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Do you have any references to this interesting result? I could not find any...

The total number of odd numbers in the first $2^n$ rows of Pascal's triangle is $3^n$, $n>=0$.

It's easy to prove by induction based on the formula for the number of odds in row $n$ ($2^m$, where $m$ is the number of ones in the binary expansion of $n$).

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  • $\begingroup$ Just saw it: the answer by @Micah to the "Odds in Pascal's Triangle" post includes a direct proof of this fact. $\endgroup$ – Mircea Draghicescu Jan 29 at 22:10
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The Online Encyclopedia of Integer Sequences lists as entry A006046

Total number of odd entries in first n rows of Pascal's triangle: a(0) = 0, a(1) = 1, a(2k) = 3*a(k), a(2k+1) = 2*a(k) + a(k+1)

which doesn't give the claim explicitly but makes it very clear implicitly. Further down, in the formula section, you can find it explicitly:

a(2^n) = 3^n. - Henry Bottomley, Apr 05 2001

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