Finding the number of even numbers in Pascal's triangle (code gives memory error)

Apparently, you want to count the number of zeroes in Pascal's triangle mod 2 with $\alpha \in \mathbb{N}$ rows. This can be done by counting the ones and subtract this number from the number of all entries of that triangle, which is $\alpha(\alpha+1)/2$.

The code that you posted has complexity $O(\alpha^2)$ and with your choice of $\alpha$, that will take forever. The key to a more efficient way of counting is to observe that Pascal's triangle mod 2 has a self-similar structure. The first $2^j$, $j\geq 1$ rows form a triangle $T_j$. The triangle $T_{j+1}$ can be obtained by gluing three copies of $T_j$ together (in the fashion of the Triforce from Zelda). So $T_{j+1}$ has 3 times as many ones than $T_j$. $T_0$ consists of a single one. By induction, the first $2^j$ rows contain $3^j$ ones. So, in fact, the number of ones can be computed from the binary represenation of the number $\alpha$. After some trial and error, I came up with this formula for the number of ones:

onecount[α_] := With[{digits = IntegerDigits[α, 2]},
  Total[
   Times[
    digits,
    3^Range[Length[digits] - 1, 0, -1],
    2^(Accumulate[digits] - 1)
    ]
   ]
  ]

I hope it is correct. A quick test:

triangle[α_] := Table[Mod[Binomial[n, k], 2], {n, 0, α - 1}, {k, 0, n}]
a = Table[Total[triangle[α], 2], {α, 1, 100}];
b = onecount /@ Range[100];
a == b

True

Also, in case $α = 10^3$, this reproduces Bob's result, which is $448363$.

So the number of zeroes in the triangle with number $\alpha = 10^6$ should be

α = 10^6;
Quotient[α (α + 1), 2] - onecount[α]

Note that this takes only $O(\log_2(\alpha))$ time and memory.

The basic approach

Clear["Global`*"]

α = 10^3; (* Reduced value *)
\[DoubleStruckCapitalT][i_, j_] := Binomial[i, j - 1];

count = 0;

Do[
  If[
   \[DoubleStruckCapitalT][n, k] != 0 && 
    Mod[\[DoubleStruckCapitalT][n, k], 2] == 0,
   count += 1],
  {n, 0, α - 1}, {k, 1, α}];

count

(* 448363 *)