Python - object and variable errors

Question

I'm attempting to program a formula that loops until it reaches the number it is asked to loop for and give an answer to it. However, I seem to be getting variable/calling errors.

def infinitesequence(n):
    a = 0
    b = 0
    for values in range(0,n+1):
        b = b + 2((-2**a)/(2**a)+2)
        a += 1
    return b

returns

TypeError: 'int' object is not callable

while

def infinitesequence(n):
    a = 0
    for values in range(0,n+1):
        b = b + 2((-2**a)/(2**a)+2)
        a += 1
    return b

returns

UnboundLocalError: local variable 'b' referenced before assignment

What is causing this error?

What is 2((-2**a)/(2**a)+2) supposed to do then? Are you forgetting a * multiplication there? — Martijn Pieters
– Martijn Pieters, Commented Apr 11, 2015 at 11:47
In your second example you simply removed the b = 0 line; that's indeed going to lead to a new error. — Martijn Pieters
– Martijn Pieters, Commented Apr 11, 2015 at 11:48
** is the exponentiation operator and not what I am talking about. I am talking about the 2 at the start. I think you meant to multiply by 2 here. — Martijn Pieters
– Martijn Pieters, Commented Apr 11, 2015 at 11:56
Yeah, your answer is correct. I'll mark it when I can. Proves how much of a rookie I am >.< — Sundrah
– Sundrah, Commented Apr 11, 2015 at 11:59

Martijn Pieters · Accepted Answer · 2015-04-11 11:53:56Z

2

2((-2**a)/(2**a)+2) is trying to use that first 2 as a function. You are asking Python to call 2() by passing in the result of the (-2**a)/(2**a)+2 expression, and that doesn't work:

>>> 2()
Traceback (most recent call last):
  File "<stdin>", line 1, in <module>
TypeError: 'int' object is not callable

Perhaps you were forgetting to use a * multiplication operator there:

2 * ((-2 ** a) / (2 ** a) + 2)

Your UnboundLocal error stems from you removing the b = 0 line, which was not the cause of your original error.

edited Apr 11, 2015 at 11:53

answered Apr 11, 2015 at 11:48

Martijn Pieters

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