hcju/proofwikips huggingface dataset:
Principle of Recursive Definition for Well-Ordered Sets
Let $J$ be a well-ordered set. Let $C$ be any set. Let $\mathcal F$ be the set of all functions that map initial segments $S_a$ of $J$ into $C$. Then for any function of the form: :$\rho: \mathcal F \to C$ there exists a unique function: :$h: J \to C$ satisfying: :$\forall \alpha \in J: h\left({\alpha}\right) = \rho\left({ h {\restriction_{S_\alpha}} }\right)$ where ${\restriction}$ denotes the restriction of a mapping.
Angles with Parallel or Perpendicular Arms are Equal or Supplementary
Let $A$ and $B$ be angles such that their corresponding arms are either parallel or perpendicular. Then $A$ and $B$ are either equal or supplementary.
Sum of External Angles of Polygon equals Four Right Angles
Let the external angles of a polygon be generated in the same direction going around the polygon. Then the sum of all these external angles equals $4$ right angles.
Bisector of Apex of Isosceles Triangle also Bisects Base
Let $\triangle ABC$ be an isosceles triangle whose apex is $A$. Let $AD$ be the bisector of $\angle BAC$ such that $AD$ intersects $BC$ at $D$. Then $AD$ bisects $BC$.
Bisector of Apex of Isosceles Triangle is Perpendicular to Base
Let $\triangle ABC$ be an isosceles triangle whose apex is $A$. Let $AD$ be the bisector of $\angle BAC$ such that $AD$ intersects $BC$ at $D$. Then $AD$ is perpendicular to $BC$.
Equilateral Triangle is Equiangular
Let $\triangle ABC$ be an equilateral triangle. Then $\triangle ABC$ is also equiangular.
Equiangular Triangle is Equilateral
Let $\triangle ABC$ be equiangular. Then $\triangle ABC$ is an equilateral triangle.
No Order Isomophism Between Distinct Initial Segments of Woset
Let $E$ be a well-ordered set. Let $S_\alpha, S_\beta$ be initial segments of $E$ that are order isomorphic. Then $S_\alpha = S_\beta$.
Perpendicular is Shortest Straight Line from Point to Straight Line
Let $AB$ be a straight line. Let $C$ be a point which is not on $AB$. Let $D$ be a point on $AB$ such that $CD$ is perpendicular to $AB$. Then the length of $CD$ is less than the length of all other line segments that can be drawn from $C$ to $AB$.
Straight Lines which make Equal Angles with Perpendicular to Straight Line are Equal
Let $AB$ be a straight line. Let $C$ be a point which is not on $AB$. Let $D$ be a point on $AB$ such that $CD$ is perpendicular to $AB$. Let $E, F$ be points on $AB$ such that $\angle DCE = \angle DCF$. Then $CE = CF$.
Straight Line making Larger Angle with Perpendicular to Straight Line is Longer
Let $AB$ be a straight line. Let $C$ be a point which is not on $AB$. Let $D$ be a point on $AB$ such that $CD$ is perpendicular to $AB$. Let $E, F$ be points on $AB$ such that $\angle DCE > \angle DCF$. Then $CE > CF$.
Two Equal Straight Lines can be Constructed from Point to Straight Line
Let $AB$ be a straight line. Let $C$ be a point which is not on $AB$. Then exactly $2$ straight lines $CD$ and $CE$ can be drawn such that $CD = CE$ and $D, E$ on $AB$.
Diagonals of Rhombus Bisect Each Other at Right Angles
Let $ABCD$ be a rhombus. The diagonals $AC$ and $BD$ of $ABCD$ bisect each other at right angles.
Quadrilateral is Parallelogram iff One Pair of Opposite Sides is Equal and Parallel
Let $ABCD$ be a quadrilateral. Then: :$ABCD$ is a parallelogram : :$AB = CD$ and $AB \parallel CD$ where $AB \parallel CD$ denotes that $AB$ is parallel to $CD$.
Quadrilateral is Parallelogram iff Both Pairs of Opposite Sides are Equal or Parallel
Let $ABCD$ be a quadrilateral. Then: :$ABCD$ is a parallelogram : :either $AB = CD$ and $AD = BC$ :or $AB \parallel CD$ and $AD \parallel BC$ where $AB \parallel CD$ denotes that $AB$ is parallel to $CD$.
Quadrilateral is Parallelogram iff Both Pairs of Opposite Angles are Equal
Let $ABCD$ be a quadrilateral. Then: :$ABCD$ is a parallelogram : :$\angle ABC = \angle ADC$ and $\angle BAD = \angle BCD$.
Quadrilateral is Parallelogram iff Diagonals Bisect each other
Let $ABCD$ be a quadrilateral. Then: : $ABCD$ is a parallelogram : :both: :: $AD$ is a bisector of $BC$ :and: :: $BC$ is a bisector of $AD$.
Parallelograms are Congruent if Two Adjacent Sides and Included Angle are respectively Equal
Let $ABCD$ and $EFGH$ be parallelograms. Then $ABCD$ and $EFGH$ are congruent if: : $2$ adjacent sides of $ABCD$ are equal to $2$ corresponding adjacent sides of $EFGH$ : the angle between those $2$ adjacent sides on both $ABCD$ and $EFGH$ are equal.
Parallel Lines which intercept Equal Segments on Transversals
Let $3$ or more parallel lines intersect equal line segments on one transversal. Then those same $3$ or more parallel lines intersect equal line segments on every transversal.
Line Parallel to Side of Triangle which Bisects One Side also Bisects Other Side
Let $ABC$ be a triangle. Let $DE$ be a straight line parallel to $BC$. Let $DE$ bisect $AB$. Then $DE$ also bisects $AC$. That is, $DE$ is a midline of $\triangle ABC$. 400px
Midline Theorem
The midline of a triangle is parallel to the third side of that triangle and half its length.
Midline and Median of Triangle Bisect Each Other
Let $\triangle ABC$ be a triangle. Let $DE$ be the midline of $\triangle ABC$ which bisects $AB$ and $AC$. Let $AF$ be the median of $ABC$ which bisects $BC$. Then $AF$ and $DE$ bisect each other.
Diagonals of Rectangle are Equal
The diagonals of a rectangle are equal.
Summation of Odd Reciprocals in terms of Harmonic Numbers
:$\displaystyle \sum_{k \mathop = 1}^n \dfrac 1 {2 k - 1} = H_{2 n} - \dfrac {H_n} 2$ where $H_n$ denotes the $n$th harmonic number.
Numerator of p-1th Harmonic Number is Divisible by Prime p
Let $p$ be an odd prime. Consider the harmonic number $H_{p - 1}$ expressed in canonical form. The numerator of $H_{p - 1}$ is divisible by $p$.
Numerator of p-1th Harmonic Number is Divisible by p^2 for Prime Greater than 3
Let $p$ be a prime number such that $p > 3$. Consider the harmonic number $H_{p - 1}$ expressed in canonical form. The numerator of $H_{p - 1}$ is divisible by $p^2$.
Summation of Power Series by Harmonic Sequence
Consider the power series: :$\map f x = \displaystyle \sum_{k \mathop \ge 0} a_k x^k$ Let $\map f x$ converge for $x = x_0$. Then: :$\displaystyle \sum_{k \mathop \ge 0} a_k {x_0}^k H_k = \int_0^1 \dfrac {\map f {x_0} - \map f {x_0 y} } {1 - y} \rd y$ where $H_n$ denotes the $n$th harmonic number.
Summation over k to n of Harmonic Numbers over n+1-k
:$\displaystyle \sum_{k \mathop = 1}^n \dfrac {H_k} {n + 1 - k} = {H_{n + 1} }^2 - H_{n + 1}^{\paren 2}$ where $H_k$ denotes the $k$th harmonic number.
Summation over k to n of Harmonic Number k by Harmonic Number n-k
:$\displaystyle \sum_{k \mathop = 1}^n H_k H_{n - k} = \paren {n + 1} \paren { {H_n}^2 - H_n^{\paren 2} } - 2 n \paren {n_n - 1}$ where $H_k$ denotes the $k$th harmonic number.
Extension of Harmonic Number to Non-Integer Argument
Let $\map H x$ be the real function defined as: :$\map H x = \gamma + \dfrac {\map {\Gamma'} {x + 1} } {\map \Gamma {x + 1} }$ where: :$\gamma$ denotes the Euler-Mascheroni constant :$\Gamma$ denotes the gamma function :$\Gamma'$ denotes the derivative of the gamma function. Then $H$ is an extension of the mapping $H: \N \to \Q$ defined as: :$\forall n \in \N: \map H n = H_n$ where $H_n$ denotes the $n$th harmonic number.
Initial Segment Determined by Smallest Element is Empty
Let $\left({S, \preceq}\right)$ be a well-ordered set, where $S$ is non-empty. Let $s_0 = \min S$, the smallest element of $S$. Then the initial segment determined by $s_0$, $S_{s_0}$, is empty.
Empty Mapping is Injective
Let $\nu: \varnothing \to T$ be an empty mapping. Then $\nu$ is an injection.
Empty Mapping to Empty Set is Bijective
Let $\nu: \varnothing \to \varnothing$ be an empty mapping. Then $\nu$ is a bijection.
Upper and Lower Bound of Fibonacci Number
For all $n \in \N_{> 0}$: :$\phi^{n - 2} \le F_n \le \phi^{n - 1}$ where: :$F_n$ is the $n$th Fibonacci number :$\phi$ is the golden section: $\phi = \dfrac {1 + \sqrt 5} 2$
Generating Function for Fibonacci Numbers
Let $\map G z$ be the function defined as: :$\map G z = \dfrac z {1 - z - z^2}$ Then $\map G z$ is a generating function for the Fibonacci numbers.
Summation over k to n of Product of kth with n-kth Fibonacci Numbers
:$\displaystyle \sum_{k \mathop = 0}^n F_k F_{n - k} = \dfrac {\left({n - 1}\right) F_n + 2n F_{n - 1} } 5$ where $F_n$ denotes the $n$th Fibonacci number.
Fibonacci Number in terms of Smaller Fibonacci Numbers/Negative Indices
Let $n \in \Z_{< 0}$ be a negative integer. Let $F_n$ be the $n$th Fibonacci number (as extended to negative integers). Then Fibonacci Number in terms of Smaller Fibonacci Numbers: :$F_{m + n} = F_{m - 1} F_n + F_m F_{n + 1}$ continues to hold, whether $m$ or $n$ are positive or negative.
Euler-Binet Formula/Negative Index
Let $n \in \Z_{< 0}$ be a negative integer. Let $F_n$ be the $n$th Fibonacci number (as extended to negative integers). Then the Euler-Binet Formula: :$F_n = \dfrac {\phi^n - \hat \phi^n} {\sqrt 5}$ continues to hold.
Fibonacci Number by Golden Mean plus Fibonacci Number of Index One Less
Let $n \in \Z$. Then: :$\phi^n = F_n \phi + F_{n - 1}$ where: :$F_n$ denotes the $n$th Fibonacci number :$\phi$ denotes the golden mean.
Fibonacci Number by Golden Mean plus Fibonacci Number of Index One Less/Positive Index
Let $n \in \Z_{\ge 0}$. Then: :$\phi^n = F_n \phi + F_{n - 1}$ where: :$F_n$ denotes the $n$th Fibonacci number :$\phi$ denotes the golden mean.
Fibonacci Number by Golden Mean plus Fibonacci Number of Index One Less/Negative Index
Let $n \in \Z_{\le 0}$. Then: :$\phi^n = F_n \phi + F_{n - 1}$ where: :$F_n$ denotes the $n$th Fibonacci number as extended to negative indices :$\phi$ denotes the golden mean.
Fibonacci Number by One Minus Golden Mean plus Fibonacci Number of Index One Less
Let $n \in \Z$. Then: :$\hat \phi^n = F_n \hat \phi + F_{n - 1}$ where: :$F_n$ denotes the $n$th Fibonacci number :$\hat \phi$ denotes the $1$ minus the golden mean: ::$\hat \phi := 1 - \phi$
Reciprocal Form of One Minus Golden Mean
:$\hat \phi = - \dfrac 1 \phi$ where: :$\phi$ denotes the golden mean :$\hat \phi$ denotes one minus the golden mean: $\hat \phi = 1 - \phi$.
Closed Form of One Minus Golden Mean
:$\hat \phi = \dfrac {1 - \sqrt 5} 2$ where: :$\hat \phi$ denotes one minus the golden mean: $\hat \phi = 1 - \phi$.
Second Order Fibonacci Number in terms of Fibonacci Numbers
The second order Fibonacci number $\mathcal F_n$ can be expressed in terms of Fibonacci numbers as: :$\dfrac {3 n + 3} 5 F_n - \dfrac n 5 F_{n + 1}$
General Fibonacci Number in terms of Fibonacci Numbers
Let $r$ and $s$ be numbers, usually integers but not necessarily so limited. Let $\left\langle{a_n}\right\rangle$ be the general Fibonacci sequence : :$a_n = \begin{cases} r & : n = 0 \\ s & : n = 1 \\ a_{n - 2} + a_{n - 1} & : n > 1 \end{cases}$ Then $a_n$ can be expressed in Fibonacci numbers as: :$a_n = F_{n - 1} r + F_n s$
Fibonacci Number plus Constant in terms of Fibonacci Numbers
Let $c$ be a number. Let $\left\langle{b_n}\right\rangle$ be the sequence defined as: :$b_n = \begin{cases} 0 & : n = 0 \\ 1 & : n = 1 \\ b_{n - 2} + b_{n - 1} + c & : n > 1 \end{cases}$ Then $\left\langle{b_n}\right\rangle$  can be expressed in Fibonacci numbers as: :$b_n = c F_{n - 1} + \left({c + 1}\right) F_n - c$
Fibonacci Number plus Binomial Coefficient in terms of Fibonacci Numbers
Let $m \in \Z_{>0}$ be a positive integer. Let $\left\langle{a_n}\right\rangle$ be the sequence defined as: :$a_n = \begin{cases} 0 & : n = 0 \\ 1 & : n = 1 \\ a_{n - 2} + a_{n - 1} + \dbinom {n - 2} m & : n > 1 \end{cases}$ where $\dbinom {n - 2} m$ denotes a binonial coefficient. Then $\left\langle{a_n}\right\rangle$  can be expressed in Fibonacci numbers as: :$a_n = F_{m + 1} F_{n - 1} + \left({F_{m + 2} + 1}\right) F_n - \displaystyle \sum_{k \mathop = 0}^m \dbinom {n + m - k} k$
Fibonacci Number plus Arbitrary Function in terms of Fibonacci Numbers
Let $f \left({n}\right)$ and $g \left({n}\right)$ be arbitrary arithmetic functions. Let $\left\langle{a_n}\right\rangle$ be the sequence defined as: :$a_n = \begin{cases} 0 & : n = 0 \\ 1 & : n = 1 \\ a_{n - 1} + a_{n - 2} + f \left({n - 2}\right) & : n > 1 \end{cases}$ Let $\left\langle{b_n}\right\rangle$ be the sequence defined as: :$b_n = \begin{cases} 0 & : n = 0 \\ 1 & : n = 1 \\ b_{n - 1} + b_{n - 2} + g \left({n - 2}\right) & : n > 1 \end{cases}$ Let $\left\langle{c_n}\right\rangle$ be the sequence defined as: :$c_n = \begin{cases} 0 & : n = 0 \\ 1 & : n = 1 \\ c_{n - 1} + c_{n - 2} + x f \left({n - 2}\right) + y g \left({n - 2}\right) & : n > 1 \end{cases}$ where $x$ and $y$ are arbitrary. Then $\left\langle{c_n}\right\rangle$ can be expressed in Fibonacci numbers as: :$c_n = x a_n + y b_n + \left({1 - x - y}\right) F_n$
Vajda's Identity/Formulation 1
:$F_{n + i} F_{n + j} - F_n F_{n + i + j} = \paren {-1}^n F_i F_j$
Vajda's Identity/Formulation 2
:$F_{n + k} F_{m - k} - F_n F_m = \left({-1}\right)^n F_{m - n - k} F_k$
Sum of Squares of Consecutive Fibonacci Numbers
:${F_n}^2 + {F_{n + 1} }^2 = F_{2 n + 1}$ where $F_n$ denotes the $n$th Fibonacci number.
Hausdorff Maximal Principle implies Well-Ordering Theorem
Let the Hausdorff Maximal Principle hold. Then the Well-Ordering Theorem holds.
Cosine of 36 Degrees
:$\cos 36 \degrees = \cos \dfrac \pi 5 = \dfrac \phi 2 = \dfrac {1 + \sqrt 5} 4$ where $\phi$ denotes the golden mean.
Sine of 36 Degrees
:$\sin 36^\circ = \sin \dfrac \pi 5 = \dfrac {\sqrt {\sqrt 5 / \phi} } 2 = \sqrt {\dfrac 5 8 - \dfrac {\sqrt 5} 8}$ where $\phi$ denotes the golden mean.
Partial Sums of Power Series with Fibonacci Coefficients
:$\displaystyle \sum_{k \mathop = 0}^n F_k x^k = \begin{cases} \dfrac {x^{n + 1} F_{n + 1} + x^{n + 2} F_n - x} {x^2 + x - 1} & : x^2 + x - 1 \ne 0 \\ \dfrac {\left({n + 1}\right) x^n F_{n + 1} + \left({n + 2}\right) x^{n + 1} F_n - 1} {2 x + 1} & : x^2 + x - 1 = 0 \end{cases}$ where $F_n$ denotes the $n$th Fibonacci number.
Sum over k of n Choose k by Fibonacci t to the k by Fibonacci t-1 to the n-k by Fibonacci m+k
:$\displaystyle \sum_{k \mathop \ge 0} \binom n k {F_t}^k {F_{t - 1} }^{n - k} F_{m + k} = F_{m + t n}$ where: :$\dbinom n k$ denotes a binomial coefficient :$F_n$ denotes the $n$th Fibonacci number.
Matrix whose Determinant is Fibonacci Number
The $n \times n$ determinant: :$D_n = \begin{vmatrix} 1 & -1 & 0 & 0 & \cdots & 0 & 0 & 0 \\ 1 & 1 & -1 & 0 & \cdots & 0 & 0 & 0 \\ 0 & 1 & 1 & -1 & \cdots & 0 & 0 & 0 \\ \vdots & \vdots & \vdots & \vdots & \ddots & \vdots & \vdots & \vdots \\ 0 & 0 & 0 & 0 & \cdots & 1 & 1 & -1 \\ 0 & 0 & 0 & 0 & \cdots & 0 & 1 & 1 \\ \end{vmatrix}$ evaluates to $F_{n + 1}$.
Determinant with Unit Element in Otherwise Zero Column
Let $D$ be the determinant: :$D = \begin{vmatrix}   1 & b_{12} & \cdots & b_{1n} \\   0 & b_{22} & \cdots & b_{2n} \\ \vdots & \vdots & \ddots & \vdots \\   0 & b_{n2} & \cdots & b_{nn} \end{vmatrix}$ Then: :$D = \begin{vmatrix}   b_{22} & \cdots & b_{2n} \\ \vdots & \ddots & \vdots \\   b_{n2} & \cdots & b_{nn} \end{vmatrix}$
Well-Ordering Theorem implies Hausdorff Maximal Principle
Let the Well-Ordering Theorem hold. Then the Hausdorff Maximal Principle holds.
Fibonacci Number with Prime Index 2n+1 is Congruent to 5^n Modulo p
Let $p = 2 n + 1$ be an odd prime. Then: :$F_p \equiv 5^n \pmod p$
Fibonacci Number n+1 Minus Golden Mean by Fibonacci Number n
:$F_{n + 1} - \phi F_n = \hat \phi^n$ where: :$F_n$ denotes the $n$th Fibonacci number :$\phi$ denotes the golden mean.
Recurrence Relation where n+1th Term is A by nth term + B to the n
Let $\left\langle{a_n}\right\rangle$ be the sequence defined by the recurrence relation: :$a_n = \begin{cases} 0 & : n = 0 \\ A a_{n - 1} + B^{n - 1} & : n > 0 \end{cases}$ for numbers $A$ and $B$. Then the closed form for $\left\langle{a_n}\right\rangle$ is given by: :$a_n = \begin{cases} \dfrac {A^n - B^n} {A - B} & : A \ne B \\ n A^{n - 1} & : A = B \end{cases}$
Fibonomial Coefficient is Integer
Let $\dbinom n k_\mathcal F$ be a Fibonomial coefficient. Then $\dbinom n k_\mathcal F$ is an integer.
Recurrence Relation for Fibonomial Coefficients
:$\dbinom n k_\mathcal F = F_{k - 1} \dbinom {n - 1} k_\mathcal F + F_{n - k + 1} \dbinom {n - 1} {k - 1}_\mathcal F$ where: :$\dbinom n k_\mathcal F$ denotes a Fibonomial coefficient :$F_{k - 1}$ etc. denote Fibonacci numbers.
Golden Mean by One Minus Golden Mean equals Minus 1
:$\phi \hat \phi = -1$ where: :$\phi$ denotes the golden mean :$\hat \phi := 1 - \phi$
Remainder of Fibonacci Number Divided by Fibonacci Number is Plus or Minus Fibonacci Number
Let $F_n$ and $F_m$ be Fibonacci numbers. By the Division Theorem, let: :$F_n = q F_m + r$ where: :$q \in \Z$ :$r \in \Z: 0 \le r < \size {F_m}$ Then either $r$ or $\size {F_m} - r$, or both, is a Fibonacci number.
Residue of Fibonacci Number Modulo Fibonacci Number
Let $F_n$ denote the $n$th Fibonacci number. Let $m, r$ be non-negative integers. Then: :$F_{m n + r} \equiv \paren {\begin{cases} F_r & : m \bmod 4 = 0 \\ \paren {-1}^{r + 1} F_{n - r} & : m \bmod 4 = 1 \\ \paren {-1}^n F_r & : m \bmod 4 = 2 \\ \paren {-1}^{r + 1 + n} F_{n - r} & : m \bmod 4 = 3 \end{cases} } \pmod {F_n}$
Sine of Multiple of Pi by 2 plus i by Natural Logarithm of Golden Mean
Let $z = \dfrac \pi 2 + i \ln \phi$. Then: :$\dfrac {\sin n z} {\sin z} = i^{1 - n} F_n$ where: :$\phi$ denotes the golden mean :$F_n$ denotes the $n$th Fibonacci number.
Golden Mean as Root of Quadratic
The golden mean $\phi$ is one of the roots of the quadratic equation: :$x^2 - x - 1 = 0$ The other root is $\hat \phi = 1 - \phi$.
Proper Subtower is Initial Segment
Let $\left({T_1,\preccurlyeq}\right)$ be a proper subtower of $\left({T_2,\preccurlyeq}\right)$. Then $\left({T_1,\preccurlyeq}\right)$ is an initial segment of $\left({T_2,\preccurlyeq}\right)$.
Power of Golden Mean as Sum of Smaller Powers
Let $\phi$ denote the golden mean. Then: :$\forall z \in \C: \phi^z = \phi^{z - 1} + \phi^{z - 2}$
100 in Golden Mean Number System is Equivalent to 011
Consider the golden mean number system. Let $p$ and $q$ be arbitrary strings in $\left\{ {0, 1}\right\}$. Let $x \in \R_{\ge 0}$ have a representation which includes the string $100$, say: :$x = p100q$ Then $x \in \R_{\ge 0}$ also has the representation: :$x = p011q$ Similarly, let $x \in \R_{\ge 0}$ have a representation which includes the string $011$, say: :$x = p011q$ Then $x \in \R_{\ge 0}$ also has the representation: :$x = p100q$ That is, any instance of $100$ appearing in a representation of a non-negative real number $x$ is equivalent to $011$, and vice versa. Note that the instance of $100$ or $011$ may also include a radix point; the instance of $011$ or $100$ to which it is equivalent will include the radix point in the same location.
Conversion of Number in Golden Mean Number System to Simplest Form
Let $x \in \R_{\ge 0}$ have a representation $S$ in the golden mean number system. Then $S$ can be converted to its simplest form as follows: :$(1): \quad$ Replace any infinite string on the right hand end of $S$ of the form $010101 \ldots$ with $100$ :$(2): \quad$ Repeatedly replace the leftmost instance of $011$ with $100$.
Simplest Form of Non-Negative Number in Golden Mean Number System is Unique
Let $x \in \R_{\ge 0}$ be represented in the golden mean number system. Let $S$ be the representation for $x$ in its simplest form. Then $S$ is unique in the sense that there exists no other representation of $x$ in simplest form.
Incidence of Double Letters in Fibonacci String
Let $S_n$ denote the $n$th Fibonacci string. Then: :$(1):\quad$ There are no instances of $2$ $\text a$'s together :$(2):\quad$ There are no instances of $3$ $\text b$'s together in $S_n$.
Fibonacci String Begins with ba
Let $S_n$ be a Fibonacci string of length $n$. Then for $n \ge 3$, $S_n$ begins with $\text{ba}$.
Fibonacci String Ends with ab or ba
Let $S_n$ be a Fibonacci string of length $n$. Then for $n \ge 3$, $S_n$ ends either with $\text{ba}$ or with $\text{ab}$.
Count of a's and b's in Fibonacci String
Let $S_n$ denote the $n$th Fibonacci string. Then for $n \ge 3$, $S_n$ has: :$F_{n - 2}$ instances of $\text a$ :$F_{n - 1}$ instances of $\text b$.
Initial Part of Fibonacci String
Let $n \in \Z_{>1}$. Let $S_n$ denote the $n$th Fibonacci string. Let $m \in \Z$ such that $1 < m \le n$. Let $F_m$ denote the $m$th Fibonacci number. The initial part of $S_n$ of length $F_m$ is the Fibonacci string $S_m$.
Positions of Instances of a in Fibonacci String
Let $S_n$ denote the $n$th Fibonacci string. Let $F_n$ denote the $n$th Fibonacci number. Let $m \in \Z$ such that $m \le F_n$. Let $m - 1$ be expressed in Zeckendorf representation as $Z_{m - 1}$. Then the $m$th letter of $S_n$ is $\text a$ : : $k_r = 2$ where $k_r$ denotes the final digit of $Z_{m - 1}$.
Positions of Instances of b in Fibonacci String
Let $S_n$ denote the $n$th Fibonacci string. Let $F_n$ denote the $n$th Fibonacci number. Let $k \in \Z$ such that $k \le F_n$. Then the $k$th letter of $S_n$ is $\text b$ : : $\left\lfloor{\left({k + 1}\right) \phi^{-1} }\right\rfloor - \left\lfloor{k \phi^{-1} }\right\rfloor = 1$ where: : $\left\lfloor{\, \cdot \,}\right\rfloor$ denotes the floor function : $\phi$ denotes the golden mean.
Optimal Strategy for Fibonacci Nim
Consider a game of Fibonacci nim with $n$ counters. Let it be the turn of player $\text A$. Let the maximum number of counters that can be taken by $\text A$ be $q$. Let $n$ be expressed in Zeckendorf representation as: :$n = F_{k_1} + F_{k_2} + \cdots + F_{k_r}$ Then $\text A$ can force a win : :$F_{k_r} \le q$ and by taking those $F_{k_r}$ counters.
Recursively Defined Sequence/Examples/Term is Term of Index less 1 plus 6 times Term of Index less 2
Consider the integer sequence $\left\langle{a_n}\right\rangle$ defined recursively as: :$a_n = \begin{cases} 0 & : n = 0 \\ 1 & : n = 1 \\ a_{n - 1} + 6 a_{n - 2} & : \text{otherwise} \end{cases}$ $a_n$ has a closed-form expression: :$a_n = \dfrac {3^n - \left({-2}\right)^n} 5$
Recursively Defined Sequence/Examples/Minimum over k of Maximum of 1 plus Function of k and 2 plus Function of n-k
Consider the integer sequence $\left\langle{f \left({n}\right)}\right\rangle$ defined recusrively as: :$f \left({n}\right) = \begin{cases} 0 & : n = 1 \\ \displaystyle \min_{0 \mathop < k \mathop < n} \max \left({1 + f \left({k}\right), 2 + f \left({n - k}\right)}\right) & : n > 1 \end{cases}$ $f \left({n}\right)$ has a closed-form expression: :$f \left({n}\right) = m$ for $F_m < n \le F_{m + 1}$ where $F_m$ denotes the $m$th Fibonacci number.
Zeckendorf Representation of Integer shifted Left
Let $\map f x$ be the real function defined as: :$\forall x \in \R: \map f x = \floor {x + \phi^{-1} }$ where: :$\floor {\, \cdot \,}$ denotes the floor function :$\phi$ denotes the golden mean. Let $n \in \Z_{\ge 0}$ be a positive integer. Let $n$ be expressed in Zeckendorf representation: :$n = F_{k_1} + F_{k_2} + \cdots + F_{k_r}$ with the appropriate restrictions on $k_1, k_2, \ldots, k_r$. Then: :$F_{k_1 + 1} + F_{k_2 + 1} + \cdots + F_{k_r + 1} = \map f {\phi n}$
Two Non-Negative Integers have Zeckendorf Representations of which one is Shifted Representation of the Other
Let $m, n \in \Z_{\ge 0}$ be non-negative integers. Then there exists a unique set of integers: :$\left\{ {k_1, k_2, \ldots, k_r}\right\}$ where: : $k_1 \gg k_2 \gg \cdots \gg k_r$ where $a \gg b$ denotes that $a - b > 1$ such that: :$m = F_{k_1} + F_{k_2} + \cdots + F_{k_r}$ and: :$n = F_{k_1 + 1} + F_{k_2 + 1} + \cdots + F_{k_r + 1}$ Note that: : each of the $k$'s may be negative and: :$r$ may equal $0$.
Generating Function by Power of Parameter
Let $G \left({z}\right)$ be the generating function for the sequence $\left\langle{a_n}\right\rangle$. Let $m \in \Z_{\ge 0}$ be a non-negative integer. Then $z^m  G \left({z}\right)$ is the generating function for the sequence $\left\langle{a_{n - m} }\right\rangle$.
Generating Function Divided by Power of Parameter
Let $G \left({z}\right)$ be the generating function for the sequence $\left\langle{a_n}\right\rangle$. Let $m \in \Z_{\ge 0}$ be a non-negative integer. Then $\dfrac 1 {z^m} \left({G \left({z}\right) - \displaystyle \sum_{k \mathop = 0}^{m - 1} a_k z^k}\right)$ is the generating function for the sequence $\left\langle{a_{n + m} }\right\rangle$.