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A beautiful aspect of Pascal's triangle modulo

Again, underneath each of these **1**'s we form a triangle whose entries
are the same as those of Pascal's triangle , while underneath
the **2** we form a triangle whose entries are twice that in Pascal's
triangle .
These three triangles meet in the **3p**th row, which thus
has **1**'s on either end, **3**'s at one--third and two--thirds of the way
across and **0**'s everywhere else.
Now underneath each of the **1**'s we again form a triangle whose entries
are the same as those of Pascal's triangle , while underneath
the **3**'s we form a triangle whose entries are three times
that in Pascal's triangle .

Continuing this process, we see that the **np**th row of Pascal's triangle
is a copy of the **n**th row, with
's placed between consecutive entries;
and that the **p-1** rows immediately beneath the **np**th row are given by
forming triangles underneath each non--zero entry of the **np**th row
(say, ),
that are times Pascal's triangle .
Thus ,
so that Lucas' Theorem may be viewed as a result about automata with
**p** possible states !

Wolfram gave an elegant proof of Glaisher's Theorem (that the number of odd
entries in a given row of Pascal's triangle is a power of **2**),
via the following induction hypothesis: For each , rows to
modulo **2** are given by taking two copies of rows **0** to
of Pascal's triangle, modulo **2**, side--by--side, and filling
the space in--between with **0**'s; moreover Glaisher's result holds for
each of these rows.
For **n=1** we observe this by computation.
For note that row must be all **1**'s
so that row has **1**'s on either end with **0**'s all the way in--between.
Thus, underneath each of these **1**'s we obtain a triangle whose entries
are the same as those of Pascal's triangle, and the triangles don't meet
until after the th row.
Therefore the th row () modulo **2** is just two
copies of the **r**th row modulo **2**, with some **0**'s in--between,
and so has twice as many odd entries as the **r**th row; this
completes the proof.

Also, as row is two copies of row **r**, whose first entries
are seperated by **0**'s, thus Roberts' integer

The above approach has a further pretty consequence (see also [15]):
If we cut Pascal's triangle modulo **p** into subtriangles whose boundaries
have entries, in the obvious way (that is, with rows **0** to
in the first such triangle, then rows to cut into
three subtriangles, two outer and one inner inverted triangle, etc. etc.),
then any given subtriangle is exactly the sum of the
two adjacent subtriangles, in the row of subtriangles immediately above.
In other words these subtriangles obey the same addition law as Pascal's
triangle itself. The behaviour of Pascal's triangle modulo higher powers of
**p** is somewhat more complicated, but still follows certain rules which
are discussed in [8].

Finally we mention a result of Trollope (1968): \
Let .
A typical integer has digits, half of
which one expects to be **1**'s, so that should be
approximately . Therefore, we compare with
, when and have the same
fractional part, by considering the function

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