Google Sheets,_{(削除) 11 (削除ここまで)} 5 Bytes

Anonymous worksheet formula that takes no input and outputs into the cell which holds the formula

=4/(2

As a single formula this evaluates to a call stack that looks a little something like

=4/(2
=4/(2)
=4/2
=2
2

However when this worksheet formula is doubled this call stack evaluates down to

=4/(2=4/(2
=4/(2=4/(2)
=4/(2=4/(2))
=4/(2=2)
=4/(True)
=4/True
=4/1
=4
4

Of course, an implication of using this method is that once this is repeated more than once, at the third and all following iterations of the problem, the call stack will reach =4/(2=4) and thus evaluate down to =4/0 and throw a #DIV/0! error

-6 bytes by switching to algebra away from the =DIVIDE(4,2 formula

Google Sheets, 5 Bytes

Anonymous worksheet formula that takes no input and outputs into the cell which holds the formula

=4/(2

As a single formula this evaluates to a call stack that looks a little something like

=4/(2
=4/(2)
=4/2
=2
2

However when this worksheet formula is doubled this call stack evaluates down to

=4/(2=4/(2
=4/(2=4/(2)
=4/(2=4/(2))
=4/(2=2)
=4/(True)
=4/True
=4/1
=4
4

Of course, an implication of using this method is that once this is repeated more than once, at the third and all following iterations of the problem, the call stack will reach =4/(2=4) and thus evaluate down to =4/0 and throw a #DIV/0! error

#Google Sheets, _{(削除) 11 (削除ここまで)} 5 Bytes

Anonymous worksheet formula that takes no input and outputs into the cell which holds the formula

=4/(2

As a single formula this evaluates to a call stack that looks a little something like

=4/(2
=4/(2)
=4/2
=2
2

However when this worksheet formula is doubled this call stack evaluates down to

=4/(2=4/(2
=4/(2=4/(2)
=4/(2=4/(2))
=4/(2=2)
=4/(True)
=4/True
=4/1
=4
4

Of course, an implication of using this method is that once this is repeated more than once, at the third and all following iterations of the problem, the call stack will reach =4/(2=4) and thus evaluate down to =4/0 and throw a #DIV/0! error

-6 bytes by switching to algebra away from the =DIVIDE(4,2 formula

Google Sheets, _{(削除) 11 (削除ここまで)} 5 Bytes

Anonymous worksheet formula that takes no input and outputs into the cell which holds the formula

=4/(2

As a single formula this evaluates to a call stack that looks a little something like

=4/(2
=4/(2)
=4/2
=2
2

However when this worksheet formula is doubled this call stack evaluates down to

=4/(2=4/(2
=4/(2=4/(2)
=4/(2=4/(2))
=4/(2=2)
=4/(True)
=4/True
=4/1
=4
4

Of course, an implication of using this method is that once this is repeated more than once, at the third and all following iterations of the problem, the call stack will reach =4/(2=4) and thus evaluate down to =4/0 and throw a #DIV/0! error

-6 bytes by switching to algebra away from the =DIVIDE(4,2 formula

#Google Sheets, _{(削除) 11 (削除ここまで)} 5 Bytes

Anonymous worksheet formula that takes no input and outputs into the cell which holds the formula

=4/(2

As a single formula this evaluates to a call stack that looks a little something like

=4/(2
=4/(2)
=4/2
=2
2

However when this worksheet formula is doubled this call stack evaluates down to

=4/(2=4/(2
=4/(2=4/(2)
=4/(2=4/(2))
=4/(2=2)
=4/True
=4/1
=4
4

Of course, an implication of using this method is that once this is repeated more than once, at the third and all following iterations of the problem, the call stack will reach =4/(2=4) and thus evaluate down to =4/0 and throw a #DIV/0! error

-6 bytes by switching to algebra away from the =DIVIDE(4,2 formula

#Google Sheets, _{(削除) 11 (削除ここまで)} 5 Bytes

Anonymous worksheet formula that takes no input and outputs into the cell which holds the formula

=4/(2

As a single formula this evaluates to a call stack that looks a little something like

=4/(2
=4/(2)
=4/2
=2
2

However when this worksheet formula is doubled this call stack evaluates down to

=4/(2=4/(2
=4/(2=4/(2)
=4/(2=4/(2))
=4/(2=2)
=4/(True)
=4/True
=4/1
=4
4

Of course, an implication of using this method is that once this is repeated more than once, at the third and all following iterations of the problem, the call stack will reach =4/(2=4) and thus evaluate down to =4/0 and throw a #DIV/0! error

-6 bytes by switching to algebra away from the =DIVIDE(4,2 formula