nyuuzyou/wikis huggingface dataset:
Triangles with One Equal Angle and Two Sides Proportional are Similar

== Theorem ==
Let two triangles have two corresponding sides which are proportional.
Let the angles adjacent to both of these sides be equal.
Then all of their corresponding angles are equal.
Thus, by definition, such triangles are similar.
== Proof ==
Let $\triangle ABC, \triangle DEF$ be triangles having $\angle BAC = \angle EDF$ and two sides proportional, so that $BA : AC = ED : DF$.
We need to show that $\triangle ABC$ is equiangular with $\triangle DEF$, such that $\angle ABC = \angle DEF$ and $\angle ACB = \angle DFE$.
:
Let $\angle FDG$ be constructed equal to either $\angle BAC$ or $\angle EDF$, and $\angle DFG = \angle ACB$.
From Sum of Angles of Triangle Equals Two Right Angles:
:$\angle ABC = \angle DGF$
Therefore $\triangle ABC$ is equiangular with $\triangle DGF$.
From Equiangular Triangles are Similar:
:$BA : AC = GD : DF$
But by hypothesis:
:$BA : AC = ED : DF$
So from Equality of Ratios is Transitive:
:$ED : DF = GD : DF$
From Magnitudes with Same Ratios are Equal:
:$ED = DG$
But $DF$ is common to $\triangle DEF$ and $\triangle DGF$, and $\angle EDF = \angle GDF$.
So from Triangle Side-Angle-Side Congruence:
:$\triangle DEF = \triangle DGF$
But $\angle DFG = \angle ACB$ and so:
:$\angle ABC = \angle DFE$
But by hypothesis:
:$\angle BAC = \angle EDF$
Therefore by Sum of Angles of Triangle Equals Two Right Angles:
:$\angle ABC = \angle DEF$
Hence the result.
== Sources ==
* : Book $\text{VI}$. Propositions
* 
* 
Category:Similar Triangles
Triangles with Proportional Sides are Similar

== Theorem ==
Let two triangles have corresponding sides which are proportional.
Then their corresponding angles are equal.
Thus, by definition, such triangles are similar.
== Proof ==
Let $\triangle ABC, \triangle DEF$ be triangles whose sides are proportional, so that:
:$ AB : BC = DE : EF$
:$ BC : CA = EF : FD$
:$ BA : AC = ED : DF$
We need to show that
:$\angle ABC = \angle DEF$
:$\angle BCA = \angle EFD$
:$\angle BAC = \angle EDF$
:
On the straight line $EF$, and at the points $E, F$ on it, construct $\angle FEG = \angle ABC$ and $\angle EFG = \angle ACB$.
From Sum of Angles of Triangle Equals Two Right Angles, the remaining angle at $A$ equals the remaining angle at $G$.
Therefore $\triangle ABC$ is equiangular with $\triangle GEF$.
From Equiangular Triangles are Similar, the sides about the equal angles are proportional, and those are corresponding sides which subtend the equal angles.
So:
:$AB : BD = GE : EF$
But by hypothesis:
:$AB : BC = DE : EF$
So from Equality of Ratios is Transitive
:$DE : EF = GE : EF$
So each of $DE, GE$ has the same ratio to $EF$.
So from Magnitudes with Same Ratios are Equal:
:$DE = GE$
For the same reason:
:$DF = GF$
So we have that $DE = EG$, $EF$ is common and $DF = FG$.
So from Triangle Side-Side-Side Congruence:
:$\triangle DEF = \triangle GEF$
That is:
:$\angle DEF = \angle GEF, \angle DFE = \angle GFE, \angle EDF = \angle EGF$
As $\angle GEF = \angle ABC$ it follows that:
:$\angle ABC = \angle DEF$
For the same reason $\angle ACB = \angle DFE$ and $\angle BAC = \angle EDF$.
Hence the result.
== Sources ==
* : Book $\text{VI}$. Propositions
* 
* 
Category:Similar Triangles
Category:Triangles
Triangles with Same Base and Same Height have Equal Area

== Theorem ==
== Proof ==
:
Let $ABC$ and $DBC$ be triangles which are on the same base $BC$ and in the same parallels $AD$ and $BC$.
Let $AD$ be produced in the directions of $E$ and $F$.
Let $BE$ through $B$ be drawn parallel to $CA$.
Let $CF$ through $C$ be drawn parallel to $BD$.
Then each of $EBCA$ and $DBCF$ are parallelograms, and by Parallelograms with Same Base and Same Height have Equal Area they have equal areas.
From Opposite Sides and Angles of Parallelogram are Equal, $ABC$ is half of $EBCA$ as $AB$ bisects it.
For the same reason, $DBC$ is half of $DBCF$.
But by Common Notion 1, $\triangle ABC = \triangle DBC$.
== Sources ==
* : Book $\text{I}$. Propositions
Category:Areas of Triangles
Triangles with Two Equal Angles are Similar

== Theorem ==
Two triangles which have two corresponding angles which are equal are similar.
== Proof ==
Let $\triangle ABC$ and $\triangle DEF$ be triangles such that $\angle ABC = \angle DEF$ and $\angle BAC = \angle EDF$.
Then from Sum of Angles of Triangle Equals Two Right Angles $\angle ACB$ is equal to two right angles minus $\angle ABC + \angle BAC$.
Also from Sum of Angles of Triangle Equals Two Right Angles $\angle DFE$ is equal to two right angles minus $\angle DEF + \angle EDF$.
That is, $\angle DFE$ is equal to two right angles minus $\angle ABC + \angle BAC$.
So $\angle DFE = \angle ACB$ and so all three corresponding angles of $\triangle ABC$ and $\triangle DEF$ are equal.
The result follows from Equiangular Triangles are Similar.
== Sources ==
* : Chapter $1$: Triangles and Polygons: Theorems and Corollaries $1.8$: Corollary $1$
* 
* 
Category:Similar Triangles
Triangles with Two Sides Parallel and Equal

== Theorem ==
== Proof ==
Let $\triangle ABC, \triangle DCE$ be two triangles such that $BA : AC = DC : DE$ be situated so that $AB \parallel DC$ and $AC \parallel DE$.
We need to show that $BC$ is in a straight line with $CE$.
:
We have that $AB \parallel DC$.
So from Parallelism implies Equal Alternate Angles we have that $\angle BAC = \angle ACD$.
For the same reason $\angle CDE = \angle ACD$.
So $\angle BAC = \angle CDE$.
From Triangles with One Equal Angle and Two Sides Proportional are Similar, it follows that $\triangle ABC$ is similar to $\triangle DCE$.
Therefore $\angle ABC = \angle DCE$.
It follows that $\angle ACE = \angle ABC + \angle BAC$.
Add $\angle ACB$ to each.
Then $\angle ACE + \angle ACB = \angle BAC + \angle ACB + \angle CBA$.
But from Sum of Angles of Triangle Equals Two Right Angles $\angle BAC + \angle ACB + \angle CBA$ equals two right angles.
So $\angle ACE + \angle ACB$ equals two right angles.
The result follows from Two Angles making Two Right Angles make Straight Line.
== Sources ==
* : Book $\text{VI}$. Propositions
Category:Triangles
Triangular Fermat Number

== Theorem ==
The only one Fermat number which is triangular is $3$.
== Proof ==
Let $F_n$ be a Fermat number which is triangular.
We use the weaker form of Divisor of Fermat Number:
:Every divisor of $F_n$ is of the form $k \times 2^{n + 1} + 1$
as this formulation is valid for all $n \in \N$.
From Closed Form for Triangular Numbers:
:$F_n = \dfrac {m \paren {m + 1} } 2$
Either $m$ or $m + 1$ is even.
Suppose $m$ is even.
Then:
As the is always odd, we must have $2^n = 1$.
This forces $n = 0$, with $F_n = 3$.
We check that $3$ is indeed a triangular number.
Suppose $m + 1$ is even.
Then:
As $k \times 2^{n + 2} + 3 > 1$ and is odd, $k \times 2^{n + 2} + 3$ cannot be a power of $2$.
So $m + 1$ cannot be even, and we have exhausted all possibilities.
== Sources ==
* : $257$
* : $257$
Category:Fermat Numbers
Category:Triangular Numbers
Triangular Fibonacci Numbers

== Theorem ==
The only Fibonacci numbers which are also triangular are:
:$0, 1, 3, 21, 55$
== Proof ==
It remains to be shown that these are the only ones.
Let $F_n$ be the $n$th Fibonacci number.
From Odd Square is Eight Triangles Plus One, $F_n$ is triangular $8 F_n + 1$ is square.
It remains to be demonstrated that $8 F_n + 1$ is square :
:$n \in \set{\pm 1, 0, 2, 4, 8, 10}$
So, let $8 F_n + 1$ be square.
Then:
:$n \equiv \begin{cases} \pm 1 \pmod {2^5 \times 5} & : n \text { odd} \\ 0, 2, 4, 8, 10 \pmod {2^5 \times 5^2 \times 11} & : n \text { even} \end{cases}$
== Sources ==
* : $55$
* 
* : $55$
Category:Fibonacci Numbers
Category:Triangular Numbers
Triangular Matrices forms Subring of Square Matrices

== Theorem ==
Let $n \in \Z_{>0}$ be a (strictly) positive integer.
Let $\map {\MM_R} n$ be the order $n$ square matrix space over a ring $R$.
Let $\struct {\map {\MM_R} n, +, \times}$ denote the ring of square matrices of order $n$ over $R$.
Let $\map {U_R} n$ be the set of upper triangular matrices of order $n$ over $R$.
Then $\map {U_R} n$ forms a subring of $\struct {\map {\MM_R} n, +, \times}$.
Similarly, let $\map {L_R} n$ be the set of lower triangular matrices of order $n$ over $R$.
Then $\map {L_R} n$ forms a subring of $\struct {\map {\MM_R} n, +, \times}$.
== Proof ==
From Negative of Triangular Matrix, if $\mathbf B \in \map {U_R} n$ then $-\mathbf B \in \map {U_R} n$.
Then from Sum of Triangular Matrices, if $\mathbf A, -\mathbf B \in \map {U_R} n$ then $\mathbf A + \paren {-\mathbf B} \in \map {U_R} n$.
From Product of Triangular Matrices, if $\mathbf A, \mathbf B \in \map {U_R} n$ then $\mathbf A \mathbf B \in \map {U_R} n$.
The result follows from the Subring Test.
The same argument can be applied to matrices in $\map {L_R} n$.
== Sources ==
* : $\S 2.1$: Subrings: Examples $3$
* : Chapter $6$: An Introduction to Groups: Exercise $4$ (limited result)
Category:Triangular Matrices
Category:Rings of Square Matrices
Category:Subrings
Triangular Number/Examples

== Examples of Triangular Numbers ==
The first few triangular numbers are as follows:
:
=== Sequence of Triangular Numbers ===
== Sources ==
* : Chapter $1$: Some Preliminary Considerations: $1.3$ Early Number Theory: Problems $1.3$: $1$
* : Fig. $160 \ \text{(a)}$
* : Chapter $\text {A}.2$: Pythagoras (ca. B.C.)
* : Chapter $\text {A}.13$: Fermat ()
* 
Category:Triangular Numbers
Triangular Number/Sequence

== Definition ==
The sequence of triangular numbers, for $n \in \Z_{\ge 0}$, begins:
:$0, 1, 3, 6, 10, 15, 21, 28, 36, 45, 55, 66, \ldots$
== Sources ==
* : $\text {1-1}$ Principle of Mathematical Induction: Table $\text{1-1}$
* : $15$
* : Tables: $1$ The First $100$ Triangular Numbers, Squares and Cubes
* : Chapter $\text {A}.2$: Pythagoras (ca. B.C.)
* : Chapter $\text {A}.13$: Fermat ()
* : $15$
* : Tables: $1$ The First $100$ Triangular Numbers, Squares and Cubes
* 
* 
* 
Category:Triangular Numbers
Triangular Number Modulo 3 and 9

== Theorem ==
Let $n$ be a triangular number.
Then one of the following two conditions applies:
:$n \equiv 0 \pmod 3$
:$n \equiv 1 \pmod 9$
== Proof ==
Let $n = T_r$.
Then from Closed Form for Triangular Numbers:
:$n = \dfrac {r \paren {r + 1} } 2$
It suffices from Euclid's Lemma to investigate the nature of $r \paren {r + 1}$ modulo $3$.
There are three cases to consider:
:$r \equiv 0 \pmod 3$
:$r \equiv 1 \pmod 3$
:$r \equiv 2 \pmod 3$
Let $r \equiv 0 \pmod 3$.
Then:
:$r \paren {r + 1} \equiv 0 \pmod 3$
and so $T_r \equiv 0 \pmod 3$.
Let $r \equiv 2 \pmod 3$.
Then:
:$r + 1 \equiv 3 \equiv 0 \pmod 3$
So:
:$r \paren {r + 1} \equiv 0 \pmod 3$
and so $T_r \equiv 0 \pmod 3$.
Let $r \equiv 1 \pmod 3$.
Then $\exists k \in \Z: r = 3 k + 1$.
So $r \paren {r + 1} = \paren {3 k + 1} \paren {3 k + 2}$
So:
:$T_r = 9 \dfrac {k \paren {k + 1} } 2 + 1$
Thus $T_r \equiv 1 \pmod 9$.
Category:Triangular Numbers
Triangular Number Pairs with Triangular Sum and Difference

== Theorem ==
The sequence of pairs of triangular numbers whose sum and difference are also both triangular begins:
:$\tuple {15, 21}, \tuple {105, 171}, \tuple {378, 703}, \tuple {780, 990}, \tuple {1485, 4186}, \tuple {2145, 3741}, \tuple {5460, 6786}, \tuple {7875, 8778}$
== Examples ==
== Sources ==
* 
* : Chapter $1$: Some Preliminary Considerations: $1.3$ Early Number Theory: Problems $1.3$: $5 \ \text {(a)}$
* : $15$
* : $15$
Category:Triangular Numbers
Category:Triangular Number Pairs with Triangular Sum and Difference
Triangular Number Pairs with Triangular Sum and Difference/Examples

== Examples of Triangular Number Pairs with Triangular Sum and Difference ==
=== $T_5$ and $T_6$ ===
=== $T_{14}$ and $T_{18}$ ===
=== $T_{27}$ and $T_{37}$ ===
=== $T_{39}$ and $T_{44}$ ===
=== $T_{1869}$ and $T_{2090}$ ===
Category:Triangular Number Pairs with Triangular Sum and Difference
Triangular Number Pairs with Triangular Sum and Difference/Examples/T1869, T2090

== Example of Triangular Number Pair with Triangular Sum and Difference ==
The triangular numbers $T_{1869}$ and $T_{2090}$ have sum and difference which are themselves both triangular:
== Sources ==
* : $1,747,515$
* : $1,747,515$
Category:Triangular Number Pairs with Triangular Sum and Difference
Triangular Number Pairs with Triangular Sum and Difference/Examples/T1869, T2090/Mistake

== Source Work ==
:
:The Dictionary
::$1,747,515$
:
:The Dictionary
::$1,747,515$
== Mistake ==
:Together with $2,185,095$ the $3$rd pair of triangular numbers whose sum and difference are also triangular.
This pair is very far from being the $3$rd such pair. In fact, ordered by the larger of the two elements, this is in fact the $26$th such pair.
== Also see ==
* Triangular Number Pairs with Triangular Sum and Difference/Mistake, where the source of the error is explained.
== Sources ==
* : $1,747,515$
* : $1,747,515$
Category:Mistakes/Curious and Interesting Numbers
Triangular Number Pairs with Triangular Sum and Difference/Examples/T39, T44

== Example of Triangular Number Pair with Triangular Sum and Difference ==
The triangular numbers $T_{39}$ and $T_{44}$ have sum and difference which are themselves both triangular:
== Sources ==
* : $780$
* : $780$
Category:Triangular Number Pairs with Triangular Sum and Difference
Triangular Number Pairs with Triangular Sum and Difference/Examples/T39, T44/Mistake

== Source Work ==
:
:The Dictionary
::$780$
:
:The Dictionary
::$780$
== Mistake ==
:$780$ and $990$ are the $2$nd smallest pair of triangular numbers whose sum and difference ($1770$ and $210$) are also triangular.
It is in fact the $4$th such pair, not the $2$nd, after $\tuple {15, 21}$, $\tuple {105, 171}$, $\tuple {378, 703}$.
== Also see ==
* Triangular Number Pairs with Triangular Sum and Difference/Mistake, where the source of the error is explained.
== Sources ==
* : $780$
* : $780$
Category:Mistakes/Curious and Interesting Numbers
Triangular Number Pairs with Triangular Sum and Difference/Mistake

== Source Work ==
:
:The Dictionary
::$15$
:
:The Dictionary
::$15$
== Mistake ==
:$15$ and $21$ are the smallest pair of triangular numbers whose sum and difference ($6$ and $36$) are also triangular. The next such pairs are $780$ and $990$, and $1,747,515$ and $2,185,095$.
== Correction ==
There are a good number of other such pairs in addition to these. It can be noted that $780$ and $990$ are the $4$th such pair, and that $1,747,515$ and $2,185,095$ are in fact the $26$th (in ascending order of the larger element of the pairs).
The pairs given are those reported by in his $1920$ work :
: found pairs$^*$ of triangular numbers $15$ and $21$, $780$ and $990$, $1747515$ and $2185095$, whose sum and difference are triangular. Their sides are $5$ and $6,$ $39$ and $44$, $1869$ and $2090$.
's work dates from the $17$th century, and has been considerably superseded by further research. In particular, many other such pairs have since been discovered, rendering 's thesis considerably out of date.
Indeed, himself goes on to say, in a footnote:
:Others are $171$ and $105$, $3741$ and $2145$.
 repeats this mistake in $2$ further places: in section $780$ and again in section $1,747,515$.
== Sources ==
* : $15$
* : $15$
Category:Mistakes/Curious and Interesting Numbers
Triangular Number as Alternating Sum and Difference of Squares

== Theorem ==
Thus the $n$th triangular number can be expressed as the alternating sum and difference of squares:
So:
and so on.
== Proof ==
The proof proceeds by induction.
For all $n \in \N_{> 0}$, let $\map P n$ be the proposition:
:$\ds \frac {n \paren {n + 1} } 2 = \sum_{j \mathop = 0}^{n - 1} \paren {-1}^j \paren {n - j}^2$
=== Basis for the Induction ===
$\map P 1$ is the case:
which is seen to hold.
This is the basis for the induction.
=== Induction Hypothesis ===
Now it needs to be shown that, if $\map P k$ is true, where $k \ge 1$, then it logically follows that $\map P {k + 1}$ is true.
So this is the induction hypothesis:
:$\ds \frac {k \paren {k + 1} } 2 = \sum_{j \mathop = 0}^{k - 1} \paren {-1}^j \paren {n - j}^2$
from which it is to be shown that:
:$\ds \frac {\paren {k + 1} \paren {k + 2} } 2 = \sum_{j \mathop = 0}^k \paren {-1}^j \paren {k - j + 1}^2$
=== Induction Step ===
This is the induction step:
So $\map P k \implies \map P {k + 1}$ and the result follows by the Principle of Mathematical Induction.
== Sources ==
* : $15$
* : $\S 1.2.1$: Mathematical Induction: Exercise $7$
* : $15$
Category:Triangular Numbers
Category:Square Numbers
Triangular Number cannot be Cube

== Theorem ==
Let $T_n$ be the $n$th triangular number such that $n > 1$.
Then $T_n$ cannot be a cube.
== Proof ==
Suppose $T_n = x^3$ for some $x \in \Z$.
Then by Odd Square is Eight Triangles Plus One:
:$\exists y \in \Z: 8 T_n + 1 = \paren {2 x}^3 + 1 = y^2$
By Cube which is One Less than a Square:
:$2 x = 2$, $y = 3$
giving the unique solution:
:$T_n = 1^3 = 1$
== Sources ==
* : $15$
* : $15$
Category:Triangular Numbers
Category:Cube Numbers
Triangular Number whose Square is Triangular

== Theorem ==
The only triangular number with less than $660$ digits, whose square is also triangular, is $6$.
== Proof ==
We have that:
::${T_3}^2 = 6^2 = 36 = 1 + 2 + 3 + 4 + 5 + 6 + 7 + 8$
To establish that it is the only one yet known can be established by an exhaustive search.
== Sources ==
* 
* : $6$
* : $15$
* : $6$
* : $15$
Category:Triangular Numbers
Category:6
Triangular Numbers are Primitive Recursive

== Theorem ==
Let $t : \N \to \N$ be defined as:
:$\map t k = T_k$
where $T_k$ is the $k$-th triangular number.
Then $t$ is a primitive recursive function.
== Proof ==
Define $f : \N^0 \to \N$ as:
:$\map f {} = 0$.
Define $g : \N^2 \to \N$ as:
$\map g x y = x + y + 1$.
By Constant Function is Primitive Recursive and Addition is Primitive Recursive:
:$f$ and $g$ are primitive recursive.
By definition of primitive recursive, it suffices to show that $t$ is obtained by primitive recursion from $f$ and $g$.
That is, to show that, for all $n \in \N$:
:$\map t n = \begin{cases}
\map f {} & : n = 0 \\
\map g {n - 1, \map t {n - 1}} & : n > 0
\end{cases}$
As:
:$\map f {} = 0$
:$\map g {n - 1, \map t {n - 1}} = n - 1 + \map t {n - 1} + 1$
:$\map t k = T_k$ for all $k \in \N$
that is equivalent to showing:
:$T_n = \begin{cases}
0 & : n = 0 \\
n + T_{n - 1} & : n > 0
\end{cases}$
which is the definition of triangular numbers.
Category:Primitive Recursive Functions
Category:Triangular Numbers
Triangular Numbers which are Product of 3 Consecutive Integers

== Theorem ==
The $6$ triangular numbers which can be expressed as the product of $3$ consecutive integers are:
:$6, 120, 210, 990, 185 \, 836, 258 \, 474 \, 216$
== Proof ==
== Sources ==
* 
* 
* : $258,474,216$
Category:Triangular Numbers
Triangular Numbers which are Product of 3 Consecutive Integers/Mistake

== Source Work ==
:
:The Dictionary
::$258,474,216$
== Mistake ==
:The largest triangular number to be the product of consecutive integers. The others are $6$, $120$, $210$, $990$ and $185 \, 136$.
This should read:
:The largest triangular number to be the product of $3$ consecutive integers. The others are $6$, $120$, $210$, $990$ and $185 \, 136$.
For example, $7140$ is a triangular number which is the product of $2$ (and not $3$) consecutive integers:
:$7140 = T_{119} = \dfrac {119 \left({119 + 1}\right)} 2 = 84 \times 85$
== Sources ==
* 
* : $258,474,216$
Category:Mistakes/Curious and Interesting Numbers
Triangular Numbers which are Sum of Two Cubes

== Theorem ==
The sequence of triangular numbers which are the sum of $2$ cubes begins:
:$28, 91, 351, 2926, 8001, 46971, 58653, 93528, 97461, \dots$
== Proof ==
Can be demonstrated by brute force.
For example:
== Sources ==
* : $28$
Category:Cube Numbers
Category:Triangular Numbers
Category:28
Triangular Numbers which are also Pentagonal

== Theorem ==
The sequence of triangular numbers which are also pentagonal begins:
:$1, 210, 40 \, 755, 7 \, 906 \, 276, 1 \, 533 \, 776 \, 805, 297 \, 544 \, 793 \, 910, \ldots$
== Proof ==
== Sources ==
* : $210$
* : $210$
Category:Triangular Numbers
Category:Pentagonal Numbers
Triangular Numbers which are also Square

== Theorem ==
Let $A_n$ be the $n$th non-negative integer whose square is also a triangular number.
Then:
:$A_n = \begin {cases} 0 & : n = 0 \\ 1 & : n = 1 \\ 6 A_{n - 1} - A_{n - 2} & : n > 1 \end {cases}$
=== Sequence of Square Triangles ===
== Proof ==
Let $n \in \Z_{>0}$ be such that $n^2$ is a triangular number.
Then we have:
This is a Pellian equation.
From Pell's Equation: $x^2 - 8 y^2 = 1$, the smallest positive integral solution is:
:$\tuple {x, y} = \tuple {3, 1}$
Thus we have that the sequence of $n$ such that $n^2$ is triangular is the sequence of numerators of the Continued Fraction Expansion of $\sqrt 8$ whose indices are even.
Let $\tuple {x, y} = \tuple {p_k, q_k}$ be a solution to $x^2 - 8 y^2 = 1$.
Then from Continued Fraction Expansion of $\sqrt 8$ we have:
Hence the result.
== Sources ==
* : $15$
* : $15$
Category:Triangular Numbers
Category:Square Numbers
Triangular Numbers which are also Square/Sequence

== Sequence of Triangular Numbers which are also Square ==
The sequence of triangular numbers which are also square begins:
:$1, 36, 1225, 41 \, 616, 1 \, 413 \, 721, \ldots$
Their indices are:
:$1, 8, 49, 288, 1681, 9800, \ldots$
Their roots are:
:$1, 6, 35, 204, 1189, \ldots$
== Sources ==
* : $15$
* : $15$
Category:Triangular Numbers
Category:Square Numbers
Tribonacci Constant is Unique Real Root of x^3-x^2-x-1

== Theorem ==
== Proof ==
=== Part 1: Reduction ===
=== Part 2: Complete the square ===
=== Part 3: Substitution ===
==== Initial substitution ====
We make a sign rectification because this cascade works with only one substitution at a time.
We choose $+$ because we know $x^3 - x^2 - x$ has a zero root and a positive root, so $x^3 - x^2 - x - 1$ can only have a positive root.
The cascade will account for the $-$ solution anyway.
==== Lambda substitution ====
This next part of the substitution occurs because of the following logic:
$P1:$ To take a cube root of a real number, there must be three possible solutions (one real, two complex).
$P2:$ This polynomial is being restricted to the real xy-plane, i.e. no complex roots are allowed.
$C1:$ To take the cube root of this real number, it needs to vary by signs as the two complex would.
$P3:$ A complex number has its own arguments with which sign works, i.e. $+i$ vs. $-i$, which does not apply on the real number line.
$C2:$ All combinations of signs have to be represented, and there must be three.
$P4:$ Since a sign has two states, $+$ and $-$, there must be at least one other entity able to be modified with a sign.
$C3:$ To take the cube root of this real number, two parts of it must be able to be modified with a sign together exactly thrice.
We can see that the first cube root is trivially determined:
$v^3 = \dfrac \Lambda {27}$
$v = \dfrac 1 3 { \Lambda }^{ \dfrac 1 3 }$
The two parts we can modify are $\dfrac 1 3$ and $\Lambda$. We already have $+$ for the first and $+$ for the second.
We know $-1 × -1 = +1$, so we can have $-$ for the first and $-$ for the second:
$v = - \dfrac 1 3 \paren { - \Lambda }^{ \dfrac 1 3 }$
There must be a third way. We can have both negative signs on the first part or on the second part (as a cube root). It's possible to write it as both at the same time if it is in the form of a third part, so that it can multiply by either:
$v = \dfrac 1 3 \paren { -1 }^{ \dfrac 2 3 } { \Lambda }^{ \dfrac 1 3 }$
Any of these three roots are able to be used in the cascade.
==== Reverse substitution ====
Substitute for $u = v + \dfrac w v$ in each of the three roots, where $w = \dfrac 4 9$.
$u = \dfrac 4 3 \paren { \Lambda }^{ - \dfrac 1 3 } + \dfrac 1 3 \paren { \Lambda }^{ \dfrac 1 3 }$
$u = - \dfrac 4 3 \paren { - \Lambda }^{ - \dfrac 1 3 } + \dfrac 1 3 \paren { -1 }^{ \dfrac 2 3 } \paren { \Lambda }^{ \dfrac 1 3 }$
$u = \dfrac 4 3 \paren { -1 }^{ \dfrac 2 3 } \paren { \Lambda }^{ - \dfrac 1 3 } - \dfrac 1 3 \paren { - \Lambda }^{ \dfrac 1 3 }$
We can pick any of these three roots. The following will use the second of these three.
For reference, let ${ \overline \Lambda } = 19 - 3 \sqrt {33}$, the negative form.
=== Part 4: Lambda cubic cascade ===
Therefore, the Tribonacci constant is equal to the preceding.
It can be evaluated manually or with a calculator to be approximately $1.83929$.
Category:Tribonacci Sequences
Category:Cubic Equations