A214656 -id:A214656 - cp's OEIS frontend

A214315 Floor of the real part of the zeros of the complex Fibonacci function on the right half-plane.

0, 1, 3, 5, 7, 9, 10, 12, 14, 16, 18, 20, 21, 23, 25, 27, 29, 31, 32, 34, 36, 38, 40, 42, 43, 45, 47, 49, 51, 53, 54, 56, 58, 60, 62, 63, 65, 67, 69, 71, 73, 74, 76, 78, 80, 82, 84, 85, 87, 89, 91, 93, 95, 96, 98, 100, 102, 104, 106, 107, 109, 111, 113, 115, 117, 118

Offset: 0

Wolfdieter Lang, Jul 24 2012

nonn

For the complex Fibonacci function and its complex zeros see the Koshy reference, pp. 523-524. See also the formula for F(z) given in the formula section of A052952. The real parts of the zeros of F are x_0(k) = alpha*k, with alpha = 2*(Pi^2)/(Pi^2 + (2*log(phi))^2), where phi = (1+sqrt(5))/2, and integer k. The corresponding imaginary parts are y_0(k) = - 4*Pi*log(phi)*k/(Pi^2 + (2*log(phi))^2). alpha is approximately 1.828404783. The zeros lie in the lower right and the upper left half-planes, and there is a zero at the origin.
a(n) = floor(alpha*n), n>=0, is a Beatty sequence with the complementary sequence b(n) = floor(beta*n), with beta = alpha/(alpha-1), approximately 2.207139336.
For the floor of the negative imaginary part see A214656.

			The complementary Beatty sequences start with:
n:    1   2 3  4  5  6   7   8   9  10  11  12  13  14  15  16
a(n): 0   1 3  5  7  9  10  12  14  16  18  20  21  23  25  27
b(n): (0) 2 4  6  8 11  13  15  17  19  22  24  26  28  30  33

Thomas Koshy, "Fibonacci and Lucas Numbers with Applications", John Wiley and Sons, 2001.

G. C. Greubel, Table of n, a(n) for n = 0..10000

Cf. A052952 (Fibonacci related formula), A214656.

R:= RealField(100); [Floor(2*n*Pi(R)^2/(Pi(R)^2 + (2*Log((1+Sqrt(5))/2))^2)) : n in [0..100]]; // G. C. Greubel, Mar 09 2024

a[n_]:= Floor[2*n*Pi^2/(Pi^2 + 4*Log[GoldenRatio]^2)]; Table[a[n], {n, 0, 65}] (* Jean-François Alcover, Jul 03 2013 *)

[floor(2*n*pi^2/(pi^2 +4*(log(golden_ratio))^2)) for n in range(101)] # G. C. Greubel, Mar 09 2024

a(n) = floor(alpha*n), n>=0, with alpha = x_0(1) given in the comment section.

A214672 Floor of the imaginary parts of the zeros of the complex Lucas function on the left half-plane.

Original entry on oeis.org

0, 0, 1, 1, 2, 3, 3, 4, 4, 5, 5, 6, 7, 7, 8, 8, 9, 9, 10, 10, 11, 12, 12, 13, 13, 14, 14, 15, 15, 16, 17, 17, 18, 18, 19, 19, 20, 21, 21, 22, 22, 23, 23, 24, 24, 25, 26, 26, 27, 27, 28, 28, 29, 29, 30, 31, 31, 32, 32, 33, 33, 34, 35, 35, 36

Offset: 0

Wolfdieter Lang, Jul 25 2012

nonn

For the complex Lucas function L(z) and its zeros see the comments in A214671 and the Koshy reference.

Thomas Koshy, "Fibonacci and Lucas Numbers with Applications", John Wiley and Sons, 2001.

G. C. Greubel, Table of n, a(n) for n = 0..10000

Cf. A214656 (Fibonacci case), A214671 (floor of real parts), A214673 (moduli).

R:= RealField(100); [Floor((2*n+1)*(2*Pi(R)*Log((1 + Sqrt(5))/2))/(Pi(R)^2 + (2*Log((1+Sqrt(5))/2))^2)) : n in [0..100]]; // G. C. Greubel, Mar 09 2024

Table[Floor[(2*n+1)*(2*Pi*Log[GoldenRatio])/(Pi^2 + (2*Log[GoldenRatio])^2)], {n, 0, 100}] (* G. C. Greubel, Mar 09 2024 *)

[floor(2*(2*n+1)*pi*log(golden_ratio)/(pi^2 +4*(log(golden_ratio))^2)) for n in range(101)] # G. C. Greubel, Mar 09 2024

a(n) = floor((2*n+1)*b/2), n >= 0, with b/2 = -y_0(0) = 2*Pi*log(phi) / (Pi^2 + (2*log(phi))^2), with phi = (1+sqrt(5))/2. Note that b/2 is approximately 0.2800649542... . The constant b appears in the corresponding Fibonacci case A214656.

A214657 Floor of the moduli of the zeros of the complex Fibonacci function.

Original entry on oeis.org

0, 1, 3, 5, 7, 9, 11, 13, 15, 17, 19, 21, 22, 24, 26, 28, 30, 32, 34, 36, 38, 40, 42, 43, 45, 47, 49, 51, 53, 55, 57, 59, 61, 63, 65, 66, 68, 70, 72, 74, 76, 78, 80, 82, 84, 86, 87, 89, 91, 93, 95, 97, 99, 101, 103, 105, 107, 108, 110, 112, 114, 116, 118, 120, 122, 124

Offset: 0

Wolfdieter Lang, Jul 25 2012

nonn

For the complex Fibonacci function F(z) and its zeros see the T. Koshy reference given in A214315. There the formula for the real and imaginary parts of the zeros is also given.
  F: C -> C, z -> F(z) with F(z) := (exp(log(phi)*z) - exp(i*Pi*z)*exp(-log(phi)*z))/(2*phi-1), with phi = (1+sqrt(5))/2 and the imaginary unit i.
The zeros in the complex plane lie on a straight line with angle Phi = -arctan(2*log(phi)/Pi). They are equally spaced with distance tau defined below. Phi is approximately -0.2972713044, corresponding to about -17.03 degrees. The moduli are |z_0(k)| = tau*k, with tau = 2*Pi/sqrt(Pi^2 + (2*log(phi))^2), which is approximately 1.912278633.
a(n) = floor(tau*n) is a Beatty sequence with the complementary sequence b(n) = floor(sigma*n), with sigma:= tau/(tau-1), approximately 2.096156332.

			The complementary Beatty sequences a(n) and  b(n) start:
n:     0 1 2 3 4  5  6  7  8  9 10 11 12 13 14 15 16 17 18 ...
a(n):  0 1 3 5 7  9 11 13 15 17 19 21 22 24 26 28 30 32 34 ...
b(n): (0)2 4 6 8 10 12 14 16 18 20 23 25 27 29 31 33 35 37 ...

Thomas Koshy, "Fibonacci and Lucas Numbers with Applications", John Wiley and Sons, 2001.

G. C. Greubel, Table of n, a(n) for n = 0..10000

Cf. A214315, A214656.

R:= RealField(100); [Floor(2*n*Pi(R)/Sqrt(Pi(R)^2 + (2*Log((1+Sqrt(5))/2))^2)) : n in [0..100]]; // G. C. Greubel, Mar 09 2024

Table[Floor[2*n*Pi/Sqrt[Pi^2 + (2*Log[GoldenRatio])^2]], {n,0,100}] (* G. C. Greubel, Mar 09 2024 *)

[floor(2*n*pi/sqrt(pi^2 +4*(log(golden_ratio))^2)) for n in range(101)] # G. C. Greubel, Mar 09 2024

a(n) = floor(n*tau), n>=0, with tau = |z_0(1)| = 2*Pi/sqrt(Pi^2 + (2*log(phi))^2).

A098295 ((3/2)^n)/2^a(n) lies in the half-open interval [1,2).

Original entry on oeis.org

0, 1, 1, 2, 2, 3, 4, 4, 5, 5, 6, 7, 7, 8, 8, 9, 9, 10, 11, 11, 12, 12, 13, 14, 14, 15, 15, 16, 16, 17, 18, 18, 19, 19, 20, 21, 21, 22, 22, 23, 23, 24, 25, 25, 26, 26, 27, 28, 28, 29, 29, 30, 31, 31, 32, 32, 33, 33, 34, 35, 35, 36, 36, 37, 38, 38, 39, 39, 40, 40, 41, 42, 42, 43, 43

Offset: 1

Wolfdieter Lang, Oct 18 2004

Stacking perfect fifths (the frequency ratio of a fifth is 3/2), a division by 2^a(n) leads the equivalent tone belonging to the first octave interval [1,2). For example, the third fifth, (3/2)^3, falls into the second octave. This means it lies in the interval [2^1,2^2)=[2,4). Hence ((3/2)^3)/2^1 belongs to the first octave, the interval [1,2).

This sequence coincides for the first 93 term with the floor of y(n)= 4*Pi*log(phi)*n/(Pi^2 + (2*log(phi)^2)), with phi:=(1+sqrt(5))/2. a(n) = floor(y(n)), for n=1..93. Note that y(n) is not the imaginary part of the zero of the Fibonacci function because of a different bracket setting. See A214656. - Wolfdieter Lang, Jul 24 2012

			(3/2)^12 lies in the eighth octave [2^7,2^8) and
((3/2)^12)/2^a(12)= ((3/2)^12)/2^7 = 3^12/2^19 = 531441/524288 = 1.01363... belongs to the first octave [1,2). This ratio is called the Pythagorean comma.

Handbook for Acoustic Ecology, Pythagorean Scale.
Eric Weisstein's World of Music, Pythagorean Scale

This sequence differs from A074840 for the first time at entry a(41)=23: A074840(41)=24.

Cf. A020857, A056576, A098294, A214656.

a(n) = logint(3^n, 2) - n; \\ Ruud H.G. van Tol, Jan 26 2024

a(n) = A098294(n)-1, n >= 1.
a(n) = ceiling(tau*n)-1 with tau = log(3)/log(2)-1 = 0.58496250072..., n >= 1.
a(n) = A056576(n) - n. - Ruud H.G. van Tol, Jan 26 2024

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A214315 Floor of the real part of the zeros of the complex Fibonacci function on the right half-plane.

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A214672 Floor of the imaginary parts of the zeros of the complex Lucas function on the left half-plane.

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A214657 Floor of the moduli of the zeros of the complex Fibonacci function.

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A098295 ((3/2)^n)/2^a(n) lies in the half-open interval [1,2).

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