but these identities appear more difficult than the original one.

esperoanow · Accepted Answer

Now,      1          (      1      −      x      )      (      1      +              x        2            )      (      1      −              x        3            )      …        =  (  1  +  x  +      x    2    +  …  )  (  1  −      x    2    +      x    4    −  …  )  (  1  +      x    3    +      x    6    +  …  )  (  1  −      x    4    +      x    8    −  …  )  …Consider a term which provides a negative coefficient to       x    n  If we pick       x          2              n        2             from the second term,       x          4              n        4             from the fourth and so on,The coefficient of   −  1 is  (  −  1      )                  n        2            +              n        4            +      …      this will be negative if and only if       n    2    +      n    4    +  … is odd which comes from   o  (  n  ).The positive ones come from   e  (  n  ).Thus we must have that  ∑  (  e  (  n  )  −  o  (  n  )  )      x    n    =      1          (      1      −      x      )      (      1      +              x        2            )      (      1      −              x        3            )      …

Let e ( n ) be the number of partitions of n with even number of even parts and

Answered question

Answer & Explanation

New Questions in Precalculus