Automatically get points in a given triangular lattice and vice versa？

Question

Given a triangular lattice which grows with number n, as follows

I want to list all the connected points in the lattice to form the pairlist. As an example:

n=1:
pairlist={{p_{1},p_{2}},{p_{1},p_{3}},{p_{2},p_{3}}}
n=2:
pairlist={{p_{1},p_{2}},{p_{1},p_{4}},{p_{2},p_{3}},{p_{2},p_{4}},{p_{2},p_{5}},{p_{3},p_{5}},{p_{4},p_{5}},{p_{4},p_{6}},{p_{5},p_{6}}}
n=3:
pairlist={{p_{1},p_{2}},{p_{1},p_{5}},{p_{2},p_{3}},{p_{2},p_{5}},{p_{2},p_{6}},{p_{3},p_{4}},{p_{3},p_{6}},{p_{3},p_{7}},{p_{4},p_{7}},{p_{5},p_{8}},{p_{5},p_{6}},{p_{6},p_{8}},{p_{6},p_{9}},{p_{6},p_{7}},{p_{7},p_{9}},{p_{8},p_{10}},{p_{8},p_{9}},{p_{9},p_{10}}}

Notice that there is no need to list the elements in pairlist in order.

Questions:

1. Is there some automatical way to do above tasks?
2. If I have points= Flatten[Table[Subscript[p, i], {i, 1, Binomial[n + 2, 2]}]];, how can I create the triangle lattice with points automatically?

For Question-2, I make one code for n=2 in the below (is there any simple and general way for different n?).

pairlists = {};
n = 2;
points = Flatten[Table[Subscript[p, i], {i, 1, Binomial[n + 2, 2]}]];

For[ii = 1, ii <= Length[points], ii++,
 For[jj = ii + 1, jj <= Length[points], jj++,
   
   If[(Abs[points[[ii]][[2]]-points[[jj]][[2]]]==1&&ii != 3) ||(Abs[points[[ii]][[2]]-points[[jj]][[2]]]==2&&Abs[ii-jj]==2&& ii!=1 )||(Abs[points[[ii]][[2]]-points[[jj]][[2]]]==3&&Abs[ii-jj]==3&&ii!=3),
      
      AppendTo[pairlists, {points[[ii]], points[[jj]]}];
    ]
   ];
 ]

Answer 1

 triangleGridGraph[n_, opts : OptionsPattern[]] := 
  Module[{vc = Join @@ Table[{j + i/2, i Sqrt[3]/2}, {i, 0, n + 1}, {j, 0,  n - i}]},
    NearestNeighborGraph[vc, VertexCoordinates -> vc, opts]]

Examples:

triangleGridGraph[4, 
  VertexLabels -> Placed["Index", Center], VertexSize -> Large]

IndexGraph[triangleGridGraph[4], 
  VertexLabels -> {v_:> Placed[Subscript[P, v], Center]},
  VertexSize->Large]

Multicolumn[Panel[
  Labeled[
    triangleGridGraph[#, GraphStyle -> "IndexLabeled", ImageSize -> 300], 
    Style[PromptForm["n", #], 16], Top]] & /@ Range[6], 
  3, Appearance -> "Horizontal"] 

To get edges as lists, you Apply List at level 1 to EdgeList of triangleGridGraph[n]. For example,

List @@@ EdgeList[triangleGridGraph[2]]

{{{0, 0}, {1, 0}}, {{0, 0}, {1/2, Sqrt[3]/2}}, {{1, 0}, {2, 0}}, 
 {{1, 0}, {1/2, Sqrt[3]/2}}, {{1, 0}, {3/2, Sqrt[3]/2}}, 
 {{2, 0}, {3/2, Sqrt[3]/2}}, {{1/2, Sqrt[3]/2}, {3/2, Sqrt[3]/2}},  
 {{1/2, Sqrt[3]/2}, {1, Sqrt[3]}}, {{3/2, Sqrt[3]/2}, {1, Sqrt[3]}}} 

Answer 2

basis = {{1, 0}, {Cos[60 Degree], Sin[60 Degree]}};
genpts[n_] :=
 # . basis & /@
  Select[
   Flatten[CoordinateBoundsArray[{{0, n}, {0, n}}], 1],
   Total[#] < n &
  ]

g = With[{pts = genpts[8]}, 
  NearestNeighborGraph[pts, VertexCoordinates -> pts]]

You can then get pairs of connected vertices as follows:

EdgeList[g] /. UndirectedEdge -> List