OFFSET
1,2
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.
LINKS
EXAMPLE
For n = 2, a(2) = 2. The interlacing triangles are given below:
2 2
1 3 and 3 1.
PROG
(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
CROSSREFS
KEYWORD
nonn,more
AUTHOR
James B. Sidoli, Sep 08 2021
EXTENSIONS
a(7)-a(9) from Dylan Nelson, May 09 2022
STATUS
approved