The way to think of this is that `##`

gets replaced by `#`

in just the
same way that `#1`

gets replaced by “whatever is the first argument”.

So if you define a macro:

- snippet.latex
\newcommand\a[1]{+#1+#1+#1+}

or (using the TeX primitive `\def`

):

- snippet.latex
\def\a#1{+#1+#1+#1+}

and use it as `\a{b}`

,
the macro expansion produces “+b+b+b+”,
as most people would expect.

However, if we now replace part of the macro:

- snippet.latex
\newcommand\a[1]{+#1+\newcommand\x[1]{xxx#1}}

then `\a{b}`

will give us the rather odd
+b+`\newcommand{x}[1]{xxxb}`

so that the new `\x`

ignores its argument.

If we use the TeX primitive:

- snippet.latex
\def\a#1{+#1+\def\x #1{xxx#1}}

`\a{b}`

will expand to “+b+`\def\x b{xxxb}`

”. This
defines `\x`

to be a macro *delimited* by `b`

, and taking no
arguments, which is surely not what was intended!

Actually, to define `\x`

to take an argument, we need

- snippet.latex
\newcommand\a[1]{+#1+\newcommand\x[1]{xxx##1}}

or, using the TeX primitive definition:

- snippet.latex
\def\a#1{+#1+\def\x ##1{xxx##1}}

and `\a{b}`

will expand to
+b+`\def\x #1{xxx#1}`

because `#1`

gets replaced by “b”
and `##`

gets replaced by `#`

.

To nest a definition inside a definition inside a definition then you
need `####1`

, doubling the number of `#`

signs; and at the next level
you need 8 `#`

s each time, and so on.