The primary motivation for this question is the following: I would like to extract some topological statistics which capture how arithmetic progressions of prime numbers "fit together" in a manner that will be made precise below. 


Setup
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Consider a nested family of simplicial complexes $K(p)$ indexed by prime $p \in \mathbb N$ defined as follows: 


1. the vertices are all odd primes less than or equal to $p$, and
2. insert a $d$-simplex ($d \geq 2$) spanning $d+1$ vertices if and only if they constitute an arithmetic progression. Of course, one must also insert all faces, and faces-of-faces etc. so that the defining property of a simplicial complex is preserved.


For instance, $K(7)$ has the vertices $3,5,7$ and a single $2$-simplex $(3,5,7)$ along with all its faces. $K(11)$ has all this, plus the vertex $11$ and the simplex $(3,7,11)$. The edge $(3,7)$ already exists so only the other two need to be added. Thus, the fact that $(3,7)$ occurs in two arithmetic progressions bounded by $11$ is encoded by placing the corresponding edge in the boundary of two 2-simplices.


Question
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Has someone already defined and studied this complex? What I am mostly interested in is



> 
> How does the homology of $K(p)$ change with $p$?
> 
> 
> 


If it helps, here are -- according to home-brew computations -- the statistics for the first few primes (Betti 0 and 1 over $\mathbb{Z}\_2$). I've already confirmed that the sequence of Betti-1's is not in Sloane's online encyclopedia of integer sequences. If an intermediate K[p] is missing in the list, that means that the homology is the same as that for the previous prime.


K [3]: 1 0  

 K [5]: 2 0  

 K [7]: 1 0  

 K [13]: 2 0  

 K [17]: 2 1  

 K [19]: 1 2  

 K [23]: 1 4  

 K [31]: 1 6  

 K [37]: 2 6  

 K [43]: 1 7  

 K [53]: 1 8  

 K [59]: 1 9  

 K [61]: 1 10  

 K [67]: 1 12  

 K [71]: 1 17  

 K [73]: 1 20  

 K [79]: 1 23  

 K [83]: 1 26  

 K [89]: 1 31  

 K [97]: 1 32  

 K [101]: 1 35  

 K [103]: 1 41  

 K [107]: 1 43  

 K [109]: 1 47  

 K [113]: 1 53  

 K [127]: 1 58  

 K [131]: 1 62  

 K [137]: 1 67  

 K [139]: 1 73  

 K [149]: 1 78 


Here's a more concrete question:



> 
> Is it true that the $d$-dimensional homology groups of $K(p)$ for $d > 1$ are always trivial?
> 
> 
> 



The topology of Arithmetic Progressions of primes
No. 
In fact, for $p = 435052917615787$, this will absolutely be false, as the second homology group will not vanish.


Note that a 2-dimensional "hole" in your complex is a 3-simplex all of whose faces are 2-simplices in $K(p)$, i.e. 4 primes such that each 3 of them lie in some arithmetic progression. 


Of course, you would suspect that such a thing is plausible, and with some effort construct an example.
Here's why this number works:


Consider $p = 398936189798617$. Then, if $d = 2124513401010$, one sees that $p+d$ is not a prime, while $p+2d,p+3d,\ldots,p+15d$ are all primes. Here $p+15d = 435052917615787$ is the prime for which we consider the complex.


In fact, the numbers were taken from the AP-k records at <http://primerecords.dk/aprecords.htm#minimal>, and $p+2d$ is just a beginning of an AP-23. 


Now, one sees immediately that $p, p+6d, p+10d$ form a simplex, as it is a subsequence of the arithmetic progression of primes $p, p+2d, p+4d, p+6d, p+10d$.


Similarly, $p, p+6d, p+15d$ form a simplex, as do $p, p+10d, p+15d$ and $p+6d, p+10d, p+15d$. These come from sequences with differences $3d$, $5d$, and $d$ respectively. 


However, $p,p+6d,p+10d,p+15d$ can only form a simplex if they all lie in some arithmetic progression of primes. But then its difference must divide $d$, contradicting the fact that $p+d$ is not prime.


