A347608 Number of interlacing triangles of size n.
1, 2, 20, 1744, 2002568, 42263042752, 21686691099024768, 344069541824691045987328, 226788686879114461294165127878656
Offset: 1
Examples
For n = 2, a(2) = 2. The interlacing triangles are given below:
2 2
1 3 and 3 1.
Links
- James B. Sidoli, On the number of interlacing triangles of size n
Crossrefs
Cf. A003121.
Programs
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Sage
def interlacing(n): C_2=[] part=[j for j in range(n-1,-1,-1)] box=[] big_box=[] pos=0 d=0 C_2_star=[] for g in Words([0,1],n*(n-1)/2).list(): C_2.append(list(g)) for h in C_2: relations=[] pos=0 big_box=[] for j in range(len(part)-1): for k in list(h)[pos:pos+part[j]]: box.append(k) big_box.append(box) box=[] pos=pos+part[j] x=0 for k in range(1,len(big_box)): for r in range(len(big_box[k])): if big_box[k][r]==1 and big_box[k-1][r]==0 and big_box[k-1][r+1]==0 or big_box[k][r]==0 and big_box[k-1][r]==1 and big_box[k-1][r+1]==1: continue else: x=x+1 if x==(n-1)*(n-2)/2: q=q+1 C_2_star.append(big_box) position=range(n*(n+1)/2) for tri in C_2_star: P=[] relations=[] counter=0 collect=[] for j in range(len(tri)): for r in range(len(tri[j])): if tri[j][r]==0: relations.append([position[counter],position[counter+n-j]]) relations.append([position[counter+n-j],position[counter+1]]) if tri[j][r]==1: relations.append([position[counter+n-j],position[counter]]) relations.append([position[counter+1],position[counter+n-j]]) counter=counter+1 counter=counter+1 P=Poset([range(n*(n+1)/2),relations]) d=d+P.linear_extensions().cardinality() return d
Extensions
a(7)-a(9) from Dylan Nelson, May 09 2022
Comments