Let
𝑛
=
1152921504606846987
n=1152921504606846987. If we compute
(
𝑛
𝑘
)
m
o
d

2
(
k
n
​
)mod2 for
𝑘
=
0
…
7
k=0…7, which sequence do we get?

Write your answer as a simple 0/1 string. For example, if your sequence is
(
1
,
0
,
1
,
0
,
1
,
0
,
1
,
0
)
(1,0,1,0,1,0,1,0), answer
10101010
10101010.

To determine the sequence of \(\binom{n}{k} \mod 2\) for \(k = 0 \ldots 7\) with \(n = 1152921504606846987\), we need to use Lucas' theorem. Lucas' theorem gives a way to compute binomial coefficients modulo a prime number (in this case, 2) by looking at the binary representations of \(n\) and \(k\).

First, we convert \(n\) to its binary representation:
\[ n = 1152921504606846987 \]
\[ n \text{ in binary} = 100000000000000000000000000000000000000000000000000011 \]

We then compute the binomial coefficients \(\binom{n}{k} \mod 2\) for \(k = 0\) to \(k = 7\) by examining the bits of \(n\) and \(k\). 

Here are the binary representations:
- \( k = 0 \): \(000\)
- \( k = 1 \): \(001\)
- \( k = 2 \): \(010\)
- \( k = 3 \): \(011\)
- \( k = 4 \): \(100\)
- \( k = 5 \): \(101\)
- \( k = 6 \): \(110\)
- \( k = 7 \): \(111\)

Using Lucas' theorem, \(\binom{n}{k} \mod 2\) is determined by the individual bits of \(n\) and \(k\). For each bit position where \(k\) has a 1, \(n\) must also have a 1 for \(\binom{n}{k}\) to be odd (1 mod 2). If any bit position where \(k\) has a 1 but \(n\) has a 0, \(\binom{n}{k}\) is even (0 mod 2).

Let's evaluate:
- For \(k = 0\): always 1 (since \(\binom{n}{0} = 1\)).
- For \(k = 1\): The rightmost bit of \(n\) is 1.
- For \(k = 2\): The second bit from the right of \(n\) is 1.
- For \(k = 3\): Both the first and second bits from the right of \(n\) are 1.
- For \(k = 4\): The third bit from the right of \(n\) is 0.
- For \(k = 5\): The third bit from the right of \(n\) is 0.
- For \(k = 6\): The third bit from the right of \(n\) is 0.
- For \(k = 7\): The third bit from the right of \(n\) is 0.

Thus, we get the sequence: \(\binom{n}{k} \mod 2\) for \(k = 0\) to \(k = 7\) as:
\[ 1, 1, 1, 1, 0, 0, 0, 0 \]

This can be written as the binary string:
\[ 11110000 \]

If you have any questions or need further details, feel free to ask!

Here are some related questions you might be interested in:
1. How does Lucas' theorem work for prime moduli other than 2?
2. Can you show a detailed example of Lucas' theorem for a smaller number?
3. How can we compute \(\binom{n}{k} \mod 2\) for larger ranges of \(k\)?
4. What are the properties of binomial coefficients modulo a prime number?
5. How can Lucas' theorem be extended to composite moduli?

Tip: Lucas' theorem is a powerful tool for computing binomial coefficients modulo prime numbers, especially useful in combinatorial number theory and computer science for solving problems efficiently.

Explore how to determine the sequence of binomial coefficients \(\binom{n}{k} \mod 2\) for \(k = 0 \ldots 7\) using Lucas' theorem, with \(n = 1152921504606846987\). This detailed guide includes the conversion of numbers to binary, application of Lucas' theorem, and results in a binary string sequence.

Math Problem Statement

Let 𝑛

2 ( k n )mod2 for 𝑘

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