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Error in rising_factorial

asked 2016-02-16 10:59:07 +0100

Peter Luschny gravatar image

updated 2016-02-16 19:11:27 +0100 

print [rising_factorial(n,n) for n in (0..5)]
print [rising_factorial(n,n) for n in range(6)]

I think both forms should give the same result. But the output is: 

[1, 1, 6, 60, 840, 15120]
AttributeError: 'int' object has no attribute 'parent'

Edit: The same error is thrown for falling_factorial. Note that the factorial function does not assume Sage-integers. This is a confusing inconsistency in the interface, at least.

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answered 2016-02-16 14:03:57 +0100

eric_g gravatar image

This is because (i) range(6) returns a list of Python integers (int), contrary to (0..5), which is a generator of Sage integers (Integer), and (ii) the function rising_factorial is expecting Sage integers. To get the expected behavior, you should replace range(6) by srange(6) (which returns Sage integers).

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But should this function take ints and coerce them?

kcrisman gravatar imagekcrisman ( 2016-02-16 18:44:14 +0100 )edit

Thanks Eric, I know the workaround as my first line shows. But this is a blatant bug because rising_factorial(0,0) = 1 no matter what technical type of integer is used.

Peter Luschny gravatar imagePeter Luschny ( 2016-02-16 18:58:07 +0100 )edit

@kcrisman: sure, I'll open a ticket for this tomorrow.

@Peter: rising_factorial(0,0) works because Sage's preparser transforms the 0's into Sage integers, as you can see by means of the function preparse:

sage: preparse("rising_factorial(0,0)")
'rising_factorial(Integer(0),Integer(0))'
eric_g gravatar imageeric_g ( 2016-02-16 21:14:27 +0100 )edit

This is now http://trac.sagemath.org/ticket/20075

eric_g gravatar imageeric_g ( 2016-02-17 10:55:15 +0100 )edit
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answered 2016-04-15 16:55:07 +0100

eric_g gravatar image

The ticket has been merged in SageMath 7.1. So if you download the latest version of SageMath, the problem has disappeared.

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Asked: 2016-02-16 10:59:07 +0100

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Last updated: Feb 16 '16

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