A125077 #4 in an infinite set of generalized Pascal's triangles with trigonometric properties.

1, 1, 3, 1, 4, 11, 1, 7, 15, 41, 1, 8, 38, 56, 153, 1, 11, 46, 186, 209, 571, 1, 12, 81, 232, 859, 780, 2131, 1, 15, 93, 499, 1091, 3821, 7953, 1, 16, 140, 592, 2774, 4912, 16556, 10864, 29681, 1, 19, 156, 1044, 3366

Offset: 1

Author

Gary W. Adamson, Nov 18 2006

Keywords

• nonn
• tabl

Comments

Row sums are powers of 4. The triangle is #4 in an infinite of generalized Pascal's triangles constrained by two rules: row sums are powers of N and upward sloping diagonals (as coefficients to polynomials with alternating signs) have roots N + 2*cos(2*Pi/Q).

Right border, A001835, and next to right border, A001353 = bisections of denominator of continued fraction [1, 2, 1, 2, 1, 2, 1, 2]; i.e., bisection of A002530. - Gary W. Adamson, Jun 21 2009

Examples

			First few rows of the triangle are:
  1;
  1,  3;
  1,  4, 11;
  1,  7, 15,  41;
  1,  8, 38,  56, 153;
  1, 11, 46, 186, 209, 571;
  1, 12, 81, 232, 859, 780, 2131;
  ...
The upward-sloping diagonal (1, 11, 38, 41) relates to the heptagon and in the form x^3 - 11x^2 + 38x - 41 has a root 5.24697960... = 4 + 2*cos(2*Pi/7). The corresponding matrix is [3, 1, 0; 1, 4, 1; 0, 1, 4]. The next upward-sloping diagonal relates to the octagon, with a characteristic polynomial x^3 - 12x^2 + 46x - 56 and a root 5.414213562... = 4 + 2*cos(2*Pi/8). The corresponding matrix is [4, 1, 0; 1, 4, 1; 0, 1, 4].
		

Crossrefs

Cf. A125076, A125078.

Cf. A001835, A001353. - Gary W. Adamson, Jun 21 2009

Formula

Upward-sloping diagonals of the triangle are derived from (alternating) characteristic polynomials of two types of matrices: those of the form: (all 1's in the super and subdiagonals and 3,4,4,4,... in the main diagonal) and (all 1's in the super and subdiagonals and 4,4,4,... in the main diagonal.