Here’s an interesting theorem I ran across recently.

The number of odd integers in the *n*th row of Pascal’s triangle equals 2^{b} where *b* is the number of 1’s in the binary representation of *n*.

Here are the first few rows of Pascal’s triangle:

1
1 1
1 2 1
1 3 3 1
1 4 6 4 1
1 5 10 10 5 1
1 6 15 20 15 6 1
1 7 21 35 35 21 7 1
...

We count rows starting from 0, so for *n* at least 1, the *n*th row has *n* in the second column.

Notice, for example, that on the 6th four of the entries are odd: 1, 15, 15, 1. The binary representation of 6 is 110, so *b* = 2, and 2² = 4. In the 7th row, all eight entries are odd. The binary representation of 7 is 111, and 2³ = 8.

There are a couple quick corollaries to the theorem above. First, the number of odd numbers in the *n*th row of Pascal’s triangle is always a power of 2. Second, in row 2^{k-1} – 1, all entries are odd.

This post is a slightly expanded version of a Twitter thread I posted on @AlgebraFact this weekend.

## More on Pascal’s triangle

Pascal’s triangle has many interesting features. For example it is not known whether a number exists that appears in it more than 8 times, or whether any number other than 3003 appears exactly 8 times. It is also known there are infinite numbers appearing exactly 6 times, the smallest one being 61 218 182 743 304 701 891 431 482 520