login
A202409
Triangle read by rows, n>=1, 1<=k<=n, T(n,k) = k*binomial(n,k)^3*(n^2+n-k*n-k+k^2)/((n-k+1)^2*n).
4
1, 4, 4, 9, 36, 9, 16, 168, 168, 16, 25, 550, 1400, 550, 25, 36, 1440, 7500, 7500, 1440, 36, 49, 3234, 30135, 61250, 30135, 3234, 49, 64, 6496, 98784, 356720, 356720, 98784, 6496, 64, 81, 11988, 278208, 1629936, 2889432, 1629936, 278208, 11988, 81
OFFSET
1,2
COMMENTS
Let a meander be defined as in the link and m = 3. Then T(n,k) counts the invertible meanders of length m(n+1) built from arcs with central angle 360/m whose binary representation have mk '1's.
LINKS
EXAMPLE
[1] 1
[2] 4, 4
[3] 9, 36, 9
[4] 16, 168, 168, 16
[5] 25, 550, 1400, 550, 25
[6] 36, 1440, 7500, 7500, 1440, 36
T(2,1) = 4 because the invertible meanders of length 9 and central angle 120 degree which have three '1's in their binary representation are {100100100, 100011000, 110001000, 111000000}.
MAPLE
A202409 := (n, k) -> k*binomial(n, k)^3*(n^2+n-k*n-k+k^2)/((n-k+1)^2*n);
seq(print(seq(A202409(n, k), k=1..n)), n=1..6);
MATHEMATICA
t[n_, k_] := k*Binomial[n, k]^3*(n^2 + n - k*n - k + k^2)/((n - k + 1)^2*n); Table[t[n, k], {n, 1, 9}, {k, 1, n}] // Flatten (* Jean-François Alcover, Jul 02 2013 *)
CROSSREFS
Row sums: A201640. Cf. A132812.
Sequence in context: A174943 A173317 A059811 * A091016 A198025 A205549
KEYWORD
nonn,tabl
AUTHOR
STATUS
approved