Opened 9 years ago
Last modified 9 years ago
#11411 closed defect
some q binomial coefficients should be zero — at Version 5
Reported by: | chapoton | Owned by: | sage-combinat |
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Priority: | minor | Milestone: | sage-4.7.2 |
Component: | combinatorics | Keywords: | |
Cc: | chapoton, sage-combinat | Merged in: | |
Authors: | Reviewers: | ||
Report Upstream: | N/A | Work issues: | |
Branch: | Commit: | ||
Dependencies: | Stopgaps: |
Description (last modified by )
I have found the following behaviour:
sage: import sage.combinat.q_analogues as qa sage: qa.q_binomial(2,-1) 1/(q^2 + q + 1) sage: binomial(2,-1) 0 sage: qa.q_binomial(2,3) 1/(q^2 + q + 1) sage: binomial(2,3) 0
I think these q-binomials should rather be zero. The q-binomial is based on the q-factorial, where one finds the following behaviour
sage: [qa.q_factorial(-i) for i in range(6)] [1, 1, 1, 1, 1, 1]
This seems to be rather wrong, as the factorial itself is infinite for negative integers.
The patch restricts the q-factorial to nonnegative arguments and defines q-binomials outside of the correct range to be zero.
Apply:
Change History (6)
Changed 9 years ago by
comment:1 Changed 9 years ago by
- Status changed from new to needs_review
comment:2 Changed 9 years ago by
- Description modified (diff)
comment:3 Changed 9 years ago by
- Cc sage-combinat added
comment:4 Changed 9 years ago by
- Description modified (diff)
comment:5 Changed 9 years ago by
- Description modified (diff)
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