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:$$\alpha(F_aF_b \cdots F_z)=\alpha([F_a,F_b, \ldots, F_z])=[\alpha(F_a), \alpha(F_b), \ldots, \alpha(F_z)]=[a,b, \ldots, z]=(a \cdot b \cdots z)=ab \cdots z$$ Type B: $$\alpha(F_2F_b \cdots F_z)=\alpha([F_2,F_b, \ldots, F_z])=[\alpha(F_2), \alpha(F_b), \ldots, \alpha(F_z)]=[1,b, \ldots, z]=(1 \cdot b \cdots z)=b \cdots z$$ 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 $$p^{y \le \lambda}|\phi_{\alpha(p)}$$ then $$\phi_{p^{y \le \lambda}}|F_{\phi_{\alpha(p)}}$$, for $$\lambda \ge 2$$, 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, $$\alpha(F_{p_{1}^{e_1}}F_{p_{2}^{e_2}} \cdots \phi_{p_{k}^{2}} \cdot \phi_{p_{k}^{ }} \cdot \phi_{\phi_{\alpha(p)}}) =\alpha([F_{p_{1}^{e_1}}, F_{p_{2}^{e_{2}}}, \ldots, \phi_{p_{k}^{2}}, \phi_{p_{k}^{ }}, \phi_{\phi_{\alpha(p)}}] ) = [ p_{1}^{e_1}, p_{2}^{e_2}, \ldots, p_{k}^{2}, p_{k}^{ }, \phi_{\alpha(p)}]=\alpha(F_{\phi_{\alpha(p)}})= \phi_{\alpha(p)}$$ $$\alpha(F_{p_{1}^{e_1}}F_{p_{2}^{e_2}} \cdots \phi_{p_{k}^{2}} \cdot \phi_{\phi_{\alpha(p)}}) =\alpha([F_{p_{1}^{e_1}}, F_{p_{2}^{e_{2}}}, \ldots, \phi_{p_{k}^{2}},  \phi_{\phi_{\alpha(p)}}] ) = [ p_{1}^{e_1}, p_{2}^{e_2}, \ldots, p_{k}^{2}, \phi_{\alpha(p)}]=\alpha(F_{\phi_{\alpha(p)}})= \phi_{\alpha(p)}$$ $$\alpha(F_{p_{1}^{e_1}}F_{p_{2}^{e_2}} \cdots  \phi_{p_{k}^{ }} \cdot \phi_{\phi_{\alpha(p)}}) =\alpha([F_{p_{1}^{e_1}}, F_{p_{2}^{e_{2}}}, \ldots,  \phi_{p_{k}^{ }}, \phi_{\phi_{\alpha(p)}}] ) = [ p_{1}^{e_1}, p_{2}^{e_2}, \ldots, p_{k}^{ }, \phi_{\alpha(p)}]=\alpha(F_{\phi_{\alpha(p)}})= \phi_{\alpha(p)}$$ $$\alpha(F_{p_{1}^{e_1}}F_{p_{2}^{e_2}} \cdots \phi_{p_{k}^{2}} \cdot \phi_{p_{k}^{ }}) = \alpha(F_{p_{1}^{e_1}}F_{p_{2}^{e_2}} \cdots \phi_{p_{k}^{}}  ) = \phi_{\alpha(p)} $$ However, we can measure that equality to verify that it is false.
 * $$\alpha(F_{p_{1}^{e_1}}F_{p_{2}^{e_2}} \cdots \phi_{p_{k}^{2}} ) = \alpha([F_{p_{1}^{e_1}}, F_{p_{2}^{e_2}}, \ldots, \phi_{p_{k}^{2}}] ) = [ p_{1}^{e_1}, p_{2}^{e_2}, \ldots, p_{k}^{2}] = \phi_{\alpha(p)} $$
 * $$\alpha(F_{p_{1}^{e_1}}F_{p_{2}^{e_2}} \cdots \phi_{p_{k}^{2}} \cdot \phi_{p_{k}^{ }}) = \alpha([F_{p_{1}^{e_1}}, F_{p_{2}^{e_2}}, \ldots, \phi_{p_{k}^{2}}, \phi_{p_{k}^{ }}] ) = [ p_{1}^{e_1}, p_{2}^{e_2}, \ldots, p_{k}^{2}, p_{k}^{ }] =\phi_{\alpha(p)} $$
 * $$\alpha(F_{p_{1}^{e_1}}F_{p_{2}^{e_2}} \cdots \phi_{p_{k}^{}} ) = \alpha([F_{p_{1}^{e_1}}, F_{p_{2}^{e_2}}, \ldots, \phi_{p_{k}^{ }}] ) = [ p_{1}^{e_1}, p_{2}^{e_2}, \ldots, p_{k}^{}] = \phi_{\alpha(p)}/p $$

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. $$n=(p_{1}^{}p_{2}^{})=[p_{1}^{},p_{2}^{}, n]=\alpha(F_{p_{1}^{}}F_{p_{2}^{}} \cdot \phi_{n})=\alpha(\phi_{p_{1}^{}} \phi_{p_{2}^{}} \cdot \phi_{n})$$ $$n=(p_{1}^{}p_{2}^{2})=[p_{1}^{},p_{2}^{},p_{2}^{2},p_{1}^{}p_{2}^{}, n]=\alpha(F_{p_{1}^{}}F_{p_{2}^{2}}\phi_{p_{1}^{}p_{2}^{}} \cdot \phi_{n})$$ $$n=(p_{1}^{\lambda}p_{2}^{\lambda})=[p_{1}^{y \le \lambda},p_{2}^{z \le \lambda},d_{1},d_{2}, \ldots, d_{j}, n]=\alpha(F_{p_{1}^{\lambda}}F_{p_{2}^{\lambda}} \cdot \phi_{d_{1}}\phi_{d_{2}} \cdots \phi_{d_{j}} \cdot \phi_{n})$$ $$n=(p_{1}^{}p_{2}^{}p_{3}^{})=$$ $$[p_{1}^{},p_{2}^{},p_{3}^{},p_{1}^{}p_{2}^{},p_{2}^{}p_{3}^{},p_{1}^{}p_{3}^{}, n]=$$ $$\alpha(F_{p_{1}^{}}F_{p_{2}^{}}F_{p_{3}^{}} \cdot \phi_{p_{1}^{}p_{2}^{}} \phi_{p_{2}^{}p_{3}^{}} \phi_{p_{1}^{}p_{3}^{}} \cdot \phi_{n})$$ $$n=(p_{1}^{}p_{2}^{}p_{3}^{}p_{4}^{})=$$ $$[p_{1}^{},p_{2}^{},p_{3}^{},p_{4}^{},p_{1}^{}p_{2}^{},p_{1}^{}p_{3}^{},p_{1}^{}p_{4}^{},p_{2}^{}p_{3}^{},p_{2}^{}p_{4}^{},p_{3}^{}p_{4}^{},p_{1}^{}p_{2}^{}p_{3}^{},p_{1}^{}p_{2}^{}p_{4}^{},p_{2}^{}p_{3}^{}p_{4}^{},p_{3}^{}p_{1}^{}p_{4}^{}, n]=$$ $$\alpha(F_{p_{1}^{}}F_{p_{2}^{}}F_{p_{3}^{}}F_{p_{4}^{}} \cdot \phi_{p_{1}^{}p_{2}^{}} \phi_{p_{1}^{}p_{3}^{}} \phi_{p_{1}^{}p_{4}^{}} \phi_{p_{2}^{}p_{3}^{}} \phi_{p_{2}^{}p_{4}^{}} \phi_{p_{3}^{}p_{4}^{}} \phi_{p_{1}^{}p_{2}^{}p_{3}^{}} \phi_{p_{1}^{}p_{2}^{}p_{4}^{}} \phi_{p_{2}^{}p_{3}^{}p_{4}^{}} \phi_{p_{3}^{}p_{1}^{}p_{4}^{}} \cdot \phi_{n})$$ $$n=(p_{1}^{}p_{2}^{} \cdots p_{k}^{})=[p_{1}^{},p_{2}^{},...,p_{k}^{}, n]=\alpha(F_{p_{1}^{}}F_{p_{2}^{}} \cdots F_{p_{k}^{}} \cdot \phi_{n})$$

=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 $$\phi=(\frac{\sqrt(125)+5}{\sqrt(125)-5})-1$$, $$\phi=(\frac{\sqrt(12500)+50}{\sqrt(12500)-50})-1, \ldots$$.