OFFSET
0,8
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
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.
EXAMPLE
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;
...
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 *)
CROSSREFS
KEYWORD
nonn,tabl
AUTHOR
Landry Salle, May 02 2017
STATUS
approved