# 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 $3$.

• Could you say precisely what is the question please ?
– user354674
Aug 11, 2016 at 21:28
• How many numbers are divisible by $3$ in any given column of the triangle? That should be clear. Aug 11, 2016 at 21:29
• @Anthony Hernandez According to your formula $f_3(1)=1$ but in the first column there is only the number 1 which is not a multiple of 3. I gave the correct formula in the answer below. Aug 11, 2016 at 22:43
• ^ I should have noticed that and I did not embarrasingly. Aug 11, 2016 at 23:32

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 $r=0,1,2,3,4,5$.

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 $f_3(n) = \Bigg\lfloor\frac{n+1}{3}\Bigg\rfloor$

• ^ +1 I suggest writing the sequence out and then comparing that to Sloan's database and finding the appropriate offset. The sequence is 0,1,1,1,2,2,2,3,3,3,4,4,4,5,5,5,.... Aug 12, 2016 at 1:03