Fibonacci numbers

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Let a1=1,a2=1,an+1=an+1+an(n⩾1)a_{1} = 1, a_{2} = 1, a_{n+1} = a_{n+1}+a_{n}(n \geqslant 1). Prove that the series ∑n=1∞1an\sum_{n = 1}^{\infty } \frac{1}{a_{n}} converges.

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The sequence of Fibonacci numbers (cf. Problem 53 of Part VII) is periodic mod mm, where mm is any integer. Calculate the length of the period for all prime numbers less than 30.

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The Fibonacci numbers 0,1,1,2,3,5,8,13,21,34,…0, 1, 1, 2, 3, 5, 8, 13, 21, 34, \ldots are defined as follows: Set u0=0u_{0} = 0, u1=1u_{1} = 1, un+un+1=un+2u_{n} + u_{n+1} = u_{n+2} for n=0,1,2,…n = 0, 1, 2, \ldots. The linear transformation

yn=u1x1+u2x2+⋯+unxn−unxn+1unun+2,n=1,2,3,… y_{n} = \frac{u_{1} x_{1} + u_{2} x_{2} + \cdots + u_{n} x_{n} - u_{n} x_{n+1}}{\sqrt{u_{n} u_{n+2}}}, \qquad n = 1, 2, 3, \ldots

is orthogonal.

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Using the method of undetermined coefficients, show that the coefficients AnA_{n} of the power series

11−z−z2=∑n=0∞Anzn \frac{1}{1-z-z^{2}} = \sum _{n = 0}^{\infty } A_{n} z^{n}

are determined by the conditions A0=1,A1=1,An+1=An+An−1,n∈NA_{0} = 1, A_{1} = 1, A_{n+1} = A_{n}+A_{n-1}, n \in \mathbb {N}. The numbers AnA_{n} are called the Fibonacci numbers. Prove that

An=15[(5+12)n+1−(1−52)n+1],n∈N. A_{n} = \frac{1}{\sqrt{5}}\left[\left(\frac{\sqrt{5}+1}{2}\right)^{n+1}-\left(\frac{1-\sqrt{5}}{2}\right)^{n+1}\right], \quad n \in \mathbb {N}.
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Problem 5D.21 from Axler S. "Linear Algebra Done Right" [Axl24]More problems in Section 5D "Diagonalizable Operators"

The Fibonacci sequence F0,F1,F2,…F_{0}, F_{1}, F_{2}, \ldots is defined by

F0=0,F1=1, and Fn=Fn−2+Fn−1 for n≥2 F_{0} = 0, F_{1} = 1, \text{ and } F_{n} = F_{n-2}+F_{n-1} \text{ for } n \geq 2

Define T∈L(R2)T \in \mathcal{L}\left(\mathbb {R}^{2}\right) by T(x,y)=(y,x+y)T(x, y) = (y, x+y).

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(a)

Show that Tn(0,1)=(Fn,Fn+1)T^{n}(0,1) = \left(F_{n}, F_{n+1}\right) for each nonnegative integer nn.

(b)

Find the eigenvalues of TT.

(c)

Find a basis of R2\mathbb {R}^{2} consisting of eigenvectors of TT.

(d)

Use the solution to (c) to compute Tn(0,1)T^{n}(0,1). Conclude that

Fn=15[(1+52)n−(1−52)n] F_{n} = \frac{1}{\sqrt{5}}\left[\left(\frac{1+\sqrt{5}}{2}\right)^{n}-\left(\frac{1-\sqrt{5}}{2}\right)^{n}\right]

for each nonnegative integer nn.

(e)

Use (d) to conclude that if nn is a nonnegative integer, then the Fibonacci number FnF_{n} is the integer that is closest to

15(1+52)n \frac{1}{\sqrt{5}}\left(\frac{1+\sqrt{5}}{2}\right)^{n}
Remark.
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Each FnF_{n} is a nonnegative integer, even though the right side of the formula in (d) does not look like an integer. The number

φ:=1+52 \varphi := \frac{1+\sqrt{5}}{2}

is called the golden ratio.

The Fibonacci numbers are 1,1,2,3,5,8,13,21,34,…1, 1, 2, 3, 5, 8, 13, 21, 34, \ldots. In general, the Fibonacci numbers are defined by f1=1f_1 = 1, f2=1f_2 = 1, and for n≥3n \geq 3, fn=fn−1+fn−2f_n = f_{n-1} + f_{n-2}. Prove that the nnth Fibonacci number fnf_n satisfies fn<2nf_n < 2^n.

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Let u(0)=0,u(1)=1u(0)=0, u(1)=1, and u(n)=u(n−1)+u(n−2)u(n)=u(n-1)+u(n-2), where n⩾2n \geqslant 2. The numbers u(n)u(n) are called Fibonacci numbers. Compute u(n)u(n).

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Let u(n)u(n) be the sequence of Fibonacci numbers. Prove that:

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(a)

u(1)+…+u(n)=u(n+2)−1u(1)+\ldots +u(n)=u(n+2)-1;

(b)

u(1)2+…+u(n)2=u(n)u(n+1)u(1)^{2}+\ldots +u(n)^{2}=u(n) u(n+1);

(c)

u(n+1)2−u(n−1)2=u(2n)u(n+1)^{2}-u(n-1)^{2}=u(2 n);

(d)

u(1)3+…+u(n)3=110[u(3n+2)+(−1)n+16u(n−1)+5]u(1)^{3}+\ldots +u(n)^{3}=\frac{1}{10}\left[u(3 n+2)+(-1)^{n+1} 6 u(n-1)+5\right];

(e)

u(m+n)=u(m)u(n−1)+u(m+1)u(n)u(m+n)=u(m) u(n-1)+u(m+1) u(n);

(f)

if nn divides mm, then u(n)u(n) divides u(m)u(m);

(g)

(u(n),u(m))=u((n,m))(u(n), u(m))=u((n, m)).

Prove that the (n+1)(n+1)-st term u(n+1)u(n+1) of the sequence of Fibonacci numbers (see 4.5) equals the determinant

[1100…00−1110…000−111…00…………………0000…−11] \left[\begin{smallmatrix} 1 & 1 & 0 & 0 & \ldots & 0 & 0 \\ -1 & 1 & 1 & 0 & \ldots & 0 & 0 \\ 0 & -1 & 1 & 1 & \ldots & 0 & 0 \\ \ldots & \ldots & \ldots & \ldots & \ldots & \ldots & \ldots \\ 0 & 0 & 0 & 0 & \ldots & -1 & 1 \end{smallmatrix}\right]

of order nn.

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Let

Cn(λ1,…,λn)=∥λ110…000−1λ21…000…………………………………000…λn−210000…−1λn−11000…0−1λn∥. C_{n}\left(\lambda _{1}, \ldots , \lambda _{n}\right)=\left\Vert \begin{smallmatrix} \lambda _{1} & 1 & 0 & \ldots & 0 & 0 & 0 \\ -1 & \lambda _{2} & 1 & \ldots & 0 & 0 & 0 \\ \ldots & \ldots \ldots & \ldots \ldots \ldots & \ldots \ldots \ldots \ldots \ldots \ldots \ldots \\ 0 & 0 & 0 & \ldots & \lambda _{n-2} & 1 & 0 \\ 0 & 0 & 0 & \ldots & -1 & \lambda _{n-1} & 1 \\ 0 & 0 & 0 & \ldots & 0 & -1 & \lambda _{n} \end{smallmatrix}\right\Vert .

Show that det⁡Cn=λndet⁡Cn−1+det⁡Cn−2\operatorname {det} C_{n}=\lambda_{n} \operatorname {det} C_{n-1}+\operatorname {det} C_{n-2}. For λ1=λ2=…=λn=1\lambda_{1}=\lambda_{2}=\ldots =\lambda_{n}=1, find the numerical value of det⁡Cn\operatorname {det} C_{n}.

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The Fibonacci numbers are defined by the recurrence relation (linear difference equation of second order with constant coefficients) sn+2=sn+1+sns_{n+2} = s_{n+1}+s_{n}, where n=0,1,…n = 0,1,\ldots and s0=0s_{0} = 0, s1=1s_{1} = 1. Write this recurrence relation in matrix form. Find s6s_{6}, s5s_{5}, and s4s_{4}.

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Let n∈Nn \in \mathbb {N}. The Fibonacci numbers are provided by the entries of the matrix

(1110)n,n=1,2,…. \begin{pmatrix} 1 & 1 \\ 1 & 0 \end{pmatrix}^{n}, \qquad n = 1, 2, \ldots .

Find the sequence provided by the 4×44 \times 4 matrix

((1110)⊗(1110))n,n=1,2,…. \left(\begin{pmatrix} 1 & 1 \\ 1 & 0 \end{pmatrix} \otimes \begin{pmatrix} 1 & 1 \\ 1 & 0 \end{pmatrix}\right)^{n}, \qquad n = 1, 2, \ldots .

Compare with the sequence of the Fibonacci numbers. Discuss.

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The Fibonacci numbers Fi,i∈ωF_{i}, i \in \omega, are defined as follows: Fi=iF_{i}=i for i<2i<2, Fi=Fi−1+Fi−2F_{i}=F_{i-1}+F_{i-2} otherwise. Prove that FiF_{i} is even if and only if ii is divisible by 3.

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Prove the identity Fi−1Fi+1=Fi2+(−1)iF_{i-1} F_{i+1}=F_{i}^{2}+(-1)^{i} for the Fibonacci numbers, for i>0i>0.

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The Fibonacci numbers are defined by the following recurrence relations: F0=0,F1=1,Fn+2=Fn+1+FnF_{0}=0, F_{1}=1, F_{n+2}=F_{n+1}+F_{n}. Prove that

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(a)

Fn=15[(1+52)n−(1−52)n];F_{n}=\frac{1}{\sqrt{5}}\left[\left(\frac{1+\sqrt{5}}{2}\right)^{n}-\left(\frac{1-\sqrt{5}}{2}\right)^{n}\right] ;

(b)

F1+F2+F3+⋯+Fn=Fn+2−1F_{1}+F_{2}+F_{3}+\cdots +F_{n}=F_{n+2}-1;

(c)

n⋅F1+(n−1)⋅F2+(n−2)⋅F3+⋯+1⋅Fn=Fn+4−n−3n \cdot F_{1}+(n-1) \cdot F_{2}+(n-2) \cdot F_{3}+\cdots +1 \cdot F_{n}=F_{n+4}-n-3 for n⩾1;n \geqslant 1 ;

(d)

Fn−1Fn+1=Fn2+(−1)nF_{n-1} F_{n+1}=F_{n}^{2}+(-1)^{n} for n⩾1n \geqslant 1;

(e)

the numbers FnF_{n} and Fn+1F_{n+1} are coprime for n⩾1n \geqslant 1;

(f)

the number FnF_{n} is even if and only if nn is divisible by 3;

(g)

Fn+m=Fn−1Fm+FnFm+1F_{n+m}=F_{n-1} F_{m}+F_{n} F_{m+1} for n⩾1n \geqslant 1.

Prove that for n⩾1n \geqslant 1

[1110]n=[Fn+1FnFnFn−1]. \left[\begin{smallmatrix} 1 & 1 \\ 1 & 0 \end{smallmatrix}\right]^{n}=\left[\begin{smallmatrix} F_{n+1} & F_{n} \\ F_{n} & F_{n-1} \end{smallmatrix}\right].
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Prove that there exist Fn+2F_{n+2} sequences of zeros and ones of length nn in which there are no two consecutive ones.

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Prove the identity ∑2m⩽n+1Cn−m+1m=Fn+2\sum_{2 m \leqslant n+1} C_{n-m+1}^{m}=F_{n+2}.

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