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I'd like to see if this concept has a name, so I will present my understanding of it (with some notation made up for readability) in the hope that someone will recognize it.

A triangle number 1+2+3+..+n can be calculated with a formula Δ(n)=n(n+1)/2 and be illustrated with a triangle figure:

1
2  3
4  5  6
7  8  9  10
11 12 13 14 15

where n is a row and Δ(n) is the sum of all elements in the triangle up to and including that row.

I've had reasons to use a triangle of the following shape when solving a problem:

1
2
3  4
5  6
7  8  9
10 11 12  

It follows a similar pattern, only that the columns expand one step further to the right only every two rows.

The sum can be calculated with ΔΔ(n)=Δ(Floor(n/2))+Δ(Ceiling(n/2)).

I.e. ΔΔ(5)=Δ(2)+Δ(3)=3+6=9

(This can be confirmed with the last number on the 5h row being 9).

The intuition being that my triangle is just two regular lower triangles weaved into each other, so there's two sums alternating in increased range every row.

Does this triangle shape have a name, and if so, what is it, and is this method the usual way of calculating what I've called ΔΔ(n)?

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Your sequence $\Delta\Delta(n):=\Delta(\lfloor n/2 \rfloor)+\Delta(\lceil n/2 \rceil) = A002620(n+1),\;$ where the OEIS sequence A002620 is called "quarter-squares" and defined by $A002620(n):= \lfloor n/2 \rfloor\lceil n/2 \rceil$.

Your triangular shape is determined by row and column numbers using OEIS sequences $\textrm{row}(n):=\lfloor\sqrt{4n-1}\rfloor=A000267(n-1)$ and $\textrm{column}(n):=\lfloor(4n+1-\lfloor\sqrt{4n-1}\rfloor^2)/4\rfloor=A122197(n)$.

For example, the number $9$ is in row $5$ and column $3$ and the number $10$ is in row $6$ and column $1$.

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  • $\begingroup$ I was hoping for this to be a much less obscure sequence, but I am thankful for it at least having some kind of name. Thank you! $\endgroup$ – Rasmus Källqvist Sep 5 '17 at 9:14
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Notice the numbers at the end of each line of your second grid of numbers are the square numbers and the oblong numbers. \begin{eqnarray*} T(n)= \frac{n(n+1)}{2}= 1,3,6,10,15,\cdots \text{Triangle numbers} \\ S(n)= n^2 = 1,4,9,16,25,\cdots \text{Square numbers} \\ O(n)= n(n+1)= 2,6,12,20,30\cdots \text{Oblong numbers} \\ \end{eqnarray*}

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  • $\begingroup$ This didn't answer the question of what the sequence was called, but did give a very valuable insight in what was going on. That even indexes (starting at 0) give square numbers and odd give oblong numbers was much more elegant than the sum formula I presented. Thank you! $\endgroup$ – Rasmus Källqvist Sep 5 '17 at 9:13

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