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 3pth 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 npth row of Pascal's triangle is a copy of the nth row, with 's placed between consecutive entries; and that the p-1 rows immediately beneath the npth row are given by forming triangles underneath each non--zero entry of the npth 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 rth row modulo 2, with some 0's in--between, and so has twice as many odd entries as the rth 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 ): 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 .
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 functionfor each . One can easily show that this limit exists and that the function is continuous. However Trollope proved the surprising result that is nowhere differentiable. For more on such questions see .