I've just finished the recent Numberphile video on Pascal's Triangle and learned about a new-to-me property of the triangle; the Fermat primes $F_0 = 3, F_1 = 5, F_2 = 17, \dots$ are encoded in it somehow!
Take the standard Pascal's Triangle $(\mod 2)$ and read out each row $r_n$ as binary, $x_n = \sum_{i=0}^n 2^{i}r_{n,i}$ then the first few rows read as 1, 3, 5, 15, 17, 51, 85, 255, 257, … These are either Fermat primes or products of Fermat primes!
I wrote a short Python Script to explore a bit. The first few rows present an elegant binary pattern, $x_n = \prod_{i=0}^n F_{i}^{r_{n,i}}$ selecting $3, 5, 3 \times 5, 17, 3 \times 17$, etc. This pattern is striking in its precision.
I wanted to know what happens around row 32/33, when we reach the Fermat non-prime $F_5 = 2^{2^5}+1 = 4294967297 = 641 \times 6700417$. The pattern breaks, and $F_5$ shows up. However, on the next few rows, we have $12884901891 = 3 \times 641 \times 6700417, 21474836485 = 5 \times 641 \times 6700417, \dots$ So this means that Fermat numbers specifically, rather than just some certain primes, are part of this overall correspondence. Additionally, we can see that the binary pattern repeats here, and it goes through all of the same permutations as earlier, but with the two new factors from $F_5$. $F_6$ also adds two more factors and repeats the pattern. I haven't yet explored higher.
I would like to know why this correspondence exists and what it might tell us about Fermat numbers. Personally, I found this jaw-dropping.
Best Answer
It is indeed true that each $x_{n}$ is a product of integers of the form $2^{2^{m}}+1$ (although not of the ones you have stated).
To prove this, we fix an $n\in\mathbb{N}$. Your definition of $x_{n}$ rewrites as $x_{n}=\sum\limits\limits_{\substack{i\in\left\{ 0,1,\ldots,n\right\} ;\\\dbinom{n}{i}\text{ is odd}}}2^{i}$.
Write $n$ in the form $n=a_{k}2^{k}+a_{k-1}2^{k-1}+\cdots+a_{0}2^{0}$ for some $k\in\mathbb{N}$ and $a_{0},a_{1},\ldots,a_{k}\in\left\{ 0,1\right\} $. (This is just the base-$2$ representation of $n$, possibly with leading zeroes.)
Lucas's theorem tells you that if $i=b_{k}2^{k}+b_{k-1}2^{k-1}+\cdots+b_{0}2^{0}$ for some $b_{0} ,b_{1},\ldots,b_{k}\in\left\{ 0,1\right\} $, then
$\dbinom{n}{i}\equiv\dbinom{a_{k}}{b_{k}}\dbinom{a_{k-1}}{b_{k-1}} \cdots\dbinom{a_{0}}{b_{0}}=\prod\limits_{j=0}^{k}\underbrace{\dbinom{a_{j}}{b_{j}} }_{\substack{= \begin{cases} 1, & \text{if }b_{j}\leq a_{j}\\ 0, & \text{if }b_{j}>a_{j} \end{cases} \\\text{(since }a_{j}\text{ and }b_{j}\text{ lie in }\left\{ 0,1\right\} \text{)}}}$
$=\prod\limits_{j=0}^{k} \begin{cases} 1, & \text{if }b_{j}\leq a_{j}\\ 0, & \text{if }b_{j}>a_{j} \end{cases} = \begin{cases} 1, & \text{if }b_{j}\leq a_{j}\text{ for all }j\text{;}\\ 0, & \text{otherwise} \end{cases} \mod 2$.
Hence, the $i\in\mathbb{N}$ for which $\dbinom{n}{i}$ is odd are precisely the numbers of the form $b_{k}2^{k}+b_{k-1}2^{k-1} +\cdots+b_{0}2^{0}$ for $b_{0},b_{1},\ldots,b_{k}\in\left\{ 0,1\right\} $ satisfying $\left( b_{j}\leq a_{j}\text{ for all }j\right) $. Since all these $i$ satisfy $i \in \left\{ 0,1,\ldots,n\right\}$ (because otherwise, $\dbinom{n}{i}$ would be $0$ and therefore could not be odd), we can rewrite this as follows: The $i \in \left\{ 0,1,\ldots,n\right\}$ for which $\dbinom{n}{i}$ is odd are precisely the numbers of the form $b_{k}2^{k}+b_{k-1}2^{k-1} +\cdots+b_{0}2^{0}$ for $b_{0},b_{1},\ldots,b_{k}\in\left\{ 0,1\right\} $ satisfying $\left( b_{j}\leq a_{j}\text{ for all }j\right) $. Since these numbers are distinct (because the base-$2$ representation of any $i\in\mathbb{N}$ is unique, as long as we fix the number of digits), we thus can substitute $b_{k}2^{k}+b_{k-1}2^{k-1}+\cdots+b_{0}2^{0}$ for $i$ in the sum $\sum\limits\limits_{\substack{i\in\left\{ 0,1,\ldots,n\right\} ;\\ \dbinom{n}{i}\text{ is odd}}}2^{i}$. Thus, this sum rewrites as follows:
$\sum\limits_{\substack{i\in\left\{ 0,1,\ldots,n\right\} ;\\ \dbinom{n}{i}\text{ is odd}}}2^{i} =\underbrace{\sum\limits_{\substack{b_{0},b_{1} ,\ldots,b_{k}\in\left\{ 0,1\right\} ;\\b_{j}\leq a_{j}\text{ for all }j} }}_{=\sum\limits_{b_{0}=0}^{a_{0}}\sum\limits_{b_{1}=0}^{a_{1}}\cdots\sum\limits_{b_{k}=0}^{a_k} }\underbrace{2^{b_{k}2^{k}+b_{k-1}2^{k-1}+\cdots+b_{0}2^{0}}}_{=\left( 2^{2^{k}}\right) ^{b_{k}}\left( 2^{2^{k-1}}\right) ^{b_{k-1}}\cdots\left( 2^{2^{0}}\right) ^{b_{0}}}$
$=\sum\limits_{b_{0}=0}^{a_{0}}\sum\limits_{b_{1}=0}^{a_{1}}\cdots\sum\limits_{b_{k}=0}^{a_{k} }\left( 2^{2^{k}}\right) ^{b_{k}}\left( 2^{2^{k-1}}\right) ^{b_{k-1} }\cdots\left( 2^{2^{0}}\right) ^{b_{0}}$
$=\left( \sum\limits_{b_{k}=0}^{a_{k}}\left( 2^{2^{k}}\right) ^{b_{k}}\right) \left( \sum\limits_{b_{k-1}=0}^{a_{k-1}}\left( 2^{2^{k-1}}\right) ^{b_{k-1} }\right) \cdots\left( \sum\limits_{b_{0}=0}^{a_{0}}\left( 2^{2^{0}}\right) ^{b_{0}}\right) $
$=\left( \sum\limits_{b=0}^{a_{k}}\left( 2^{2^{k}}\right) ^{b}\right) \left( \sum\limits_{b=0}^{a_{k-1}}\left( 2^{2^{k-1}}\right) ^{b}\right) \cdots\left( \sum\limits_{b=0}^{a_{0}}\left( 2^{2^{0}}\right) ^{b}\right) $
$=\prod\limits_{g=0}^{k}\underbrace{\left( \sum\limits_{b=0}^{a_{g}}\left( 2^{2^{g} }\right) ^{b}\right) }_{\substack{= \begin{cases} 2^{2^{g}}+1, & \text{if }a_{g}=1;\\ 1 & \text{if }a_{g}=0 \end{cases} \\\text{(since }a_{g}\in\left\{ 0,1\right\} \text{)}}}$
$=\prod\limits_{g=0}^{k} \begin{cases} 2^{2^{g}}+1, & \text{if }a_{g}=1;\\ 1 & \text{if }a_{g}=0 \end{cases} $
$=\left( \prod\limits_{\substack{g\in\left\{ 0,1,\ldots,k\right\} ;\\a_{g} =1}}\left( 2^{2^{g}}+1\right) \right) \underbrace{\left( \prod\limits _{\substack{g\in\left\{ 0,1,\ldots,k\right\} ;\\a_{g}=0}}1\right) }_{=1}$
$=\prod\limits_{\substack{g\in\left\{ 0,1,\ldots,k\right\} ;\\a_{g}=1}}\left( 2^{2^{g}}+1\right) $.
Thus, $x_{n}=\sum\limits_{\substack{i\in\left\{ 0,1,\ldots,n\right\} ;\\\dbinom{n}{i}\text{ is odd}}}2^{i}=\prod\limits_{\substack{g\in\left\{ 0,1,\ldots,k\right\} ;\\a_{g}=1}}\left( 2^{2^{g}}+1\right) $. This is clearly a product of Fermat numbers.
EDIT: This result also appears as equation (10) in Andrew Granville, Binomial coefficients modulo prime powers, 1997, where it is ascribed to Larry Roberts.