A141678 Symmetrical triangle of coefficients based on invert transform of A001906.

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, 987, 1131, 1152, 1155, 1155, 1152, 1131, 987, 2584, 2961, 3016, 3024, 3025, 3024, 3016, 2961, 2584, 6765, 7752, 7896, 7917, 7920, 7920, 7917, 7896, 7752, 6765

Offset: 1

Author

Roger L. Bagula and Gary W. Adamson, Sep 07 2008

Keywords

• nonn
• tabl

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

Examples

			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; ...
		

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.

Crossrefs

Cf. A001906.

Programs

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

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 *)

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

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

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

Extensions

Edited by G. C. Greubel, Apr 02 2019