A283613 T(n,k) = number of linear arrays of n 1's, n -1's, and k 0's such that no two adjacent elements are equal.
1, 1, 2, 6, 6, 2, 2, 12, 30, 38, 24, 6, 2, 18, 74, 174, 248, 212, 100, 20, 2, 24, 138, 480, 1092, 1668, 1700, 1110, 420, 70, 2, 30, 222, 1026, 3228, 7188, 11492, 13140, 10500, 5572, 1764, 252, 2, 36, 326, 1882, 7580, 22274, 48852, 80672, 100044, 91840, 60564, 27132, 7392, 924, 2, 42, 450, 3118, 15324, 56040, 156664, 339720, 574716, 757148, 769356, 591444, 332640, 129096, 30888, 3432, 2, 48, 594, 4804, 27888, 122136, 415576, 1118268, 2403588, 4143116, 5719788, 6281856, 5416488, 3586968, 1760616, 603174, 128700, 12870
Offset: 0
Examples
The table starts with columns k=0...11 and rows n=0...5: | 0 1 2 3 4 5 6 7 8 9 10 11 ----------------------------------------------------------- 0 | 1 1 1 | 2 6 6 2 2 | 2 12 30 38 24 6 3 | 2 18 74 174 248 212 100 20 4 | 2 24 138 480 1092 1668 1700 1110 420 70 5 | 2 30 222 1026 3228 7188 11492 13140 10500 5572 1764 252 For n=2, k=4 the 24 arrays are: [-1,0,-1,0,1,0,1,0] [-1,0,1,0,-1,0,1,0] [-1,0,1,0,1,0,-1,0] [1,0,-1,0,-1,0,1,0] [1,0,-1,0,1,0,-1,0] [1,0,1,0,-1,0,-1,0] [0,-1,1,0,-1,0,1,0] [0,-1,1,0,1,0,-1,0] [0,-1,0,-1,1,0,1,0] [0,-1,0,-1,0,1,0,1] [0,-1,0,1,-1,0,1,0] [0,-1,0,1,0,-1,1,0] [0,-1,0,1,0,-1,0,1] [0,-1,0,1,0,1,-1,0] [0,-1,0,1,0,1,0,-1] [0,1,-1,0,-1,0,1,0] [0,1,-1,0,1,0,-1,0] [0,1,0,-1,1,0,-1,0] [0,1,0,-1,0,-1,1,0] [0,1,0,-1,0,-1,0,1] [0,1,0,-1,0,1,-1,0] [0,1,0,-1,0,1,0,-1] [0,1,0,1,-1,0,-1,0] [0,1,0,1,0,-1,0,-1]
Programs
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Mathematica
nmax=8; Flatten[CoefficientList[Series[CoefficientList[Series[((x + 1)^2*Sqrt[(1 - y)/(1 - (2x + 1)^2*y)] - x - 1)/x, {y, 0, nmax}], y], {x, 0, 2nmax + 1}], x]] (* Indranil Ghosh, Mar 22 2017 *)
Formula
G.f.:((x+1)^2*sqrt((1-y)/(1-(2*x+1)^2*y))-x-1)/x.
T(n,0) G.f.: (1+y)/(1-y).
T(n,1) G.f.: (y^2 + 4*y + 1)/(1-y)^2.
T(n,2) G.f.: 2*y*(y^2 + 6*y + 3)/(1-y)^3.
T(n,3) G.f.: 2*y*(2*y^3 + 17*y^2 + 15*y + 1)/(1-y)^4.
T(n,4) G.f.: 4*y^2*(2*y^3 + 23*y^2 + 32*y + 6)/(1-y)^5.
T(n,5) G.f.: 2*y^2*(8*y^4 + 120*y^3 + 243*y^2 + 88*y + 3)/(1-y)^6.
T(n,2*n+1) = binomial(2*n,n).
T(n,2*n) = (n+2)*binomial(2*n,n).
T(n,n) = A110706(n) n > 0.
Sum_{2*n+k = m} T(n,k) = A199697(m).