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A125077
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#4 in an infinite set of generalized Pascal's triangles with trigonometric properties.
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3
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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
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OFFSET
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1,3
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COMMENTS
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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
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LINKS
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FORMULA
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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.
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EXAMPLE
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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].
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CROSSREFS
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KEYWORD
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AUTHOR
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STATUS
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approved
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