OFFSET
1,2
COMMENTS
Row sums are {1, 6, 25, 90, 300, 954, 2939, 8850, 26195, 76500, ...}.
It can be noticed that the interior of the triangle is relatively "flat", which is a smaller variation than in most symmetrical triangles of this type.
16*T(n,k) is the number of Boolean (equivalently, lattice, modular lattice, distributive lattice) intervals of the form [s_{k+1},w] in the Bruhat order on S_{n+3}, for the simple reflection s_{k+1}. - Bridget Tenner, Jan 16 2020
LINKS
G. C. Greubel, Rows n=1..101 of triangle, flattened
Matthew Blair, Rigoberto Flórez, and Antara Mukherjee, Honeycombs in the Pascal triangle and beyond, arXiv:2203.13205 [math.HO], 2022. See p. 5.
B. E. Tenner, Interval structures in the Bruhat and weak orders, arXiv:2001.05011 [math.CO], 2020.
FORMULA
Let b(n) = Sum_{k=1..n} k*b(n - k), then T(n, m) = b(n-m+1)*b(m+1).
Alternatively, let f(n) = Fibonacci(2*n) with f(0)=1, then T(n, k) = f(n-k+1)*f(k+1). - G. C. Greubel, Apr 06 2019
EXAMPLE
Triangle begins as:
1;
3, 3;
8, 9, 8;
21, 24, 24, 21;
55, 63, 64, 63, 55;
144, 165, 168, 168, 165, 144;
377, 432, 440, 441, 440, 432, 377; ...
MATHEMATICA
b[0]=1; b[n_]:= Sum[k*b[n-k], {k, 1, n}];
Table[b[n-m+1]*b[m+1], {n, 0, 10}, {m, 0, n}]//Flatten
f[n_]:= If[n == 0, 1, Fibonacci[2*n]]; Table[f[n-k+1]*f[k+1], {n, 0, 10}, {k, 0, n}]//Flatten (* G. C. Greubel, Apr 06 2019 *)
PROG
(PARI) {b(n) = if(n==0, 1, fibonacci(2*n))};
for(n=0, 10, for(k=0, n, print1(b(n-k+1)*b(k+1), ", "))) \\ G. C. Greubel, Apr 06 2019
(Magma) b:= func< n| n eq 0 select 1 else Fibonacci(2*n) >; [[b(n-k+1)*b(k+1): k in [0..n]]: n in [0..10]]; // G. C. Greubel, Apr 06 2019
(Sage)
@CachedFunction
def b(n):
if n==0: return 1
return fibonacci(2*n)
[[b(n-k+1)*b(k+1) for k in (0..n)] for n in (0..10)] # G. C. Greubel, Apr 06 2019
CROSSREFS
KEYWORD
nonn
AUTHOR
Roger L. Bagula and Gary W. Adamson, Sep 07 2008
EXTENSIONS
Edited by G. C. Greubel, Apr 02 2019
STATUS
approved