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A106408 Triangle, read by rows, where T(1,1) = 1; T(2,1) = T(2,2) = 2; for n > 2, T(n,n) = T(n-1,n-1) + T(n-2,n-2); T(n+1,n) = 2 * T(n,n); for all other entries, T(n,k) = T(n-1,k) + T(n-2,k).

%I A106408 #16 Jan 04 2021 19:58:33
%S A106408 1,2,2,3,4,3,5,6,6,5,8,10,9,10,8,13,16,15,15,16,13,21,26,24,25,24,26,
%T A106408 21,34,42,39,40,40,39,42,34,55,68,63,65,64,65,63,68,55,89,110,102,105,
%U A106408 104,104,105,102,110,89,144,178,165,170,168,169,168,170,165,178,144
%N A106408 Triangle, read by rows, where T(1,1) = 1; T(2,1) = T(2,2) = 2; for n > 2, T(n,n) = T(n-1,n-1) + T(n-2,n-2); T(n+1,n) = 2 * T(n,n); for all other entries, T(n,k) = T(n-1,k) + T(n-2,k).
%C A106408 Row sums are A004798 (convolution of Fibonacci numbers 1,2,3,5,... with themselves). Central numbers of the rows are A006498 (a(n) = a(n-1)+a(n-3)+a(n-4)). First column and main diagonal are Fibonacci numbers 1,2,3,5,... First subdiagonal are 2*Fibonacci numbers. T(n,k) = F(n-k+2)*F(k+1) where F(m) is the m-th Fibonacci number. For the antidiagonal sums b(n): b(1) = 1, b(2) = 2, then b(n) = b(n-1) + b(n-2) + F(floor((n+3)/2)).
%C A106408 T(n,k) is the number of Boolean intervals of the form [s_k,w] in the weak order on S_n, for a fixed simple reflection s_k. - _Bridget Tenner_, Jan 16 2020
%H A106408 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 A106408 G.f.: (1+x+y+x*y)/((1-x-x^2)*(1-y-y^2)) [U coordinates] - _N. J. A. Sloane_, Jun 01 2005
%e A106408 Triangle begins
%e A106408    1;
%e A106408    2,  2;
%e A106408    3,  4,  3;
%e A106408    5,  6,  6,  5;
%e A106408    8, 10,  9, 10,  8;
%Y A106408 Cf. A000045, A004798, A006498.
%K A106408 nonn,tabl
%O A106408 1,2
%A A106408 _Gerald McGarvey_, May 28 2005