关于Fibonacci number的一些comment


关于Tribonacci constant…

对于Fibonacci number, 它的discriminant \( \sqrt{5}\)起作用,对于Tribonacci series来说,它就变成了\( \sqrt{11}\)

Sequences

Given three sequences with recurrence \( s_n = s_{n-1}+s_{n-2}+s_{n-3}\)but different initial values as,
\[
\begin{array}{|c|c|c|c|c|c|c|c|c|}
\hline
\text{Name} & \text{Formula} & 0 & 1 & 2 & 3 & 4 & 5 & 6 & 7 & OEIS\\
\hline
R_n & x_1^n+x_2^n+x_3^n &3 &1 &3 &7 &11 &21 &39 & 71 & A001644 \\
\hline
S_n &\frac{x_1^n}{y_1}+\frac{x_2^n}{y_2}+\frac{x_3^n}{y_3}&3 &2 &5 &10 &17 &32 &59 &108 &(none)\\
\hline
T_n &\frac{x_1^n}{z_1}+\frac{x_2^n}{z_2}+\frac{x_3^n}{z_3}&0 &1 &1 &2 &4 &7 &13 &24& A000073 \\
\hline
\end{array}
\\
y_i =\tfrac{1}{19}(x_i^2-7x_i+22)
\\
z_i =-x_i^2+4x_i-1
\]
\( x_i \)是方程\(x^3-x^2-x-1=0\)的根,其中\(T_n\)是Tribonacci series,它的实根\(T = x_1 \approx 1.83929\)近似于Tribonacci constant

Powers

令\( a=\dfrac{1}{3}(19+3\sqrt{33})^{1/3},\; b=\dfrac{1}{3}(19-3\sqrt{33})^{1/3} \), then powers of the tribonacci constant \(T\) can be expressed in terms of those three sequences as,
\[ 3T^n = R_{n}+(a+b)S_{n-1}+3(a^2+b^2)T_{n-1}\]

q-Continued fraction

令\( q = -1/(e^{\pi\sqrt{11}})\)则
\[ \frac{(e^{\pi\sqrt{11}})^{1/24}}{\frac{1}{T}+1} = 1 + \cfrac{q}{1-q + \cfrac{q^3-q^2}{1 + \cfrac{q^5-q^3}{1 + \cfrac{q^7-q^4}{1 + \ddots }}}}\]

Snub cube

https://en.wikipedia.org/wiki/Snub_cube

Pi Formula

令\( v = 1/T\)则

\[ \sum_{n=0}^\infty \frac{(2n)!^3}{n!^6}\frac{2(4v+1)(2v+1)n+(4v^2+2v-1)}{(v+1)^{24n}} =\frac{4}{\pi}\]

Infinite Nested Radical

\[ \frac{1}{T-1} = \sqrt[3]{\frac{1}{2}+ \sqrt[3]{\frac{1}{2}+ \sqrt[3]{\frac{1}{2}+ \sqrt[3]{\frac{1}{2}+\dots}}}}\]

Complete elliptic integral of the first kind

它的精确值是

\[
K(k_{11}) = \frac{1}{11^{1/4}(4\pi)^2} \bigl(\tfrac{T+1}{T}\bigr)^2\; \Gamma\bigl(\tfrac{1}{11}\bigr) \Gamma\bigl(\tfrac{3}{11}\bigr) \Gamma\bigl(\tfrac{4}{11}\bigr) \Gamma\bigl(\tfrac{5}{11}\bigr) \Gamma\bigl(\tfrac{9}{11}\bigr) = 1.570983\dots
\]

Cubic Pell-Type Equation

丢番图三次Pell型方程
\[
a^3 – 2 a^2 b + 2 b^3 – a^2 c – 2 a b c + 2 b^2 c + a c^2 + 2 b c^2 + c^3=1
\]
有无穷多组整数解
\[ a,\;b,\;c = T_{n-1},\;T_{n-2},\;T_{n-3}\]

0