A204201 Triangle based on (0,1/3,1) averaging array.
1, 1, 4, 1, 5, 10, 1, 6, 15, 22, 1, 7, 21, 37, 46, 1, 8, 28, 58, 83, 94, 1, 9, 36, 86, 141, 177, 190, 1, 10, 45, 122, 227, 318, 367, 382, 1, 11, 55, 167, 349, 545, 685, 749, 766, 1, 12, 66, 222, 516, 894, 1230, 1434, 1515, 1534, 1, 13, 78, 288, 738, 1410
Offset: 1
Examples
The (0,1/3,1) averaging array has these first four rows: 1/3 1/6....2/3 1/12...5/12...5/6 1/24...1/4....5/8...11/12. Multiplying those rows by 3,6,12,24, respectively: 1 1...4 1...5...10 1...6...15...22 The first nine rows: 1 1...4 1...5...10 1...6...15...22 1...7...21...37...46 1...8...28...58...83...94 1...9...36...86...141..177..190 1...10..45...122..227..318..367..382 1...11..55...167..349..545..685..749..766
Crossrefs
Cf. A204202.
Programs
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Mathematica
a = 0; r = 1/3; b = 1; t[1, 1] = r; t[n_, 1] := (a + t[n - 1, 1])/2; t[n_, n_] := (b + t[n - 1, n - 1])/2; t[n_, k_] := (t[n - 1, k - 1] + t[n - 1, k])/2; u[n_] := Table[t[n, k], {k, 1, n}] Table[u[n], {n, 1, 5}] (* averaging array *) u = Table[(1/2) (1/r) 2^n*u[n], {n, 1, 12}]; TableForm[u] (* A204102 triangle *) Flatten[u] (* A204201 sequence *)
Formula
From Philippe Deléham, Dec 24 2013: (Start)
T(n,n) = A033484(n-1).
Sum{k=1..n} T(n,k) = A053220(n).
T(n,k) = T(n-1,k)+3*T(n-1,k-1)-2*T(n-2,k-1)-2*T(n-2,k-2), T(1,1)=1, T(2,1)=1, T(2,2)=4, T(n,k)=0 if k<1 or if k>n. (End)
Comments