Counting multiple of $3$ in a triangular arrangement of the integers. I write 
:$$\mathbb{N}_\triangle = \begin{matrix}
    &&&&&21&\ldots         \\
    &&&&15&20&\ldots       \\
    &&&10&14&19&\ldots     \\
    &&6&9&13&18&\ldots   \\
    &3&5&8&12&17&\ldots \\
    1&2&4&7&11&16&\ldots 
    \end{matrix}$$
Can we determine how many numbers in any given column of $\mathbb{N}_\triangle$ are divisible by $3$. I need help proving the following claim: 

If $f_3(n)$ counts the numbers in the $n^{th}$ column of
  $\mathbb{N}_\triangle$ that are divisible by $3$ then $$f_3(n)
 =\Bigg\lfloor{n+2 \above 1.5pt 3}\Bigg\rfloor$$ 

For example $$f_3(6) = \Bigg\lfloor{6+2 \above 1.5pt 3}\Bigg\rfloor=\Bigg\lfloor{8 \above 1.5pt 3}\Bigg\rfloor=\Bigg\lfloor 2.66666\Bigg\rfloor =2$$ and surely there are $2$ numbers in column six that are divisible by $3$. I suspect there are no multiples of $3$ on the base of the triangle since possibly $$\Bigg(3, 1+ {n(n-1) \above 1.5pt 2}\Bigg)=1$$ In fact if $(3,T_n)=1$ then I suspect that there are no numbers in the $n^{th}$ row that are divisible by $3$. Surely there are consecutive triangular numbers divisible by $3$. 
 A: The number of multiples of $3$ in the interval $[1,N]$ is $\lfloor N/3\rfloor$.
Moreover the $n^{th}$ column of $\mathbb{N}_\triangle$ is the set of consecutive integers from $T_{n-1}+1$ to $T_n$ with $T_n=n(n+1)/2$.
Hence, the number of multiples of $3$ in the $n^{th}$ column of
$\mathbb{N}_\triangle$ is 
$$f_3(n)=\lfloor T_{n}/3\rfloor-\lfloor T_{n-1}/3\rfloor=
\left\lfloor \frac{n(n+1)}{6}\right\rfloor-
\left\lfloor \frac{n(n-1)}{6}\right\rfloor
=\left\lfloor \frac{n+1}{3}\right\rfloor.$$ 
The last equality can be proven easily by letting $n=6k+r$. In fact, after the substitution, we obtain 
$$\left\lfloor \frac{r(r+1)}{6}\right\rfloor-
\left\lfloor \frac{r(r-1)}{6}\right\rfloor
=\left\lfloor \frac{r+1}{3}\right\rfloor$$
which can be verified by hand for $r=0,1,2,3,4,5$.
A: The $n$ th column is the sequence of $n$ consecutive numbers starting at $1+\frac{n (n-1)}{2}$. Let's call $b$ this base value.
The sequence length is $n$. $n$ starts at 1 like the height of its column. $k$ is an integer.
Each 3 consecutive numbers sequence contains exactly one multiple of 3.
When starting with a base number like $3k+2$ , $n$ covers $\Bigg\lfloor\frac{n+1}{3}\Bigg\rfloor$ 3-multiples.
If starting with $3k$, $n$ covers $\Bigg\lfloor\frac{n+2}{3}\Bigg\rfloor$ 3-multiples and with $3k+1$ , $\Bigg\lfloor\frac{n}{3}\Bigg\rfloor$ .
It appears that $1+\frac{n (n-1)}{2}$ may be 1 or 2 modulo 3 and is never 0 :
if $n=0$ modulo 3 , we have $1+\frac{n (n-1)}{2} = 1$ modulo 3 , then $f_3(n) = \Bigg\lfloor\frac{n}{3}\Bigg\rfloor$
if $n=1$ modulo 3 , we have
$1+\frac{n (n-1)}{2} = 1$ modulo 3 , then $f_3(n) = \Bigg\lfloor\frac{n}{3}\Bigg\rfloor$
if $n=2$ modulo 3 , we have
$1+\frac{n (n-1)}{2} = 2$ modulo 3 , then $f_3(n) = \Bigg\lfloor\frac{n+1}{3}\Bigg\rfloor$
