Parity Pattern of Triangle Numbers

Question

I noticed that the triangle numbers given by $\frac{n(n+1)}{2}$ have a parity pattern of odd, odd, even, even, ... and I'm curious as to why.

E.g:

n       T(n)    P(T(n))
1        1       1
2        3       1
3        6       0
4        10      0
5        15      1
6        21      1
7        28      0
8        36      0
9        45      1
10       55      1
11       66      0
12       78      0
13       91      1
14       105     1

I can see that the parities for $n^2+n$ are all 0, as in:

n       F(n)    P(F(n))
1        2       0
2        6       0
3        12      0
4        20      0
5        30      0
6        42      0
7        56      0

but I'm having trouble seeing why halving these values leads to the odd, odd, even, even,... pattern for the triangle numbers.

Can anyone shed some light please?

Answer 1

If $n\equiv 1 \mod 4$ or $n\equiv 2 \mod 4$, then out of $n$ and $(n+1),$ one of them is odd, and the other is divisible by $2$ but not $4,$ and so $n(n+1)/2$ is not divisible by $2.$

If $n\equiv 3 \mod 4$ or $n\equiv 4 \mod 4$, then out of $n$ and $(n+1),$ one of them is odd, and the other is divisible by $4,$ and so $n(n+1)/2$ is even.

Answer 2

You can divide $n$ by $4$ and write $n=4a+k$ where $a$ is some integer and $k=0, 1, 2,$ or $3$. Then

$$\frac{n(n+1)}{2} = \frac{(4a+k)(4a+k+1)}{2} = \frac{16a^2+8ak+4a+k^2+k}{2} $$

$$= 8a^2+4ak+2a +\frac{k^2+k}{2}$$

Which is an even integer plus $\frac{k^2+k}{2}$.

If we plug in the values $0, 1, 2, 3$ for $k$, we see the pattern:

$$0, 1, 3, 6$$

which goes even odd odd even.

Answer 3

First is odd, we add an even number,that gives us odd again, we add an odd number we get even number, we add an even number we get an even number, continues starting from the next one.