User talk:Primedivine

Main idea

Let $n=p\cdot \alpha (p)$.
For all divisors $d$ , where $d_{z}=\alpha (p)$.
For proper divisors $d$, where $d_{1},...,d_{y}\neq \alpha (p)$.
$\varphi _{n}=LCM[F_{d_{1}},F_{d_{2}},...,F_{d_{z}},F_{p},F_{p\cdot d_{1}},F_{p\cdot d_{2}},...,F_{p\cdot d_{y}}]$
$p^{3}\nmid \varphi _{n}$
$p^{2}\nmid F_{\alpha (p)}$
$p\parallel F_{\alpha (p)}$

Greatest common divisor

$\gcd(F_{m},F_{n})=F_{\gcd(m,n)}$ - Lucas
For $n<p$, $\gcd(F_{n},F_{p})=F_{\gcd(n,p)}=F_{1}=1$. - SF

For all divisors $d$ of $n$, the antecedent can be swapped with the consequent.
If $d\mid n$ then $F_{d}\mid F_{n}$.
If $F_{d}\not \mid F_{n}$ then $d\not \mid n$.

Entry point of divisors

All positive divisors $d$ divide some Fibonacci number.
Let $F_{\alpha (d)}$ denote the least positive Fibonacci number divisible by $d$, such that $d|F_{n}$.
Let $F_{F_{\alpha (d)}}$ denote the least positive Fibonacci number in the sub sequence that is divisible by $F_{d}$, such that $F_{d}|F_{F_{n}}$.
For $n\neq 2$, the entry point of a positive Fibonacci number $F_{n}$ is simply the subscript $n$, ie $\alpha (F_{n})=n$.

Methods

Primitive prime powers

For $n>12$, each Fibonacci number $F_{n}$ will have at least one primitive prime divisor, by Carmichael's theorem. By the Wall-Sun-Sun prime conjecture, $F_{n}$ could have at least one primitive prime power. Let $\phi _{\alpha (p)}=(p_{1}^{e_{1}}p_{2}^{e_{2}}\cdots p_{k}^{e_{k}})$ denote the full product of primitive prime powers (one or more) that divide $F_{\alpha (p)}$. By definition, this product of primitive prime powers always has an equal entry point to the whole Fibonacci number itself, ie $\alpha (\phi _{n})=\alpha (F_{n})=n$.

Lowest common multiple

For any positive integers a and b, let [a,b] denote the least common multiple of a and b.
$\alpha ([a,b])=[\alpha (a),\alpha (b)]$ - D. W. Robinson April 1963

If $a$ through $z$ are relatively prime then $F_{a}$ through $F_{z}$ are also relatively prime.
$\alpha (ab\cdots z)=\alpha ([a,b,\ldots ,z])=[\alpha (a),\alpha (b),\ldots ,\alpha (z)]$

If $F_{a}$ through $F_{z}$ are relatively prime then we have the following.
Type A: $2\not \mid n$ or else $2\not \parallel n$
Type B: $2\parallel n$ and also $4\not \parallel n$ Twice the odd numbers, also called singly even numbers.
Type A:$\alpha (F_{a}F_{b}\cdots F_{z})=\alpha ([F_{a},F_{b},\ldots ,F_{z}])=[\alpha (F_{a}),\alpha (F_{b}),\ldots ,\alpha (F_{z})]=[a,b,\ldots ,z]=(a\cdot b\cdots z)=ab\cdots z$
Type B: $\alpha (F_{2}F_{b}\cdots F_{z})=\alpha ([F_{2},F_{b},\ldots ,F_{z}])=[\alpha (F_{2}),\alpha (F_{b}),\ldots ,\alpha (F_{z})]=[1,b,\ldots ,z]=(1\cdot b\cdots z)=b\cdots z$

The fundamental theorem of arithmetic is bi-conditional with prime powered Fibonacci numbers. Let $n=p_{1}^{e_{1}}p_{2}^{e_{2}}\cdots p_{k}^{e_{k}}$.
Type A: $\alpha (F_{p_{1}^{e_{1}}}F_{p_{2}^{e_{2}}}\cdots F_{p_{k}^{e_{k}}})=\alpha ([F_{p_{1}^{e_{1}}},F_{p_{2}^{e_{2}}},\ldots ,F_{p_{k}^{e_{k}}}])=[\alpha (F_{p_{1}^{e_{1}}}),\alpha (F_{p_{2}^{e_{2}}}),\ldots ,\alpha (F_{p_{k}^{e_{k}}})]=[p_{1}^{e_{1}},p_{2}^{e_{2}},\ldots ,p_{k}^{e_{k}}]=(p_{1}^{e_{1}}p_{2}^{e_{2}}\cdots p_{k}^{e_{k}})=n=\alpha (F_{n})$.
Type B:

• $2\alpha (F_{2}F_{p_{2}^{e_{2}}}\cdots F_{p_{k}^{e_{k}}})=2\alpha ([F_{2},F_{p_{2}^{e_{2}}},\ldots ,F_{p_{k}^{e_{k}}}])=2[\alpha (F_{2}),\alpha (F_{p_{2}^{e_{2}}}),\ldots ,\alpha (F_{p_{k}^{e_{k}}})]=2[1,p_{2}^{e_{2}},\ldots ,p_{k}^{e_{k}}]=2(1\cdot p_{2}^{e_{2}}\cdots p_{k}^{e_{k}})=n=\alpha (F_{n})$.
• $\alpha (F_{F_{3}}F_{p_{2}^{e_{2}}}\cdots F_{p_{k}^{e_{k}}})=\alpha ([F_{F_{3}},F_{p_{2}^{e_{2}}},\ldots ,F_{p_{k}^{e_{k}}}])=[\alpha (F_{F_{3}}),\alpha (F_{p_{2}^{e_{2}}}),\ldots ,\alpha (F_{p_{k}^{e_{k}}})]=[2,p_{2}^{e_{2}},\ldots ,p_{k}^{e_{k}}]=(2\cdot p_{2}^{e_{2}}\cdots p_{k}^{e_{k}})=n=\alpha (F_{n})$.
• $\alpha (F_{F_{F_{4}}}F_{p_{2}^{e_{2}}}\cdots F_{p_{k}^{e_{k}}})=\alpha ([F_{F_{F_{4}}},F_{p_{2}^{e_{2}}},\ldots ,F_{p_{k}^{e_{k}}}])=[\alpha (F_{F_{F_{4}}}),\alpha (F_{p_{2}^{e_{2}}}),\ldots ,\alpha (F_{p_{k}^{e_{k}}})]=[2,p_{2}^{e_{2}},\ldots ,p_{k}^{e_{k}}]=(2\cdot p_{2}^{e_{2}}\cdots p_{k}^{e_{k}})=n=\alpha (F_{n})$.

Wall Sun Sun prime conjecture

Let $\phi _{\alpha (p)}=(p_{1}^{e_{1}}p_{2}^{e_{2}}\cdots p_{k}^{e_{k}})$.
Suppose $\alpha (p^{2})=\alpha (p)\neq \alpha (p^{3})$.
Suppose $p_{i}=p$, for one or more Wall-Sun-Sun primes. In this particular instance, take $p_{k}=p$ for the sake of notation below.
Suppose $p^{2}\parallel F_{\alpha (p)}$ and also $p^{3}\not \parallel F_{\alpha (p)}$, Type A.
If $p^{2}|F_{\alpha (p)}$ then $F_{p^{2}}|F_{F_{\alpha (p)}}$.
If $p^{y\leq \lambda }|\phi _{\alpha (p)}$ then $\phi _{p^{y\leq \lambda }}|F_{\phi _{\alpha (p)}}$, for $\lambda \geq 2$, where $p^{\lambda +1}\nmid \phi _{\alpha (p)}$.

Claim 1 (Right side b)

If $F_{p^{2}}|F_{F_{\alpha (p)}}$ then $\alpha (F_{p_{1}^{e_{1}}}F_{p_{2}^{e_{2}}}\cdots F_{p_{k}^{2}}\cdot b)=\alpha (F_{p_{1}^{e_{1}}}F_{p_{2}^{e_{2}}}\cdots F_{p_{k}^{}}\cdot b)=F_{\alpha (p)}$.

Proof 1 (Right side b)

$\alpha (\phi _{F_{\alpha (p)}})=F_{\alpha (p)}$. Solve for the products with the Robinson equality.
$\alpha (F_{p_{1}^{e_{1}}}F_{p_{2}^{e_{2}}}\cdots F_{p_{k}^{2}}\cdot \phi _{F_{\alpha (p)}})=[p_{1}^{e_{1}},p_{2}^{e_{2}},\ldots ,p_{k}^{2},F_{\alpha (p)}]=F_{\alpha (p)}$
$\alpha (F_{p_{1}^{e_{1}}}F_{p_{2}^{e_{2}}}\cdots F_{p_{k}^{}}\cdot \phi _{F_{\alpha (p)}})=[p_{1}^{e_{1}},p_{2}^{e_{2}},\ldots ,p_{k}^{},F_{\alpha (p)}]=F_{\alpha (p)}$

If $\phi _{F_{\alpha (p)}}|b$, then $\alpha (b)=\alpha (\phi _{F_{\alpha (p)}})$, for divisors $b_{i}$ of $F_{F_{\alpha (p)}}$.

$\alpha (F_{p_{1}^{e_{1}}}F_{p_{2}^{e_{2}}}\cdots F_{p_{k}^{2}}\cdot \phi _{F_{\alpha (p)}})=\alpha (F_{p_{1}^{e_{1}}}F_{p_{2}^{e_{2}}}\cdots F_{p_{k}^{}}\cdot \phi _{F_{\alpha (p)}})=F_{\alpha (p)}$

Claim 2 (Left side a)

If $F_{p^{2}}|F_{F_{\alpha (p)}}$ then $\alpha (a\cdot F_{p_{1}^{e_{1}}}F_{p_{2}^{e_{2}}}\cdots F_{p_{k}^{2}})=\alpha (a\cdot F_{p_{1}^{e_{1}}}F_{p_{2}^{e_{2}}}\cdots F_{p_{k}^{}})=F_{\alpha (p)}$.

Proof 2 (Left side a)

If $\phi _{\alpha (p)}|F_{\alpha (p)}$ then $F_{\phi _{\alpha (p)}}|F_{F_{\alpha (p)}}$.

$\alpha (F_{p_{1}^{e_{1}}}F_{p_{2}^{e_{2}}}\cdots F_{p_{k}^{e_{k}}}\cdot \phi _{\phi _{\alpha (p)}})=\alpha (F_{\phi _{\alpha (p)}})=\phi _{\alpha (p)}$.

Establish the hypothetical equality conjectured by Wall-Sun-Sun.
$\alpha (F_{p_{1}^{e_{1}}}F_{p_{2}^{e_{2}}}\cdots F_{p_{k}^{2}}\cdot \phi _{\phi _{\alpha (p)}})=\alpha (F_{p_{1}^{e_{1}}}F_{p_{2}^{e_{2}}}\cdot F_{p_{k}^{}}\cdot \phi _{\phi _{\alpha (p)}})$?

Solve for the products with the Robinson formula to prove that hypothetically a Wall Sun Sun prime would cause this equality to be true.
$\alpha (F_{p_{1}^{e_{1}}}F_{p_{2}^{e_{2}}}\cdots F_{p_{k}^{2}}\cdot \phi _{\phi _{\alpha (p)}})=[p_{1}^{e_{1}},p_{2}^{e_{2}},\ldots ,p_{k}^{2},\phi _{\alpha (p)}]=\phi _{\alpha (p)}$
$\alpha (F_{p_{1}^{e_{1}}}F_{p_{2}^{e_{2}}}\cdots F_{p_{k}^{}}\cdot \phi _{\phi _{\alpha (p)}})=[p_{1}^{e_{1}},p_{2}^{e_{2}},\ldots ,p_{k}^{},\phi _{\alpha (p)}]=\phi _{\alpha (p)}$

Claim 3 (Invalidate the conditional of Claim 2)

$\alpha (\phi _{\phi _{\alpha (p)}}\cdot F_{p_{1}^{e_{1}}}F_{p_{2}^{e_{2}}}\cdots F_{p_{k}^{2}})\neq \alpha (\phi _{\phi _{\alpha (p)}}\cdot F_{p_{1}^{e_{1}}}F_{p_{2}^{e_{2}}}\cdots F_{p_{k}^{}})$
$\alpha (F_{p_{1}^{e_{1}}}F_{p_{2}^{e_{2}}}\cdots F_{p_{k}^{2}})\neq \alpha (F_{p_{1}^{e_{1}}}F_{p_{2}^{e_{2}}}\cdots F_{p_{k}^{}})$

Proof 3 (by contradiction)

By the greatest common divisor, we have
$F_{p}=\phi _{p},F_{p^{2}}=\phi _{p}\phi _{p^{2}}$.

By Wall's hypothesis,
$\alpha (F_{p_{1}^{}}F_{p_{2}^{}}\cdots F_{p_{k}^{}}\cdot \phi _{\phi _{\alpha (p)}})=\alpha ([F_{p_{1}^{}},F_{p_{2}^{}},\ldots ,F_{p_{k}^{}},\phi _{\phi _{\alpha (p)}}])=[p_{1}^{},p_{2}^{},\ldots ,p_{k}^{},\phi _{\alpha (p)}]=\alpha (F_{\phi _{\alpha (p)}})=\phi _{\alpha (p)}$

By the Wall-Sun-Sun prime conjecture,
$\alpha (F_{p_{1}^{e_{1}}}F_{p_{2}^{e_{2}}}\cdots \phi _{p_{k}^{2}}\cdot \phi _{p_{k}^{}}\cdot \phi _{\phi _{\alpha (p)}})=\alpha ([F_{p_{1}^{e_{1}}},F_{p_{2}^{e_{2}}},\ldots ,\phi _{p_{k}^{2}},\phi _{p_{k}^{}},\phi _{\phi _{\alpha (p)}}])=[p_{1}^{e_{1}},p_{2}^{e_{2}},\ldots ,p_{k}^{2},p_{k}^{},\phi _{\alpha (p)}]=\alpha (F_{\phi _{\alpha (p)}})=\phi _{\alpha (p)}$
$\alpha (F_{p_{1}^{e_{1}}}F_{p_{2}^{e_{2}}}\cdots \phi _{p_{k}^{2}}\cdot \phi _{\phi _{\alpha (p)}})=\alpha ([F_{p_{1}^{e_{1}}},F_{p_{2}^{e_{2}}},\ldots ,\phi _{p_{k}^{2}},\phi _{\phi _{\alpha (p)}}])=[p_{1}^{e_{1}},p_{2}^{e_{2}},\ldots ,p_{k}^{2},\phi _{\alpha (p)}]=\alpha (F_{\phi _{\alpha (p)}})=\phi _{\alpha (p)}$
$\alpha (F_{p_{1}^{e_{1}}}F_{p_{2}^{e_{2}}}\cdots \phi _{p_{k}^{}}\cdot \phi _{\phi _{\alpha (p)}})=\alpha ([F_{p_{1}^{e_{1}}},F_{p_{2}^{e_{2}}},\ldots ,\phi _{p_{k}^{}},\phi _{\phi _{\alpha (p)}}])=[p_{1}^{e_{1}},p_{2}^{e_{2}},\ldots ,p_{k}^{},\phi _{\alpha (p)}]=\alpha (F_{\phi _{\alpha (p)}})=\phi _{\alpha (p)}$

• $\alpha (F_{p_{1}^{e_{1}}}F_{p_{2}^{e_{2}}}\cdots \phi _{p_{k}^{2}})=\alpha ([F_{p_{1}^{e_{1}}},F_{p_{2}^{e_{2}}},\ldots ,\phi _{p_{k}^{2}}])=[p_{1}^{e_{1}},p_{2}^{e_{2}},\ldots ,p_{k}^{2}]=\phi _{\alpha (p)}$
• $\alpha (F_{p_{1}^{e_{1}}}F_{p_{2}^{e_{2}}}\cdots \phi _{p_{k}^{2}}\cdot \phi _{p_{k}^{}})=\alpha ([F_{p_{1}^{e_{1}}},F_{p_{2}^{e_{2}}},\ldots ,\phi _{p_{k}^{2}},\phi _{p_{k}^{}}])=[p_{1}^{e_{1}},p_{2}^{e_{2}},\ldots ,p_{k}^{2},p_{k}^{}]=\phi _{\alpha (p)}$

$\alpha (F_{p_{1}^{e_{1}}}F_{p_{2}^{e_{2}}}\cdots \phi _{p_{k}^{2}}\cdot \phi _{p_{k}^{}})=\alpha (F_{p_{1}^{e_{1}}}F_{p_{2}^{e_{2}}}\cdots \phi _{p_{k}^{}})=\phi _{\alpha (p)}$

However, we can measure that equality to verify that it is false.

• $\alpha (F_{p_{1}^{e_{1}}}F_{p_{2}^{e_{2}}}\cdots \phi _{p_{k}^{}})=\alpha ([F_{p_{1}^{e_{1}}},F_{p_{2}^{e_{2}}},\ldots ,\phi _{p_{k}^{}}])=[p_{1}^{e_{1}},p_{2}^{e_{2}},\ldots ,p_{k}^{}]=\phi _{\alpha (p)}/p$

Proper divisors of the product of primitive prime powers

By Carmichael's theorem, for $n>12,F_{n}$ will have at least one primitive prime divisor that has not appeared as a divisor of an earlier Fibonacci number. By the Wall-Sun-Sun prime conjecture, let $\phi _{\alpha (p)}=(p_{1}^{e_{1}}p_{2}^{e_{2}}\cdots p_{k}^{e_{k}})$ denote the full product of primitive prime powers (one or more) that divide $F_{\alpha (p)}$.

For proper divisors $d_{i}$ of $\phi _{\alpha (p)}$,
$\alpha (\phi _{d_{1}}\phi _{d_{2}}\cdots \phi _{d_{j}}\cdot \phi _{\phi _{\alpha (p)}})=\alpha ([\phi _{d_{1}},\phi _{d_{2}},\ldots ,\phi _{d_{j}},\phi _{\phi _{\alpha (p)}}])=[\alpha (\phi _{d_{1}}),\alpha (\phi _{d_{2}}),\ldots ,\alpha (\phi _{d_{j}}),\alpha (\phi _{\phi _{\alpha (p)}})]=[d_{1},d_{2},\ldots ,d_{j},\phi _{\alpha (p)}]=\alpha (F_{\phi _{\alpha (p)}})=\phi _{\alpha (p)}$.

For $d_{i}\neq 1,2,6,12$,
$n=(p_{1}^{e_{1}}p_{2}^{e_{2}}\cdots p_{k}^{e_{k}})=[d_{1},d_{2},\ldots ,d_{j},n]=[\alpha (\phi _{d_{1}}),\alpha (\phi _{d_{2}}),\ldots ,\alpha (\phi _{d_{j}}),\alpha (\phi _{n})]=\alpha ([\phi _{d_{1}},\phi _{d_{2}},\ldots ,\phi _{d_{j}},\phi _{n}])=\alpha (\phi _{d_{1}}\phi _{d_{2}}\cdots \phi _{d_{j}}\cdot \phi _{n})$.

For example, if $\phi _{\alpha (p)}=p^{2}\cdot r$ then
$\alpha (\phi _{p}\phi _{p^{2}}\phi _{r}\phi _{pr}\cdot \phi _{\phi _{\alpha (p)}})=\alpha ([\phi _{p},\phi _{p^{2}},\phi _{r},\phi _{pr},\phi _{\phi _{\alpha (p)}}])=[\alpha (\phi _{p}),\alpha (\phi _{p^{2}}),\alpha (\phi _{r}),\alpha (\phi _{pr}),\alpha (\phi _{\phi _{\alpha (p)}})]=[p,p^{2},r,pr,\phi _{\alpha (p)}]=\alpha (F_{\phi _{\alpha (p)}})=\phi _{\alpha (p)}$.
$\alpha (\phi _{p}\phi _{r}\phi _{pr}\cdot \phi _{\phi _{\alpha (p)}})=\alpha ([\phi _{p},\phi _{r},\phi _{pr},\phi _{\phi _{\alpha (p)}}])=[\alpha (\phi _{p}),\alpha (\phi _{r}),\alpha (\phi _{pr}),\alpha (\phi _{\phi _{\alpha (p)}})]=[p,r,pr,\phi _{\alpha (p)}]=\alpha (F_{\phi _{\alpha (p)}})=\phi _{\alpha (p)}$.

$\alpha (\phi _{p}\phi _{p^{2}}\phi _{r}\phi _{pr}\cdot \phi _{\phi _{\alpha (p)}})=\alpha (\phi _{p}\phi _{r}\phi _{pr}\cdot \phi _{\phi _{\alpha (p)}})$
$\alpha (F_{pr}\cdot \phi _{p^{2}}\cdot \phi _{\phi _{\alpha (p)}})=\alpha (F_{pr}\cdot \phi _{\phi _{\alpha (p)}})$
$\alpha (F_{p^{2}}\cdot \phi _{r}\cdot \phi _{pr}\cdot \phi _{\phi _{\alpha (p)}})=\alpha (F_{pr}\cdot \phi _{\phi _{\alpha (p)}})$
$F_{\phi _{\alpha (p)}}=(\prod _{p_{i}^{\lambda _{i}}\mid \phi _{\alpha (p)}}F_{p_{i}^{\lambda _{i}}})\cdot (\prod _{d_{i}\mid \phi _{\alpha (p)}}\phi _{d_{i}})$

FTA

$n=(p_{1}^{\lambda _{1}}p_{2}^{\lambda _{2}}\cdots p_{k}^{\lambda _{k}})=[p_{1}^{\lambda _{1}},p_{2}^{\lambda _{2}},...,p_{k}^{\lambda _{k}},n]=\alpha (F_{p_{1}^{\lambda _{1}}}F_{p_{2}^{\lambda _{2}}}\cdots F_{p_{k}^{\lambda _{k}}}\cdot \phi _{n})$
$n=(p_{1}^{\lambda _{1}}p_{2}^{\lambda _{2}}\cdots p_{k}^{\lambda _{k}})=[p_{1}^{\lambda _{1}},p_{2}^{\lambda _{2}},...,p_{k}^{},n]=\alpha (F_{p_{1}^{\lambda _{1}}}F_{p_{2}^{\lambda _{2}}}\cdots F_{p_{k}^{}}\cdot \phi _{n})$

$n=(p_{1}^{\lambda _{1}}p_{2}^{\lambda _{2}}\cdots p_{k}^{\lambda _{k}})=[p_{1}^{\lambda _{1}},p_{2}^{\lambda _{2}},...,p_{k}^{\lambda _{k}}]=\alpha (F_{p_{1}^{\lambda _{1}}}F_{p_{2}^{\lambda _{2}}}\cdots F_{p_{k}^{\lambda _{k}}})$
$n/(p_{k}^{\lambda _{k}-1})=(p_{1}^{\lambda _{1}}p_{2}^{\lambda _{2}}\cdots p_{k}^{})=[p_{1}^{\lambda _{1}},p_{2}^{\lambda _{2}},...,p_{k}^{}]=\alpha (F_{p_{1}^{\lambda _{1}}}F_{p_{2}^{\lambda _{2}}}\cdots F_{p_{k}^{}})$

Constructing Fibonacci numbers

Let $n=\phi _{\alpha (p)}$.
Let $d_{1},d_{2},\ldots ,d_{j}$ be proper divisors of n, composed of at least two distinct prime divisors.

$n=(p_{1}^{}p_{2}^{})=[p_{1}^{},p_{2}^{},n]=\alpha (F_{p_{1}^{}}F_{p_{2}^{}}\cdot \phi _{n})=\alpha (\phi _{p_{1}^{}}\phi _{p_{2}^{}}\cdot \phi _{n})$

$n=(p_{1}^{}p_{2}^{2})=[p_{1}^{},p_{2}^{},p_{2}^{2},p_{1}^{}p_{2}^{},n]=\alpha (F_{p_{1}^{}}F_{p_{2}^{2}}\phi _{p_{1}^{}p_{2}^{}}\cdot \phi _{n})$

$n=(p_{1}^{\lambda }p_{2}^{\lambda })=[p_{1}^{y\leq \lambda },p_{2}^{z\leq \lambda },d_{1},d_{2},\ldots ,d_{j},n]=\alpha (F_{p_{1}^{\lambda }}F_{p_{2}^{\lambda }}\cdot \phi _{d_{1}}\phi _{d_{2}}\cdots \phi _{d_{j}}\cdot \phi _{n})$

$n=(p_{1}^{}p_{2}^{}p_{3}^{})=$
$[p_{1}^{},p_{2}^{},p_{3}^{},p_{1}^{}p_{2}^{},p_{2}^{}p_{3}^{},p_{1}^{}p_{3}^{},n]=$
$\alpha (F_{p_{1}^{}}F_{p_{2}^{}}F_{p_{3}^{}}\cdot \phi _{p_{1}^{}p_{2}^{}}\phi _{p_{2}^{}p_{3}^{}}\phi _{p_{1}^{}p_{3}^{}}\cdot \phi _{n})$

$n=(p_{1}^{}p_{2}^{}p_{3}^{}p_{4}^{})=$
$[p_{1}^{},p_{2}^{},p_{3}^{},p_{4}^{},p_{1}^{}p_{2}^{},p_{1}^{}p_{3}^{},p_{1}^{}p_{4}^{},p_{2}^{}p_{3}^{},p_{2}^{}p_{4}^{},p_{3}^{}p_{4}^{},p_{1}^{}p_{2}^{}p_{3}^{},p_{1}^{}p_{2}^{}p_{4}^{},p_{2}^{}p_{3}^{}p_{4}^{},p_{3}^{}p_{1}^{}p_{4}^{},n]=$
$\alpha (F_{p_{1}^{}}F_{p_{2}^{}}F_{p_{3}^{}}F_{p_{4}^{}}\cdot \phi _{p_{1}^{}p_{2}^{}}\phi _{p_{1}^{}p_{3}^{}}\phi _{p_{1}^{}p_{4}^{}}\phi _{p_{2}^{}p_{3}^{}}\phi _{p_{2}^{}p_{4}^{}}\phi _{p_{3}^{}p_{4}^{}}\phi _{p_{1}^{}p_{2}^{}p_{3}^{}}\phi _{p_{1}^{}p_{2}^{}p_{4}^{}}\phi _{p_{2}^{}p_{3}^{}p_{4}^{}}\phi _{p_{3}^{}p_{1}^{}p_{4}^{}}\cdot \phi _{n})$

$n=(p_{1}^{}p_{2}^{}\cdots p_{k}^{})=[p_{1}^{},p_{2}^{},...,p_{k}^{},n]=\alpha (F_{p_{1}^{}}F_{p_{2}^{}}\cdots F_{p_{k}^{}}\cdot \phi _{n})$

Dirichlet

$|\phi -{\frac {[16;5,1,1,5,22...]}{10}}|\leqslant {\frac {1}{?}}$, ie ${\frac {F_{21}}{F_{20}}},{\frac {F_{36}}{F_{35}}},{\frac {F_{51}}{F_{50}}},...$

Continued fractions for phi (golden ratio)

It is well known that,
$\phi ={\frac {1+{\sqrt {(}}5)}{2}}$.
However,
$\phi ={\frac {5\cdot 10^{n}+{\sqrt {(}}5^{3}\cdot 10^{2n})}{10^{n+1}}}$.
Yielding,
$\phi ={\frac {5+{\sqrt {(}}125)}{10}}={\frac {10}{{\sqrt {(}}125)-5}}$,
$\phi ={\frac {50+{\sqrt {(}}12500)}{100}}={\frac {100}{{\sqrt {(}}12500)-50}}$,
$\phi ={\frac {500+{\sqrt {(}}1250000)}{1000}}={\frac {1000}{{\sqrt {(}}1250000)-500}}$,
$\phi ={\frac {5000+{\sqrt {(}}125000000)}{10000}}={\frac {10000}{{\sqrt {(}}125000000)-5000}}$, and so on.

Let $n=-1$.
$\phi ={\frac {5\cdot 10^{-1}+{\sqrt {(}}125\cdot 10^{-1})}{10^{-1+1}}}$ yields
$\phi ={\frac {1/2+{\sqrt {(}}125\cdot 1/10)}{1}}$.

Let $n=-2$.
$\phi ={\frac {5\cdot 10^{-2}+{\sqrt {(}}125\cdot 10^{-4})}{10^{-2+1}}}$ yields
$\phi ={\frac {1/20+{\sqrt {(}}125\cdot 1/10000)}{1/10}}$.

Observe the related terms for $(1+\phi )$ and $\phi ^{2}$.
For all n, $\phi =({\frac {{\sqrt {(}}5^{3}\cdot 10^{2n})+(5\cdot 10^{n})}{{\sqrt {(}}5^{3}\cdot 10^{2n})-(5\cdot 10^{n})}})-1$ yields
$\phi =({\frac {{\sqrt {(}}125)+5}{{\sqrt {(}}125)-5}})-1$,
$\phi =({\frac {{\sqrt {(}}12500)+50}{{\sqrt {(}}12500)-50}})-1,\ldots$.