A286096 Triangle read by rows giving numerators of the Fourier expansion of cos^n(x).

1, 0, 1, 1, 0, 1, 0, 3, 0, 1, 3, 0, 4, 0, 1, 0, 10, 0, 5, 0, 1, 10, 0, 15, 0, 6, 0, 1, 0, 35, 0, 21, 0, 7, 0, 1, 35, 0, 56, 0, 28, 0, 8, 0, 1, 0, 126, 0, 84, 0, 36, 0, 9, 0, 1, 126, 0, 210, 0, 120, 0, 45, 0, 10, 0, 1, 0, 462, 0, 330, 0, 165, 0, 55, 0, 11, 0, 1, 462, 0, 792, 0, 495, 0, 220, 0, 66, 0, 12, 0, 1

Offset: 0

Author

Landry Salle, May 02 2017

Keywords

• nonn
• tabl

Comments

Doubling the initial term of each line and dropping the 0's transforms this triangle to the right half of Pascal's triangle (A007318).

Row sums are A011782. - Omar E. Pol, May 02 2017

Examples

			Triangle begins:
1;
0,   1;
1,   0,   1;
0,   3,   0,   1;
3,   0,   4,   0,   1;
0,  10,   0,   5,   0,   1;
10,  0,  15,   0,   6,   0,   1;
0,  35,   0,  21,   0,   7,   0,   1;
35,  0,  56,   0,  28,   0,   8,   0,   1;
0, 126,   0,  84,   0,  36,   0,   9,   0,   1;
126, 0, 210,   0, 120,   0,  45,   0,  10,   0,   1;
0, 462,   0, 330,   0, 165,   0,  55,   0,  11,   0,   1;
462, 0, 792,   0, 495,   0, 220,   0,  66,   0,  12,   0,   1;
...
		

Crossrefs

Cf. A007318, A100257 (same sequence with rows reversed).

Programs

Mathematica

row[n_] := If[n==0, {1}, 2^(n-1)*TrigReduce[Cos[x]^n] /. Cos[Times[k_., x]] -> x^k // CoefficientList[#, x]&]; Table[row[n], {n, 0, 12}] // Flatten
(* Second program: *)
T[n_, n_] = 1; T[n_, k_] /; k == n-1 || k>n = 0; T[n_, 1] := 2 T[n-1, 0] + T[n-1, 2]; T[n_, 0] := T[n-1, 1]; T[n_, k_] /; 1 < k <= n := T[n, k] = T[n-1, k-1] + T[n-1, k+1]; T[, ] = 0; Table[T[n, k], {n, 0, 12}, {k, 0, n}] // Flatten (* Jean-François Alcover, May 02 2017 *)

Formula

cos^n(x) = (1/2^(n-1)) * Sum_{k=0..n} T(n,k) * cos(k*x).

T(n,k) = T(n-1,k-1) + T(n-1,k+1) if k != 1, T(n,1) = 2*T(n-1,0) + T(n-1,2), T(n,k) = 0 if k < 0 or k > n.