Additional comments from Gary W. Adamson (qntmpkt(AT)yahoo.com), Aug 22 2009 on A014664
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 Similarly, the entry for prime 7 = 3 since 1/7 in base 2 = .001001001...;
 (cycle length 3) or 2^3 == 1 MOD 7, where 1/7 = (1/8 + 1/64 + 1/512 + ...)
 Using the example in [Conway & Guy] using the seed constant 1.24697...
 = 2 * Cos 2Pi/7, and f(x) = x^2 - 2, we obtain the "3" cycle: 1.24697;...->
 -.445041...-> -1.80193...; then back to 1.24697...; noting that -.445041...=
 2 * Cos 2*2Pi/7, and -1.801937...= 2 * Cos 3*2Pi/7.
 Generally, we may obtain the cyclic constants for any odd integer as done for
 "7", then extract the data pertaining to the primes, where in the following
 set of polynomials, n-th degree pertains to 2*n + 1; e.g. 3-rd degree
 polynomial x^3 + x^2 - 2x - 1 = 0 has roots: (-1.801..., 1.24697..., &
 -.445041...). Since the next prime = 11, we would access (11-1)/2 = 5-th
 degree polynomial: x^5 + x^4 - 4x^3 - 3x^2 + 3x + 1 = 0; noting that
 using f(x) = x^2 - 2, the cycle length = 12 for those constants.
 The set of polynomials with cyclical roots = a signed version of A065941:
 with rows signed (++--++--++...) and Pascal's triangle columns are repeated:
 1
 1, 1;
 1, 1, 1;
 1, 1, 2, 1;
 1, 1, 3, 2, 1;
 1, 1, 4, 3, 3, 1;
 1, 1, 5, 4, 6, 3, 1;
 ...
 The polynomials are then:
 1;
 x + 1;
 x^2 + x - 1;
 x^3 + x^2 - 2x - 1;
 (with the "3" cycle roots pertaining to the Heptagon)
 x^4 + x^3 - 3x^2 - 2x + 1;
 x^5 + x^4 - 4x^3 - 3x^2 + 3x + 1;
 ...
 with the following trigonometric identities holding: (given x = 2*Cos 2A)
 then:
 .
 Sin A / Sin A = .(1) 1;
 Sin 3A / Sin A = (x + 1);
 Sin 5A / Sin A = (x^2 + x - 1);
 Sin 7A / Sin A = (x^3 + x^2 - 2x - 1);
 ...
 (same set of polynomials as before)...
 Besides f(x) = x^2 - 2, corresponding to the 2*Cos 2Pi/N case, (N = prime
 N-gons with associated trigonometric constants), we may manipulate x^2 - 2,
 obtaining the following set with the same cycle lengths (e.g. for N = 7 the
 cycle length for the constants, seed 2*2Cos 2Pi/7 = 3, f(x) = x^2 - 2.
 Similarly,
 .
 Using seed Cos 2*Pi/N, the cyclic f(x) = 2x^2 - 1
 ......seed Sin^2 2*Pi/N, the cyclic f(x) = 4x*(1-x), the logistic map.
 ......seed Tan 2*Pi/N, the cyclic f(x) = 2x / (1-x^2).
 ......seed Sec 2*Pi/N, the cyclic f(x) = x^2 / (2 - x^2).
 ......seed Csc 2*Pi/N, the cyclic f(x) = x^2 / (x^2 - 2).
 ......seed Cot 2*Pi/N, the cyclic f(x) = (x^2 - 1)/ 2x.
 .
 Furthermore, we may obtain the Cot formula using Newton's method on sqrt(-1)
 .All of the formulas for example, have a "3" cycle with N=7, and similarly
 we may obtain a(n) given N, primes.
 .
 The Mandelbrot set may be mapped using A014664 along with the "least
 exponents" for the odd integers, using "Lavaurs Algorithm" (Cf. the
 homepage website of Benoit Mandelbrot). Accessing the chart showing the
 M-set, we note that (for example), the 7-ths and 15-ths are connected by
 "3" and "4" respectively, since 2^3 == 1 MOD 7 and 2^4 == 1 MOD 15/
 The set of polynomials with cyclical constants (derived from repeated
 columns of Pascal's triangle, Cf. A065941) are charpolys of a special set
 of Cartan-like matrices with (1,1,1,...) in the super and subdiagonals
 and (-1, 0, 0, 0,...) as the main diagonal. For example, the matrix:
 [ -1, 1, 0;
 ..1, 0, 1;
 ..0, 1, 0;]
 .
 has the 3-rd degree charpoly used in previous examples, relating to the
 Heptagon: x^3 + x^2 - 2x - 1 = 0, (roots 1.24697...; -1.801937...; and
 -.445041...) and the 3-cycle given f(x) = x^2 - 2 or the other trigonometric
 equivalents for Cos 2*Pi/7, Sin^2 2Pi/7,...etc.