Partitions with numbers in their gaps by
Pełen tekst
a n1
where x > m y means that x ≥ y + m. We now fix a positive integer j and insert up to g l parts j in the gap between a nl
Q n = 1 + q j + q a2
q a3
q an+1
Q 0 n = 1 + q j + q a1
−q j 1 + q j + q an+1
P r o o f. The standard recurrences for P n and P n 0 are (1) P n = P n−1 + q j (P n−1 − P n−2 ) + q an+1
(2) P n 0 = P n−1 0 + q an+1
C a s e 2a. Consider any partition π enumerated by A n containing the part a n+1 . Since m = 2, the part with the next largest subscript is a k with k < n. Then the gapspace between a n+1 and a k is n − k − 1. Hence removing a n+1 from π leaves a partition π 0 beginning with at most n − k − 1 j’s followed by a k , k < n. By induction π 0 is enumerated by P n−2 . Hence π is enumerated by the term q an+1
C a s e 2b. Consider any partition π enumerated by A 0 n containing the part a n+1 . Since m = 1, the part with the next largest subscript is a k with k ≤ n. Then the gapspace between a n+1 and a k is n − k. Hence removing a n+1 from π leaves a partition π 0 beginning with at most n − k j’s followed by a k , k ≤ n. By induction π 0 is enumerated by P n−1 0 . Hence π is enumerated by the term q an+1
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