A181365 Triangle read by rows: T(n,k) is the number of 2-compositions of n having least entry equal to k (n >= 1; 0 <= k <= floor(n/2)).
2, 6, 1, 22, 2, 78, 3, 1, 272, 6, 2, 940, 13, 2, 1, 3232, 28, 2, 2, 11080, 58, 3, 2, 1, 37920, 118, 6, 2, 2, 129648, 239, 12, 2, 2, 1, 443008, 484, 22, 2, 2, 2, 1513248, 979, 37, 3, 2, 2, 1, 5168000, 1976, 60, 6, 2, 2, 2, 17647552, 3980, 97, 12, 2, 2, 2, 1, 60258304, 8004
Offset: 1
Examples
T(4,1) = 3 because we have (1/3), (3/1), and (1,1/1,1) (the 2-compositions are written as (top row / bottom row)). Triangle starts: 2; 6, 1; 22, 2; 78, 3, 1; 272, 6, 2; 940, 13, 2, 1;
Links
- Alois P. Heinz, Rows n = 1..200, flattened
- G. Castiglione, A. Frosini, E. Munarini, A. Restivo and S. Rinaldi, Combinatorial aspects of L-convex polyominoes, European J. Combin. 28 (2007), no. 6, 1724-1741.
Programs
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Maple
h := proc (k) if k = 0 then (1-z)^2/(1-4*z+2*z^2) else (1-z)^2/(1-2*z+z^2-z^(2*k)) end if end proc: f := proc (k) options operator, arrow: h(k)-h(k+1) end proc; G := f(0)+sum(f(k)*t^k, k = 1 .. 30): Gser := simplify(series(G, z = 0, 20)): for n to 15 do P[n] := sort(coeff(Gser, z, n)) end do: for n to 15 do seq(coeff(P[n], t, k), k = 0 .. floor((1/2)*n)) end do; # yields sequence in triangular form # second Maple program: A:= proc(n, k) option remember; `if`(n=0, 1, add(add( `if`(i=0 and j=0, 0, A(n-i-j, k)), i=k..n-j), j=k..n)) end: T:= (n, k)-> A(n, k) -A(n, k+1): seq(seq(T(n, k), k=0..n/2), n=1..15); # Alois P. Heinz, Mar 16 2014
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Mathematica
A[n_, k_] := A[n, k] = If[n == 0, 1, Sum[Sum[If[i == 0 && j == 0, 0, A[n-i-j, k]], {i, k, n-j}], {j, k, n}]]; T[n_, k_] := A[n, k] - A[n, k+1]; Table[Table[T[n, k], {k, 0, n/2}], {n, 1, 15}] // Flatten (* Jean-François Alcover, May 28 2015, after Alois P. Heinz *)
Formula
G.f. for 2-compositions with all entries >= k is h(k,z) = (1-z)^2/(1-2*z+z^2-z^(2*k)) if k>0 and h(0,z) = (1-z)^2/(1-4*z+2*z^2) if k=0.
G.f. for 2-compositions with least entry k is f(k,z) = h(k,z)-h(k+1,z) (these are the column g.f.'s).
G.f.: G(t,z) = f(0,z) + Sum_{k>=1} f(k,z)*t^k.
Sum_{k >= 0} T(n,k) = A003480(n).
T(n,1) = A181367(n).
Sum_{k >= 0} k*T(n,k) = A181366(n).
Comments