binary value

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One can convert this "long" to a "Sage integer", and use the method "bits".

sage: a = 189866136719308462018271159242437168532L
sage: b = ZZ(a).bits()

How many bits do we have:

sage: len(b)
128

Show the first eight bits

sage: b[:8]
[0, 0, 1, 0, 1, 0, 0, 1]

and the last eight bits

sage: b[-8:]
[0, 1, 1, 1, 0, 0, 0, 1]

Alternatively, if we want the binary representation as a p h o n e number:

sage: a = 189866136719308462018271159242437168532L
sage: bin(a)
'0b10001110110101101110001101000111101110110011101001111001010101010010100111110111010100011010100111001000110100000000110110010100'

sage: type(bin(a))
<type 'str'>

This is similar to

sage: hex(a)
'0x8ed6e347bb3a795529f751a9c8d00d94L'
sage: oct(a)
'02166556150756635171252247672432471064006624L'

We can check that this matches the bits solution from the answer of slelievre:

sage: ZZ(a).bits() == [ int(s) for s in bin(a)[:1:-1] ]
True

Of course.

Note: Repeated times i tried to post versions of this answer. (The solution was to take a shower, then i wrote the word phone in an "unexpected" way. It seems that we cannot post the string pho.. number followed by an obvious such number here.)