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In this MSE post, user GeMir noticed that,

where the green dots are the *triangular numbers*,

$$T_n = \frac{n(n+1)}{2} = 1,3,6,10,15,21,28,36,45,55,66,78,91,105,120,136,\dots$$

in the *Ulam spiral* with center $c=1$. If you look closely, the center seems to have a *triskelion* (a three-curve figure). GeMir says one "arm" (blue numbers below) starts as,

$$T_n = \frac{n(n+1)}{2} = 1,3,\color{brown}{6},\color{blue}{10},15,\color{brown}{21},\color{blue}{28},36,\color{brown}{45},\color{blue}{55},66,\color{brown}{78},\color{blue}{91},105,\color{brown}{120},\color{blue}{136},\dots$$

with a second "arm" (in brown). As pointed out by B. Cipra, the sequences (at least initially) look like oeis.org/A060544 and oeis.org/A081266, respectively.

But that can't be right since the figure apparently has *seventeen* arms, not just three.

Questions:

- It seems
*the triskelion splits into $17$ arms*. Using a Mathematica code, can we "zoom out" and see if the $17$ armsinto sub-arms? (That is, make a bigger Ulam spiral that ends with a higher bound B than used by GeMir.)*split further* - If indeed it stabilizes into just $17$, can we use Mathematica to find a polynomial formula for each arm? (
*InterpolatingPolynomial*[] would do, but one would need the values, and if I make this spiral with pen-and-paper it would be too tedious.)

**P.S.** The post, *Generating an Ulam spiral*, contains various codes. However, since my Mathematica coding is very basic, I do not know how to tweak them to answer my questions.

Thanks, especially for establishing the stability of the 17 arms. I had considered the possibility that maybe they would split further. – Tito Piezas III – 2015-07-12T03:29:27.050