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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). 3
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 (list; table; graph; refs; listen; history; text; internal format)
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

Table of n, a(n) for n=1..45.

Peter Luschny, Meanders.

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-Fran├žois Alcover, Jul 02 2013 *)

CROSSREFS

Row sums: A201640. Cf. A132812.

Sequence in context: A174943 A173317 A059811 * A091016 A198025 A205549

Adjacent sequences:  A202406 A202407 A202408 * A202410 A202411 A202412

KEYWORD

nonn,tabl

AUTHOR

Peter Luschny and Susanne Wienand, Dec 19 2011

STATUS

approved

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Last modified February 19 09:33 EST 2020. Contains 332041 sequences. (Running on oeis4.)