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%I #34 Oct 05 2020 17:47:50
%S 0,1,0,1,0,1,0,1,2,4,4,3,0,3,2,0,0,4,8,11,0,11,14,16,16,80,80,76,68,
%T 57,0,57,46,32,16,0,0,80,160,236,304,361,0,361,418,464,496,512,512,
%U 3904,3904,3824,3664,3428,3124,2763,0,2763,2402,1984,1520,1024,512,0
%N Triangle read by rows: number of type B alternating permutations according to their last value.
%C "The table counting type B alternating permutations by their last value is obtained by the following algorithm: first separate the picture by the column p = 0 and then compute two triangles. Put 1 at the top of each triangle and compute the rest as follows: fill the second row of the left (resp. right) triangle as the sum of the elements of the first row (resp. strictly) to their left. Then fill the third row of the right (resp. left) triangle as the sum of the elements of the previous row (resp. strictly) to their right. Compute all rows successively by reading from left to right and right to left alternately." [Joshua-Verges et al.]
%H M. Josuat-Vergès, J.-C. Novelli and J.-Y. Thibon, <a href="http://arxiv.org/abs/1110.5272">The algebraic combinatorics of snakes</a>, arXiv preprint arXiv:1110.5272 [math.CO], 2011.
%e Triangle begins:
%e 0
%e 1 0 1
%e 0 1 0 1 2
%e 4 4 3 0 3 2 0
%e 0 4 8 11 0 11 14 16 16
%e 80 80 76 68 57 0 57 46 32 16 0
%e 0 80 160 236 304 361 0 361 418 464 496 512 512
%o (PARI) T(n,k) = {if ((k==0), return(0)); if (n==1, if (abs(k)==1, return(1))); if (n%2, if (k<0, sum(j=k+1, n-1, T(n-1,j)), sum(j=k, n-1, T(n-1,j))), if (k<0, sum(j=-n+1, k, T(n-1,j)), sum(j=-n+1, k-1, T(n-1,j))));}
%o tabf(nn) = {for (n=0, nn, for (k=-n, n, print1(T(n, k), ", ");); print;);} \\ _Michel Marcus_, Jun 03 2020
%Y See A202690 for another version.
%Y See A010094 and A008281 for type A permutations.
%Y Cf. A000831 (row sums, for n>0).
%Y Cf. A001586 for the middle coefficients.
%K nonn,tabf
%O 0,9
%A _N. J. A. Sloane_, Dec 22 2011
%E More terms from _Michel Marcus_, Jun 03 2020