Like Erdos-Straus conjecture, another result, which is very simple to state and understand and yet a proof remains elusive, is the Collatz conjecture.

If the function $f(n)$ is applied recursively enough number of times on any positive integer $n$, then unity will always be reached. \begin{align*} f(n) &= \left\{ \begin{array}{ll} n/2 &\text{if }n \bmod2=0 \\ 3n+1 &\text{if }n \bmod2=1 \end{array} \right.\\ \strut\\ \end{align*}

Some mathematicians have commented on the difficulty level of this problem, which makes it more worthy of collaborative effort.

Paul Erdős said about the Collatz conjecture: "Mathematics may not be ready for such problems."[8] He also offered $500 for its solution.[9] Jeffrey Lagarias in 2010 claimed that based only on known information about this problem, "this is an extraordinarily difficult problem, completely out of reach of present day mathematics." -Source

I believe this contribution might fight the comment below

http://idrissaberkane.org/wp-content/uploads/2017/08/Aberkane_Syracuse_2017.pdf

Like Erdos-Straus conjecture, another result, which is very simple to state and understand and yet a proof remains elusive, is the Collatz conjecture.

If the function $f(n)$ is applied recursively enough number of times on any positive integer $n$, then unity will always be reached. \begin{align*} f(n) &= \left\{ \begin{array}{ll} n/2 &\text{if }n \bmod2=0 \\ 3n+1 &\text{if }n \bmod2=1 \end{array} \right.\\ \strut\\ \end{align*}

Some mathematicians have commented on the difficulty level of this problem, which makes it more worthy of collaborative effort.

Paul Erdős said about the Collatz conjecture: "Mathematics may not be ready for such problems."[8] He also offered $500 for its solution.[9] Jeffrey Lagarias in 2010 claimed that based only on known information about this problem, "this is an extraordinarily difficult problem, completely out of reach of present day mathematics." -Source

I believe this contribution might fulfil the comment below

http://idrissaberkane.org/wp-content/uploads/2017/08/Aberkane_Syracuse_2017.pdf

Like Erdos-Straus conjecture, another result, which is very simple to state and understand and yet a proof remains elusive, is the Collatz conjecture.

If the function $f(n)$ is applied recursively enough number of times on any positive integer $n$, then unity will always be reached. \begin{align*} f(n) &= \left\{ \begin{array}{ll} n/2 &\text{if }n \bmod2=0 \\ 3n+1 &\text{if }n \bmod2=1 \end{array} \right.\\ \strut\\ \end{align*}

Some mathematicians have commented on the difficulty level of this problem, which makes it more worthy of collaborative effort.

Paul Erdős said about the Collatz conjecture: "Mathematics may not be ready for such problems."[8] He also offered $500 for its solution.[9] Jeffrey Lagarias in 2010 claimed that based only on known information about this problem, "this is an extraordinarily difficult problem, completely out of reach of present day mathematics." -Source

Like Erdos-Straus conjecture, another result, which is very simple to state and understand and yet a proof remains elusive, is the Collatz conjecture.

If the function $f(n)$ is applied recursively enough number of times on any positive integer $n$, then unity will always be reached. \begin{align*} f(n) &= \left\{ \begin{array}{ll} n/2 &\text{if }n \bmod2=0 \\ 3n+1 &\text{if }n \bmod2=1 \end{array} \right.\\ \strut\\ \end{align*}

Some mathematicians have commented on the difficulty level of this problem, which makes it more worthy of collaborative effort.

Paul Erdős said about the Collatz conjecture: "Mathematics may not be ready for such problems."[8] He also offered $500 for its solution.[9] Jeffrey Lagarias in 2010 claimed that based only on known information about this problem, "this is an extraordinarily difficult problem, completely out of reach of present day mathematics." -Source

I believe this contribution might fight the comment below

http://idrissaberkane.org/wp-content/uploads/2017/08/Aberkane_Syracuse_2017.pdf