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Colin Barker, <a href="/A328604/b328604.txt">Table of n, a(n) for n = 0..1000</a>

<a href="/index/Rec#order_02">Index entries for linear recurrences with constant coefficients</a>, signature (2,9).

(PARI) Vec((1 + 7*x) / (1 - 2*x - 9*x^2) + O(x^30)) \\ Colin Barker, Dec 13 2019

nonn,new,easy

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a(n)/a(n-1) ~ 1 + sqrt(10).

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Numerators of metallic fraction convergents to the positive root of 4*x^2 - 4*x - 9.

G.f.: (1 + 7*x) / (1 - 2*x - 9*x^2).

Quantitatively, a metallic fraction convergence is a pattern defined by a rational sequence that approximates the positive real root of a*(x^0) = x^(y - 1)*(ax + 2b), recognizing: that a^2 + b^2 = c^2, that b + c = ax + 2b, that a*(x^0) = 2b*(x^(y - 1)) + a*(x^y), that a*(x^1) = 2b*(x^y) + a*(x^(y + 1)) toward positive y vanishing point convergence and that a*(x^-1) = 2b*(x^(y - 2)) + a*(x^(y - 1)) toward negative y vanishing point convergence.

If y is positive, the metallic fraction sequence is recursively defined by b(n) = (2b/a)*b(n-(y-1)) + b(n-y) with initial seeding inputs numbering from b(0) to b(y-1). If y is negative, the metallic fraction sequence is recursively defined by b(n) = (2b/a)*b(n+y-1) + b(n+y) with initial seeding inputs numbering from b(0) to b(-y).

This particular A328604 metallic fraction convergence uses b(0) = 1/2 and b(1) = 9/4 but, provided that b(n) = (9/4)*b(n-2) + b(n-1), the generated sequence will converge to the x = (1/2)*(1 + 10^(1/2)) root of 4 = x^(-2)*(4x + 9), no matter the b(0) and b(1) initial seeding input values.

More broadly, the positive real root of x^(y - 1)*(ax + 2b) - a = 0 is approximated by recursive b(n) as long as recursive b(n) is given the right number of initial seeding inputs to start its recursions. As such, a "metallic fraction" is variable because its name references the generating methodology, not the generated sequence per se, though A328604 is an excellent canonical representation.

Along the 2x^2 - 2x = b metallic fraction convergences, two different b(0) input values recursively output relatively simple sequences: if 2b is odd, b(0) = 1/2; b(1) = 2b/a outputs a powers-of-two denominator and, if 2b is even, b(0) = 1; b(1) = 2b/a outputs either no denominator, if 2b is divisible by 4, or else a powers-of-two doubled up denominator, if 2b is not divisible by 4.

Extrapolating from Colin Barker's conjectures and this sequence's cross-references, the numerators of these relatively simple a = 4 and y = -1 metallic fraction sequences can be recursively generated: by a(0) = 1; a(1) = 2b; a(n) = 2*a(n-1) + 2b*a(n-2) when 2b is odd, by a(0) = 1; a(1) = b/2; a(n) = a(n-1) + (b/2)*a(n-2) when 2b is a multiple of 4 or else by a(0) = 1; a(1) = b; a(2) = 2b; a(3) = b*(b + 4); a(n) = (2b + 2)*a(n-2) - b^2*a(n-4).

Moreover, these relatively simple x*(x - 1) = b/2 metallic fraction sequence are explicitly defined as b(n) = (((2b + 1) + ((2b - 2)/2)*sqrt(2b + 1))/(2(2b + 1)))*((1 + sqrt(2b + 1))/2)^n + (((2b + 1) - ((2b - 2)/2)*sqrt(2b + 1))/(2*(2b + 1)))*((1 - sqrt(2b + 1))/2)^n, if 2b is even, and b(n) = (((((2b - 1)/2) + 1) + ((2b - 1)/2)*sqrt(2b + 1))/(2*(2b + 1)))*((1 + sqrt(2b + 1))/2)^n + (((((2b - 1)/2) + 1) - ((2b - 1)/2)*sqrt(2b + 1))/(2*(2b + 1)))*((1 - sqrt(2b + 1))/2)^n, if 2b is odd.

Qualitatively, a metallic fraction convergence is a pattern defined by merging Euclid's algorithm with the golden ratio rectangle (and its Fibonacci spiral approximation) by constructing a |Pi/y| generalization of the metallic mean family (by playing with mathematical aesthetics based on the spiral of Theodorus, pinwheel tilings, Thales's theorem, the Pythagorean theorem, Gabriel's horn, the cornucopia and perspective graphic arts).

Kyle MacLean Smith, <a href="https://youtu.be/cjwzEsK37tc">PowerLawGeometry.com Classical Construction</a>, Bestape YouTube video (2019).

Kyle MacLean Smith, <a href="https://youtu.be/eq9AbcirsLk">Rectangular *Unity - PowerLawGeometry.com</a>, Bestape YouTube video (2019).

Kyle MacLean Smith, <a href="https://youtu.be/I84qsslRH8w">The Hyper *Metal Multi-Dimension</a>, Bestape YouTube video (2019).

Kyle MacLean Smith, <a href="https://youtu.be/s2_8grzsYvw">Thalesian Pinwheel Rotation</a>, Bestape YouTube video (2019).

Kyle MacLean Smith, <a href="https://community.wolfram.com/groups/-/m/t/1719229">Translating Between Iterations and Vanishing Point Recursions</a>, Wolfram Community discussion (2019).

b(0) = 1/2; b(1) = 9/4; b(n) = (9/4)*b(n-2) + b(n-1).

b(n) = ((5 + 4*sqrt(10))/20)*((1 + sqrt(10))/2)^n + ((5 - 4*sqrt(10))/20)*((1 - sqrt(10))/2)^n.

a(n)/ = 2*a(n-1) ~ + 9*a(n-2) for n>1 + sqrt(10). - _Colin Barker_, Oct 21 2019

Conjectures from Colin Barker, Oct 21 2019: (Start)

G.f.: (1 + 7*x) / (1 - 2*x - 9*x^2).

a(n) = 2*a(n-1) + 9*a(n-2) for n>1.

(End)

RecurrenceTable[{f[n] == (9/4)*f[n-2] + f[n-1], f[0] == 1/2, f[1] == 9/4}, f, {n, 0, 25}]

Counting the integers from 2b = 1 to 2b = 12, these "metallic fraction" numerators converge to the (1/2)*(1 + sqrt(2b + 1)) positive root of 4x^2 - 4x - 2b, using b(0) = 1/2; b(1) = 2b/a for odd 2b and b(0) = 1; b(1) = 2b/a for even 2b: A001333, A002531, A000244, A000045, A123011, A297189, A164544, A000079, Cf. A328604, Cf. A328605, Cf. A328606, A105476.

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Edited by N. J. A. Sloane, Dec 05 2019

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