A347608 Number of interlacing triangles of size n.

1, 2, 20, 1744, 2002568, 42263042752, 21686691099024768, 344069541824691045987328, 226788686879114461294165127878656

Offset: 1

Author

James B. Sidoli, Sep 08 2021

Keywords

• nonn
• more

Comments

An interlacing triangle of size n is a triangular array of the numbers 1, 2, ..., (n+1)*n/2 such that if T(i,j) denotes the j-th number in the i-th row then either T(i-1,j+1) < T(i,j) < T(i-1,j) or T(i-1,j) < T(i,j) < T(i-1,j+1) for 1 < i <= n and 1 <= j <= n-i+1.

Generalizes A003121 for the case when rows are not strictly increasing. See comment from Mar 25 2012 and comment from Dec 02 2014.

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

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