A362582 Triangular array read by rows. T(n,k) is the number of alternating permutations of [2n+1] having exactly 2k elements to the left of 1, n >= 0, 0 <= k <= n.
1, 1, 1, 5, 6, 5, 61, 75, 75, 61, 1385, 1708, 1750, 1708, 1385, 50521, 62325, 64050, 64050, 62325, 50521, 2702765, 3334386, 3427875, 3438204, 3427875, 3334386, 2702765, 199360981, 245951615, 252857605, 253708455, 253708455, 252857605, 245951615, 199360981
Offset: 0
Examples
T(2,1) = 6 because we have: {2, 3, 1, 5, 4}, {2, 4, 1, 5, 3}, {2, 5, 1, 4, 3}, {3, 4, 1, 5, 2}, {3, 5, 1, 4, 2}, {4, 5, 1, 3, 2}. Triangle begins 1; 1, 1; 5, 6, 5; 61, 75, 75, 61; 1385, 1708, 1750, 1708, 1385; 50521, 62325, 64050, 64050, 62325, 50521; ...
Links
- Alois P. Heinz, Rows n = 0..140, flattened
Programs
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Maple
b:= proc(u, o) option remember; `if`(u+o=0, 1, add(b(o-1+j, u-j), j=1..u)) end: T:= (n, k)-> binomial(2*n, 2*k)*b(2*k, 0)*b(2*(n-k), 0): seq(seq(T(n, k), k=0..n), n=0..8); # Alois P. Heinz, Apr 25 2023
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Mathematica
nn = 6; B[n_] := (2 n)!/2^n; e[z_] := Sum[z^n/B[n], {n, 0, nn}]; Map[Select[#, # > 0 &] &,Table[B[n], {n, 0, nn}] CoefficientList[Series[1/e[-u z]*1/e[-z], {z, 0, nn}], {z, u}]] // Grid
Comments