Function[n,
In[]:=
NestGraph[n{2n,n+1},0,11]
Out[]=
In[]:=
NestGraph[n{2n,n+1,n/2,n-1},1,3,VertexLabelsAutomatic]
Out[]=
In[]:=
NestGraph[n{n+1,n-1},1,3,VertexLabelsAutomatic]
Out[]=
In[]:=
NestGraph[n{n+1,n-1,n+5,n-5},1,3,VertexLabelsAutomatic]
Out[]=
In[]:=
NestGraph[n{n+1,n-1,n+5,n-5},1,6]
Out[]=
In[]:=
Graph3D[%]
Out[]=
In[]:=
NestGraph[n{n+4,n-4,n+7,n-7},1,10]
Out[]=
In[]:=
Graph3D[%]
Out[]=
In[]:=
NestGraph[n{n+4,n+7},0,5,VertexLabelsAutomatic]
Out[]=
The grid is just commutativity
In[]:=
FrobeniusNumber[{4,7}]
Out[]=
17
In[]:=
NestGraph[n{n+4,n+7},18,7,VertexLabelsAutomatic]
Out[]=
In[]:=
Graph[MultiwayFunctionSystem[n{n+4,n+7},0,10,"StatesGraphStructure"],VertexLabelsAutomatic]
Out[]=
How is the distribution of path weights determined by a and b?
After a certain transient, every value can be reached by the iteration.
Branch pair convergence
Branch pair convergence
a n+b, c n+d
a n+b, c n+d
The symbolic case will not crosslink:
Look at all numbers mod k
Look at all numbers mod k
A necessary condition for merging is the last step not go astray by having numbers that don’t match mod k
n = 1 mod 3 and n = 0 mod 2
Chinese remainder estimation
Chinese remainder estimation
Tilga’s claim
Tilga’s claim