A125077 - OEIS

%I #16 Jun 05 2021 01:37:43

%S 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,

%T 859,780,2131,1,15,93,499,1091,3821,7953,1,16,140,592,2774,4912,16556,

%U 10864,29681,1,19,156,1044,3366

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

%C 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).

%C 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

%F 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.

%e First few rows of the triangle are:

%e   1;

%e   1,  3;

%e   1,  4, 11;

%e   1,  7, 15,  41;

%e   1,  8, 38,  56, 153;

%e   1, 11, 46, 186, 209, 571;

%e   1, 12, 81, 232, 859, 780, 2131;

%e   ...

%e 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].

%Y Cf. A125076, A125078.

%Y Cf. A001835, A001353. - _Gary W. Adamson_, Jun 21 2009

%K nonn,tabl

%O 1,3

%A _Gary W. Adamson_, Nov 18 2006