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

%I A141678 #44 Feb 19 2025 17:31:53
%S A141678 1,3,3,8,9,8,21,24,24,21,55,63,64,63,55,144,165,168,168,165,144,377,
%T A141678 432,440,441,440,432,377,987,1131,1152,1155,1155,1152,1131,987,2584,
%U A141678 2961,3016,3024,3025,3024,3016,2961,2584,6765,7752,7896,7917,7920,7920,7917,7896,7752,6765
%N A141678 Symmetrical triangle of coefficients based on invert transform of A001906.
%C A141678 Row sums are {1, 6, 25, 90, 300, 954, 2939, 8850, 26195, 76500, ...}.
%C A141678 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.
%C A141678 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
%H A141678 G. C. Greubel, <a href="/A141678/b141678.txt">Rows n=1..101 of triangle, flattened</a>
%H A141678 Matthew Blair, Rigoberto Flórez, and Antara Mukherjee, <a href="https://arxiv.org/abs/2203.13205">Honeycombs in the Pascal triangle and beyond</a>, arXiv:2203.13205 [math.HO], 2022. See p. 5.
%H A141678 B. E. Tenner, <a href="https://arxiv.org/abs/2001.05011">Interval structures in the Bruhat and weak orders</a>, arXiv:2001.05011 [math.CO], 2020.
%F A141678 Let b(n) = Sum_{k=1..n} k*b(n - k), then T(n, m) = b(n-m+1)*b(m+1).
%F A141678 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
%e A141678 Triangle begins as:
%e A141678     1;
%e A141678     3,   3;
%e A141678     8,   9,   8;
%e A141678    21,  24,  24,  21;
%e A141678    55,  63,  64,  63,  55;
%e A141678   144, 165, 168, 168, 165, 144;
%e A141678   377, 432, 440, 441, 440, 432, 377; ...
%t A141678 b[0]=1; b[n_]:= Sum[k*b[n-k], {k, 1, n}];
%t A141678 Table[b[n-m+1]*b[m+1], {n, 0, 10}, {m, 0, n}]//Flatten
%t A141678 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 *)
%o A141678 (PARI) {b(n) = if(n==0, 1, fibonacci(2*n))};
%o A141678 for(n=0, 10, for(k=0, n, print1(b(n-k+1)*b(k+1), ", "))) \\ _G. C. Greubel_, Apr 06 2019
%o A141678 (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
%o A141678 (Sage)
%o A141678 @CachedFunction
%o A141678 def b(n):
%o A141678     if n==0: return 1
%o A141678     return fibonacci(2*n)
%o A141678 [[b(n-k+1)*b(k+1) for k in (0..n)] for n in (0..10)] # _G. C. Greubel_, Apr 06 2019
%Y A141678 Cf. A001906.
%K A141678 nonn,tabl
%O A141678 1,2
%A A141678 _Roger L. Bagula_ and _Gary W. Adamson_, Sep 07 2008
%E A141678 Edited by _G. C. Greubel_, Apr 02 2019