David DiRico
Pascal’s Triangle mod(3): The number of 1’s in row r can be modeled by this function:
N1(r) = 2n-1(3m + 1)
and the number of 2’s in row r by:
N2(r) = 2n-1(3m - 1)
Where n = the number of 1’s in the ternary expansion of row r, and
m = the number of 2’s in the ternary expansion of row r.
Figure 1.1
Geometrically, our base case for this triangle is going to be in the first three rows of
Pascal’s triangle mod(3), so if we have T1 as our first triangle, then we know that two duplicates
will be underneath it and three more beneath them with the center bottom triangle inverted.
If a triangle is inverted, then all of what normally would be ones compared to the rest of the
triangle would now be 2’s and vice versa. Any triangle in Figure 1.1 with blue outlining it is
inverted. These six triangles then come to form a larger triangle called T2.
Figure 1.2
Any of the white space is some number of zeros separating each T k. (To clarify, T2 is going to be
just a 1, two duplicates of that 1, and three more duplicates with an inverse in the center of the
bottom triangles. There will be no zeros separating the triangles in T2. T1 is just the very first
triangle). So the base case works for both formulas, therefore assume that the same
relationship works up to Tk.
What makes the functions initially stated in this proof more complex than the mod 2
case is that the greater the triangle we are looking is, then the more inverses it will carry inside
it. Figure 1.1 is really T3, and there are three primary inverses – the second entry in the third
row, the blue triangle in the middle in rows 6-8, and the blue triangle in the middle in rows 1826. Furthermore, there are even some inverses in the non-inverse triangles (such as the two 2’s
in row 5), which make the counting more difficult. Therefore, this cannot be solved the same
way that mod(2) Pascal’s Triangle was solved.
In the ternary expansion of row r, having a 1 in front of the ternary expansion will
indicate that the row we are looking at is located in the first set of duplicated Tk’s. Having a 2 in
front of the ternary expansion will indicate that that the row we are looking at is located in the
second set of duplicated Tk’s (see Figure 1.3 below). Having a 2 in front will also result in there
being an inverse in that row. The place of any 2 in the ternary expansion of r will tell us which
kind of inverse triangles we are looking at. For instance, 212 means that we have an inverse
triangle that is similar to T3 and another that is similar to T1.
Let’s look at the number of 1’s in row r now (where row r is in triangle Tk):
N1(r) = 2n-1(3m + 1)
So what does this all mean? Because r is located somewhere in Tk, then adding a 1 to
the ternary expansion will have the exact same effect on r as the case proven in mod(2), except
this time, 3k-1 will be added to the row number. So, here we find that, if r1=r+3k-1:
N1(r1) = (2n-1(3m+1))(2) = 2n(3m+1)
and we have n+1 1’s in the ternary expansion and m 2’s in the ternary expansion.
Adding a 2 will have the same effect as adding a 1, in that everything will be doubled,
but there is also an inverse of Tk (Ik) that needs to be added to the construction. This means
that we double the number of 1’s and also add the number of 2’s from row r as well, since each
of the 2’s in triangle Tk is inverted to become a 1 in triangle Ik. To find which row we are in, we
must add (2)3k-1 to row r. So, here we let r2=r+(2)3k-1:
N1(r2) = (2n-1(3m+1))(2) + 2n-1(3m - 1) = 2n-1(3m)(2)+ 2n+ 2n-1(3m)- 2n-1
=2n-1(3m+1)+ 2n-1 = 2n-1(3m+1 + 1)
and we have n 1’s in the ternary expansion and m+1 2’s in the ternary expansion.
Figure 1.3
To show that r and the first new r (or r1) are exactly 3k-1 apart: The remaining rows of
the top Tk can be found by 3k-1 – r, and the first rows of the next Tk before r1 can be found by r1
– 3k-1, so the 3k-1’s cancelling out leave just r1 – r = 3k-1.
To show that r1 and the newer r (or r2) are exactly 3k-1 apart: The remaining rows of the
second Tk’s can be found by 3k-1 – r1, and the first rows of the third Tk’s and Ik before r2 can be
found by r2 – 3k-1, so the 3k-1’s cancelling out leave just r2 – r1 = 3k-1.
And if the adding of a 1 or 2 to the ternary expansion of row r adds 3 k-1 or (2)3k-1, then in
Tk+1, we will be looking at rows all relative to Tk, as shown on the right side of figure 1.3. If this
is true, then
N1(r) = 2n-1(3m + 1)■
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To see if the structure of the triangle also works for the number of 2’s in any given row,
we must check the formula: N2(r) = 2n-1(3m - 1). Although the proof is almost identical to that
of N1(r) there are some differences which must be noted. The only difference between the two
formulas is that there is a sign change inside the parentheses. What this means is that the
number of 2’s will always be less than the number of 1’s in any given row, for:
2n-1(3m + 1) > 2n-1(3m - 1).
Even if the ternary expansion consists only of 2’s (which would produce the highest
possible number of 2’s in the row) we would have (3m + 1)/2 > (3m - 1)/2.
So, if the number of 2’s is always smaller than the number of 1’s and the number of 2’s follows
the same pattern in the triangles as the number of ones, then
N2(r) = 2n-1(3m - 1)■