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This seems like a combinatorial problem so there might be something simple that hasn't struck me. Although I do have an idea but I am unable to proceed from it. The statement of the question is:

Prove that in any row of the pascal's triangle, the number of odd elements is $2^k$, for some $k \in N$.

I was working on a Pascal's triangle but in a binary format, where two adjacent 1's add up to 0. Something of the sort:

                 1
               1   1
             1   0   1
           1   1   1   1
         1   0   0   0   1
       1   1   0   0   1   1

It is a definitive sequence and the thing I liked about it was that you can add up the adjacent 1's in a row to get the number of odd elements, but I haven't been able to generalize this information. I was thinking along the lines of a recursive relation in polynomials whose coefficients would represent the rows of this triangle, then I can just feed 1 into the equation and inductively prove that it is a perfect power of 2.

Can someone help me out on a proof along these lines, if I'm going in the right direction here?

P.S. I know there is the modulo 2 proof but can someone help me generalize that binary pascal's triangle?

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    This will be helpful: https://www.math.hmc.edu/funfacts/ffiles/30001.4-5.shtml – 2017-01-11
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    Okay, cool. But I want to try a proof along the lines of what I was trying. That is a good solution, but I wanna see where this goes. – 2017-01-11
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    To get an inductive proof, you may want to use the results from the answers based on Kummer's Theorem. That is, try to show that the number of $1$s in a row of Pascal's Triangle is $2$ to the power of the number of $1$ bits in the binary representation of $n$. – 2017-01-11
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    You may be interested by the fact that Pascal's triangle modulo 2 generates Sierpinski's sieve (http://mathworld.wolfram.com/SierpinskiSieve.html) – 2017-01-11

3 Answers 3

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Similar idea to Andreas Caranti's answer:

Kummer's Theorem says that the number of factors of $p$ in $\binom{n}{k}$ is the number of carries when adding $k$ and $n-k$. There are no carries when adding $k$ and $n-k$ if and only if $k\veebar n-k=0$; when that is true, $n=k\lor n-k$. This means that $\binom{n}{k}$ is odd when the $1$ bits of $k$ are a subset of the $1$ bits of $n$. That is, there are $2^{\sigma_2(n)}$ odd entries in row $n$ of Pascal's Triangle, where $\sigma_2(n)$ is the sum of the bits in the binary representation of $n$.

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Here is a different direction.

Consider modulo 2 the expansion $$(a+b)^n=\sum_{k=0}^n \binom{n}{k}a^kb^{n-k}$$

Among the 4 cases that may occur:

$$a \equiv 0, b \equiv 0, \ \ \ \ \ \ a \equiv 1, b \equiv 0, \ \ \ \ \ \ a \equiv 0, b \equiv 1, \ \ \ \ \ \ a \equiv 1, b \equiv 1.$$

use the last one.

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Will delete because this is not what OP intends, but a proof via Kummer's Theorem looks rather neat to me.

Write $n, m$ in base $2$ $$ n = 2^{k} n_{k} + \dots + 2 n_{1} + n_{0}, \qquad m = 2^{k} m_{k} + \dots + 2 m_{1} + m_{0}, $$ with $n_{i}, m_{i} \in \{ 0, 1 \}$.

Then

$\dbinom{n}{m}$ is odd iff there are no carries in the sum in base $2$ $$ m + (n - m) = n, $$ that is, iff $m_{i} \le n_{i}$ for each $i$.

Given $n$, there are thus $$ \prod_{i=0}^{k} (n_{i} + 1) $$ possibilities for $m$, and each $n_{i} + 1 \in \{ 1, 2 \}$.

This also tells you what is the exponent $k$ in the question - it is the number of $1$'s in the diadic expansion of $n$, or equivalently, the number of distinct powers of $2$ that sum up to $n$.