In this Numberphile video (from 3:36 to 7:41), Neil Sloane explains an amazing sequence:

It is the lexicographically first among the sequences of positive integers without triple in arithmetic progression (i.e., such that for any two distinct points in the graph, there is no point in their middle, as for $a_n=n^2$ or $b_n = 2^n$). It is due to Jack W. Grahl (2013): A229037. $$1, 1, 2, 1, 1, 2, 2, 4, 4, 1, 1, 2, 1, 1, 2, 2, 4, 4, 2, 4, 4, 5, 5, 8, 5, 5, 9, 1, 1, 2, 1, 1, 2, 2, 4, 4, \dots$$ Its graph is surprising:

The following colored plot of 16 million terms is due to reddit user garnet420 (horizontal divisions are 1000000; vertical divisions are 25000).

This sequence reveals many possible questions (like the existence of a convergent pattern, a fractal structure)). The following one is due to Charles R. Greathouse (and deserves to be posted here):

**Question**: Does the term $1$ appear infinitely many times in this sequence?

*Remark*: the first numbers $n$ for which the $n^{th}$ term is $1$ are given by A236246.

$$1, 2, 4, 5, 10, 11, 13, 14, 28, 29, 31, 32, 37, 38, 40, 41, 82, 83, 85, 86, 92, 93, 96, 105, \dots$$

In addition, we observe that the succession $1, 1, 2, 1, 1, 2, 2, 4, 4$ appears several times, so we could also ask whether it appears infinitely many times.

Finally, we can create variations of the first sequence, replacing triple by $r$-tuple for a fixed $r \ge 3$, and ask the same questions.

Let us consider the variation suggested by Richard Stanley in comment where we only exclude *weakly increasing* arithmetic progressions. The first terms are the same up to the $54^{th}$ term.

For the former sequence, the $20$ next terms are
$$ 9, 4, 4, 5, 5, 10, 5, 5, 10, 2, 10, 13, 11, 10, 8, 11, 13, 10, 12, 10, 10 $$
whereas for the variation they are
$$ 2, 4, 4, 5, 5, 10, 5, 5, 10, 10, 11, 13, 10, 11, 10, 11, 13, 10, 10, 12, 13 $$
It is *now* available on OEIS at A309890. Listen it played with a marimba, it's incredibly pleasant!

Its graph is completely different, but also amazing:

Finally the numbers $n$ for which the $n^{th}$ term is $1$ are given by A003278 (see Richard's comment).

A SageMath code computing the variation:

```
# %attach SAGE/ThreeFree.spyx
from sage.all import *
cpdef ThreeFree(int n):
cdef int i,j,k,s,Li,Lj
cdef list L,Lb
cdef set b
L=[1,1]
for k in range(2,n):
b=set()
for i in range(k):
if 2*((i+k)/2)==i+k:
j=(i+k)/2
Li,Lj=L[i],L[j]
s=2*Lj-Li
if s>0 and Li<=Lj: # this second assumption provides the variation
b.add(s)
if 1 not in b:
L.append(1)
else:
Lb=list(b)
Lb.sort()
for t in Lb:
if t+1 not in b:
L.append(t+1)
break
return L
```

For having the graph, do `points([(i+1,L[i]) for i in range(n)],size=1,alpha=0.04,dpi=1000)`

weakly increasingarithmetic progressions, i.e., $a(j),a(j+k),a(j+2k)$ are in arithmetic progression (with $k\geq 1$) and $a(j)\leq a(j+k)\leq a(j+2k)$, then it is easy to see that terms equal to 1 are indexed by oeis.org/A236246. This suggests looking some more at this modified sequence. Could it coincide with the original sequence? I haven't tried to check this experimentally. $\endgroup$3more comments