Pascal's Pyramid

Let $A(x,y,z)$ be the number in column $x$ , row $y$ of layer $z$ . Then $A(x,y,z) = \binom {z-1}{x-1}\binom{z-1}{y-1}.$ It follows then that $A(3,4,100) = \binom{99}{2}\binom{99}{3} = \frac{99\cdot 98\cdot 99\cdot 98\cdot 97}{2\cdot 6} = 33\cdot 49\cdot 99\cdot 98\cdot 97 = \boxed{760\,874\,499}.$

The idea for this solution comes from observing that the rule for determining the numbers is a 3D-version of the rule for Pascal's triangle.

In fact, the pyramid is the product, both geometrically and numerically, of two copies of Pascal's triangle.

Proof by induction : Suppose the statement holds true for all blocks in layer $z$ . Then we are told that $A(x,y,z+1) = A(x-1,y-1,z) + A(x-1,y,z) + A(x,y-1,z) + A(x,y,z);$ with the induction assumption, $A(x,y,z+1) = \binom{z-1}{x-2}\binom{z-1}{y-2} + \binom{z-1}{x-1}\binom{z-1}{y-2} + \binom{z-1}{x-2}\binom{z-1}{y-1} + \binom{z-1}{x-1}\binom{z-1}{y-1} \\ = \left\{\binom{z-1}{x-2} + \binom{z-1}{x-1}\right\}\left\{\binom{z-1}{y-2} + \binom{z-1}{y-1}\right\},$ which with the familiar addition property of binomial coefficients becomes $= \binom{z}{x-1}\binom{z}{y-1}.$

It can proved inductively that the block in column $c$ , row $r$ , and layer $n$ would have the number ${n - 1 \choose c - 1}{n - 1 \choose r - 1}$ written on it (see proof below). Therefore, the block in column $3$ , row $4$ , and layer $100$ would have the number ${100 - 1 \choose 3 - 1}{100 - 1 \choose 4 - 1} = \boxed{760874499}$ written on it.

To prove that each block in column $c$ , row $r$ , and layer $n$ would have the number ${n - 1 \choose c - 1}{n - 1 \choose r - 1}$ written on it, we must examine each type of block, namely the non-edge blocks, the left edge blocks, the back edge blocks, the right edge blocks, the front edge blocks, and the corner blocks, and show that the equation holds for the first case and for each case after it.

The first non-edge block appears on layer $n = 3$ , column $c = 2$ , and row $r = 2$ , and is $4$ , and ${3 - 1 \choose 2 - 1}{3 - 1 \choose 2 - 1} = 4$ . All non-edge blocks in layer $n$ , column $c$ , and row $r$ are the sum of the $4$ blocks above it, which is ${n - 2 \choose c - 1}{n - 2 \choose r - 1} + {n - 2 \choose c - 2}{n - 2 \choose r - 1} + {n - 2 \choose c - 1}{n - 2 \choose r - 2} + {n - 2 \choose c - 2}{n - 2 \choose r - 2}$ , and simplifies (with some algebraic manipulation) to ${n - 1 \choose c - 1}{n - 1 \choose r - 1}$ .

All left edge blocks are when $c = 1$ , and the first one appears on layer $n = 3$ , column $c = 1$ , and row $r = 2$ , and is $2$ , and ${3 - 1 \choose 1 - 1}{3 - 1 \choose 2 - 1} = 2$ . All left edge blocks in layer $n$ , column $c$ , and row $r$ are the sum of the $2$ blocks above it, which is ${n - 2 \choose c - 1}{n - 2 \choose r - 1} + {n - 2 \choose c - 1}{n - 2 \choose r - 2}$ , and also simplifies to ${n - 1 \choose c - 1}{n - 1 \choose r - 1}$ .

All right edge blocks are when $c = n$ , and the first one appears on layer $n = 3$ , column $c = 3$ , and row $r = 2$ , and is $2$ , and ${3 - 1 \choose 3 - 1}{3 - 1 \choose 2 - 1} = 2$ . All right edge blocks in layer $n$ , column $c$ , and row $r$ are the sum of the $2$ blocks above it, which is ${n - 2 \choose c - 2}{n - 2 \choose r - 1} + {n - 2 \choose c - 2}{n - 2 \choose r - 2}$ , and also simplifies to ${n - 1 \choose c - 1}{n - 1 \choose r - 1}$ .

All back edge blocks are when $r = 1$ , and the first one appears on layer $n = 3$ , column $c = 2$ , and row $r = 1$ , and is $2$ , and ${3 - 1 \choose 2 - 1}{3 - 1 \choose 1 - 1} = 2$ . All back edge blocks in layer $n$ , column $c$ , and row $r$ are the sum of the $2$ blocks above it, which is ${n - 2 \choose c - 1}{n - 2 \choose r - 1} + {n - 2 \choose c - 2}{n - 2 \choose r - 1}$ , and also simplifies to ${n - 1 \choose c - 1}{n - 1 \choose r - 1}$ .

All front edge blocks are when $r = n$ , and the first one appears on layer $n = 3$ , column $c = 2$ , and row $r = 3$ , and is $2$ , and ${3 - 1 \choose 2 - 1}{3 - 1 \choose 3 - 1} = 2$ . All front edge blocks in layer $n$ , column $c$ , and row $r$ are the sum of the $2$ blocks above it, which is ${n - 2 \choose c - 1}{n - 2 \choose r - 2} + {n - 2 \choose c - 2}{n - 2 \choose r - 2}$ , and also simplifies to ${n - 1 \choose c - 1}{n - 1 \choose r - 1}$ .

All the corner blocks are equal to the one block above it, which is always $1$ , and for each corner case (when $c = 1$ and $r = 1$ , when $c = 1$ and $r = n$ , when $c = n$ and $r = 1$ , and when $c = n$ and $r = n$ ), the first ones appear on layer $n = 2$ , and ${2 - 1 \choose 1 - 1}{2 - 1 \choose 1 - 1} = 1$ , ${2 - 1 \choose 1 - 1}{2 - 1 \choose 2 - 1} = 1$ , ${2 - 1 \choose 2 - 1}{2 - 1 \choose 1 - 1} = 1$ , and ${2 - 1 \choose 2 - 1}{2 - 1 \choose 2 - 1} = 1$ . Also, for each corner case, ${n - 1 \choose c - 1}{n - 1 \choose r - 1} = 1$ .

Therefore, each block in column $c$ , row $r$ , and layer $n$ would have the number ${n - 1 \choose c - 1}{n - 1 \choose r - 1}$ written on it.

You might notice that the first row and the first column of the $n^{\text{th}}$ layer is the $n^{\text{th}}$ row in Pascal's Triangle .

We can also observe that the point at coordinates $(i, j)$ on the $n^{\text{th}}$ layer is $(l_{1,\space j}) \cdot (l_{i,\space 1})$ , whereas $l$ is the layer represented using matrix notation .

Therefore, a general formula for the $3^{\text{rd}}$ column and the $4^{\text{th}}$ row must be $P_n(3)\cdot P_n(4)$ , where $P_a(b)$ represents the $b^{\text{th}}$ element of the $a^{\text{th}}$ row in Pascal's triangle. Now we know that Pascal's triangle is the table of binomial coefficients between its row indices and its column indices, namely $\binom{r}{c}$ for the $r^{\text{th}}$ row and the $c^{\text{th}}$ column.

That being said, we ought to find $\binom{100-1}{4-1}\cdot \binom{100-1}{3-1}=\boxed{760874499}$ .

Extra: The sequence of numbers on the 3 rd column and 4 th row is listed in OEIS A004282 .

Notice that the numbers along the n'th edge of a layer are of the form

1 x (n-1) x (n-2)/2 x ... x (n-i)/i on the i'th place from the corners (symmetrical from both sides),

n > 2; i < n.

Notice also that the elements in layer n in the i'th row and the j'th column just are the product of the i'th and j'th edge values

In layer 100 the 3rd edge element is therefore 1 x 99 x 98/2 = 4.851 and the 4th edge element is 1 x 99 x 98/2 x 97/3 = 156.849

The (3, 4)th element in layer 100 is therefore 4.851 x 156.849 = 760.874.499

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**CC PROGRAM WRITTEN IN PYTHONISTA ON A IPAD PRO**
old = [0, 1, 4, 6, 4]
new = [0, 0, 0, 0, 0]
beg = 6
end = 100
for i in range(  beg, end + 1 ):
   new[1] = old[1]
   new[2] = old[2] + 1
   new[3] = old[2] + old[3]
   old[2] = new[2]
   new[4] = old[3] + old[4]
   old[3] = new[3]
   old[4] = new[4]
   print( new, i)
**CC End For Loop**
answer = new[4]*new[3]
print('layer 100 ans = row 4 times col 3',answer)
input(' end block program ')

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**CC Partial Program Run**
First 4 nums in row1 Layer
[1, 89, 3916, 113564] 90
[1, 90, 4005, 117480]  91
[1, 91, 4095, 121485] 92
[1, 92, 4186, 125580] 93
[1, 93, 4278, 129766] 94
[1, 94, 4371, 134044] 95
[1, 95, 4465, 138415] 96
[1, 96, 4560, 142880] 97
[1, 97, 4656, 147440] 98
[1, 98, 4753, 152096] 99
[1, 99, 4851, 156849] 100
layer 100 ans = row 4 times col 3  = 760874499
end block program 

``` Partial Layout of layer 100 [156849 * 4851 = 760874499]

000001, 99, 4851, 000000156849,

000099, ..... , ......., ......................,

004851, ....., ......., ......................,

156849, ....., 760874499, .........., `

The number in the 3rd column 4th row (or 4th column and 3rd row) is the product of the 3rd and 4th numbers in the 1st row. The 3rd number is 2 less than the layer number triangulated (98 x 49.5=4851). The 4th number is the sum of all the 3rd numbers in the previous layers. So (97 x 1)+(96 x 2)+(95 x 3) ... (3 x 95)+(2 x 96)+(1 x 97)=156849. The answer is 156849 x 4851=760874499

For row x, column y, and layer z, it is simply ${z-1 \choose x-1}$ ${z-1 \choose y-1}$ When the variables are plugged in, we get \({99 \choose 2} ) ${99 \choose 3}$ , which equals our answer, 760874499.

My solution was to have my friend go on his computer and tell me all the answers 🤣