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Here is the proposition I'm assessing:

If $n = 2^k - 1$ for $k \in \mathbb{N}$, then every entry in row $n$ of pascal's triangle is odd.

While I've seen many valid answers, I really like this answer: https://math.stackexchange.com/a/1002167/367034

However, can someone clarify for me the validity of the following claim within the answer cited above:

Now, if $k$ is either odd or even, $\displaystyle\frac{k+1}{2^n-k}$ is odd.

I understand there are stipulations on the values of $2^n$ and $k$ (i.e. $k\ne 0,2^n$) but I've been able to derive both even simplifications of this expression as well as some values for $k$ that produce non-integer values. Can someone let me know if I'm missing something here?

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ClownInTheMoon
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2 Answers2

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It seems you are correct: For small values of $k$, $\frac{k+1}{2^n-k}$ is not an integer. Let's try to fix this and show that $\binom{2^n}{k}$ is always even directly (the rest of the argument in the linked answer is fine). We have:

$$\binom{2^n}{k}=\frac{2^n!}{k!(2^n-k)!}=\frac{2^n(2^n-1)\cdots(2^n-(k-1))}{k!}=\frac{2^n}{k}\prod_{j=1}^{k-1}\frac{2^n-j}{j}$$

Let's look at how many $2$s appear in the prime decomposition of the denominator and the numerator:

The power of $2$ in the factorization of $2^n-j$ is the same as the power of $2$ in the factorization of $j$. Thus all the $2$s that appear in the numerator, from terms $2^n-j$, cancel out with all the $2$s that appear in the denominator, from terms $j$, for $1\leq j\leq k-1$. Since we are assuming $k<2^n$ then the power of $2$ in $k$ will be strictly smaller than $n$.

Therefore, there will be some $2$s left after simplifying the fraction, and so $\binom{2^n}{k}$ is even.

Luiz Cordeiro
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  • can you explain a bit more about why the prime factorization of $2^n-j$ and $j$ cancel out? I see it's true but trying to wrap my head around why this works – ClownInTheMoon Sep 24 '16 at 13:35
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    We are just looking at the exponent of $2$ in the prime factorizations, that is, the greatest $k$ for which $2^k$ divides either number. Call $k_1$ and $k_2$ the exponents for $2^n-j$ and $j$, respectively. Let's show that $k_1=k_2$. Since $2^n-j,j\leq 2^n$, we have $k_i\leq n$. Then $2^{k_2}$ divides $j$ and $2^n$, so it divides $2^n-j$ and therefore $k_2\leq k_1$. The reverse inequality is similar: $2^{k_1}$ divides $2^n-j$ and $2^n$, so it divides $j=2^n-(2^n-j)$, hence $k_1\leq k_2$. – Luiz Cordeiro Sep 24 '16 at 20:48
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    Note that the same argument can be used for any prime $p$: $\binom{p^n}{k}$ is a multiple of $p$ for $1\leq k\leq p^n-1$. – Luiz Cordeiro Sep 24 '16 at 20:48
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About the first property you mention, there is a short intuitive explanation (I don't say a true "proof").

We are looking at lines with all their coefficients equal to 1 in Pascal triangle modulo 2.

But Pascal's triangle modulo 2 is known to modelize in certain way the fractal known as Sierpinki triangle (http://mathworld.wolfram.com/SierpinskiSieve.html).

In this structure, lines where all coefficients are ones are found by the (self-repeating) homothety with ratio 2 that corresponds to the doubling $2^k \rightarrow 2^{k+1}$ if the lines had been numbered beginning at 1 instead of zeros.

Jean Marie
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