Legendary Wall-Sun-Sun primes

Let $$\alpha(p)$$ denote the least positive subscript, where $$p|F_{\alpha(p)}$$. A prime p is a Wall Sun Sun prime if $$\alpha(p^2) = \alpha(p)$$.
Then $F_{F_{\alpha(p^2)}} = F_{F_{\alpha(p)}}$ , where $F_{p^2}|F_{F_{\alpha(p)}}$ and $\alpha(F_{F_{\alpha(p^2)}}) = \alpha(F_{F_{\alpha(p)}})$. Let $n=p_{1}^{e_{1}}p_{2}^{e_{2}} \cdots p_{k}^{e_{k}}=F_{\alpha(p)}$. Then

(i)

$\text{If } p_{k}=p, e_{k} \ge 2 \text{, then } \alpha{(F_{p_{1}^{e_{1}}}F_{p_{2}^{e_{2}}} \cdots F_{p_{k}^{e_{k}}})} = \alpha{(F_{p_{1}^{e_{1}}}F_{p_{2}^{e_{2}}} \cdots F_{p_{k}})} \text{, for } n \ne p_{1}^{e_{1}}, n=p_{1}^{e_{1}}p_{2}^{e_{2}} \cdots p_{k}^{e_{k}} \text{, where } 2 \nparallel n \text{, or }2 \nmid n \text{.}$

$\text{If } p_{1}=p, e_{1} \ge 2 \text{, then } \alpha{(F_{p_{1}^{e_{1}}}F_{p_{2}^{e_{2}}} \cdots F_{p_{k}^{e_{k}}})} = \alpha{(F_{p_{1}}F_{p_{2}^{e_{2}}} \cdots F_{p_{k}^{e_{k}}})} \text{, for } n \ne p_{1}^{e_{1}}, n=p_{1}^{e_{1}}p_{2}^{e_{2}} \cdots p_{k}^{e_{k}} \text{, where } 2 \nparallel n \text{, or }2 \nmid n \text{.}$

$\text{If } p_{i}=p, e_{i} \ge 2 \text{, then } \alpha{(F_{p_{1}^{e_{1}}} \cdots F_{p_{i}^{e_{i}}} \cdots F_{p_{k}^{e_{k}}})} = \alpha{(F_{p_{1}^{e_{1}}} \cdots F_{p_{i}}\cdots F_{p_{k}^{e_{k}}})} \text{, for } n \ne p_{1}^{e_{1}}, n=p_{1}^{e_{1}}p_{2}^{e_{2}} \cdots p_{k}^{e_{k}} \text{, where } 2 \nparallel n \text{, or }2 \nmid n \text{.}$

(ii)

$\text{If } p_{k}=p, e_{k} \ge 2 \text{, then } \alpha{(F_{2}F_{p_{2}^{e_{2}}}F_{p_{3}^{e_{3}}} \cdots F_{p_{k}^{e_{k}}})} = \alpha{(F_{2}F_{p_{2}^{e_{2}}}F_{p_{3}^{e_{3}}} \cdots F_{p_{k}})} \text{, for } n \ne p_{1}^{e_{1}}, n \ne p_{1}^{e_{1}}p_{2}^{e_{2}}, n=2p_{2}^{e_{2}}p_{3}^{e_{3}} \cdots p_{k}^{e_{k}} \text{, where } 2 \parallel n \text{.}$

$\text{If } p_{2}=p, e_{2} \ge 2 \text{, then } \alpha{(F_{2}F_{p_{2}^{e_{2}}}F_{p_{3}^{e_{3}}} \cdots F_{p_{k}^{e_{k}}})} = \alpha{(F_{2}F_{p_{2}}F_{p_{3}^{e_{3}}} \cdots F_{p_{k}^{e_{k}}})} \text{, for } n \ne p_{1}^{e_{1}}, n \ne p_{1}^{e_{1}}p_{2}^{e_{2}}, n=2p_{2}^{e_{2}}p_{3}^{e_{3}} \cdots p_{k}^{e_{k}} \text{, where } 2 \parallel n \text{.}$

$\text{If } p_{i}=p, e_{i} \ge 2 \text{, then } \alpha{(F_{2}F_{p_{2}^{e_{2}}} \cdots F_{p_{i}^{e_{i}}} \cdots F_{p_{k}^{e_{k}}})} = \alpha{(F_{2}F_{p_{2}^{e_{2}}} \cdots F_{p_{i}}\cdots F_{p_{k}^{e_{k}}})} \text{, for } n \ne p_{1}^{e_{1}}, n \ne p_{1}^{e_{1}}p_{2}^{e_{2}}, n=2p_{2}^{e_{2}}p_{3}^{e_{3}} \cdots p_{k}^{e_{k}} \text{, where } 2 \parallel n \text{.}$

(iii)

$\text{If } p_{2}=p, e_{2} \ge 2 \text{, then } \alpha{(F_{2}F_{p_{2}^{e_{2}}})} = \alpha{(F_{2}F_{p_{2}})} \text{, for } n=2p_{2}^{e_{2}} \text{, where } 2 \parallel n \text{.}$

(iv)

$\text{If } p_{1}=p, e_{1} \ge 2 \text{, then } \alpha{(F_{p_{1}^{e_{1}}})} = \alpha{(F_{p_{1}})} \text{, for } n=p_{1}^{e_{1}} \text{, where } 2 \nparallel n \text{, or }2 \nmid n \text{.}$

We know that:

$\text{If }d|n \text{, then }F_{d}|F_{n} \text{, for }d \le n, n \ge 1 \text{.}$

$\text{If }F_{d} \nmid F_{n}, \text{ then }d \nmid n \text{, for }d \le n, n \ge 1 \text{.}$

$\alpha(F_n) = n \text{, for }n\ne 2 \text{.}$

$\alpha(F_{F_n}) = F_n \text{, for }n \ne 2,n\ne 3 \text{.}$

$\alpha(\phi_{n} )=\alpha(F_n) \text{, for } n \ne 2 \text{.}$

$\phi_{n}= \prod_{d \mid n} F_{d}^{\mu(n/d)} \text{, for all divisors } d \text{ of } n \text{.}$ {T.M. Apostol 1976}

$\beta_{n}= \prod_{d \mid n} F_{d}^{-\mu(n/d)} \text{, for proper divisors } d \text{ of } n \text{.}$

$F_{n}= \beta_{n} \phi_{n} \text{.}$

$\text{We can prove }F_{p_{i}^{e_{i}}} \nmid F_{F_{\alpha(p_{i})}} \text{, for } \lambda \ge e_{i} \ge 2 \text{. Then } p_{i}^{e_{i}} \nmid F_{\alpha(p_{i})} \text{, for } \lambda \ge e_{i} \ge 2 \text{.}$

$\gcd(F_{p_{1}^{e_{1}}}, F_{p_{2}^{e_{2}}})=F_{\gcd(p_{1}^{e_{1}}, p_{2}^{e_{2}})}=F_{1}=1 \text{, for distinct primes } p_{1}, p_{2}, e_{1} \ge 1, e_{2} \ge 1 \text{.}$ {E. Lucas}

Let $[a,b]$ denote the least common multiple of $a$ and $b$. Then $\alpha([a, b]) = [\alpha(a), \alpha(b)]$. {D. W. Robinson 1963}

$(i) \text{ } \alpha(F_{p_{1}^{e_{1}}}F_{p_{2}^{e_{2}}}\cdots F_{p_k^{e_k}}) = \alpha([F_{p_{1}^{e_{1}}},F_{p_{2}^{e_{2}}},...,F_{p_k^{e_k}}]) = [\alpha(F_{p_{1}^{e_{1}}}), \alpha(F_{p_{2}^{e_{2}}}),..., \alpha(F_{p_k^{e_k}})] = [p_{1}^{e_{1}},p_{2}^{e_{2}},...,p_{k}^{e_{k}}] = (p_{1}^{e_{1}}p_{2}^{e_{2}}\cdots p_{k}^{e_{k}}) = n \text{, for } n \ne p_{1}^{e_{1}},n = p_{1}^{e_{1}}p_{2}^{e_{2}} \cdots p_{k}^{e_{k}} \text{, where }2 \nparallel n \text{, or }2 \nmid n \text{.}$

$(ii )\text{ } \alpha(F_{2}F_{p_{2}^{e_{2}}}F_{p_{3}^{e_{3}}}\cdots F_{p_k^{e_k}}) = \alpha([F_{2},F_{p_{2}^{e_{2}}},F_{p_{3}^{e_{3}}},...,F_{p_k^{e_k}}]) = [\alpha(F_{2}),\alpha(F_{p_{2}^{e_{2}}}), \alpha(F_{p_{3}^{e_{3}}}),..., \alpha(F_{p_k^{e_k}})]=[1,p_{2}^{e_{2}},p_{3}^{e_{3}},...,p_{k}^{e_{k}}]=(p_{2}^{e_{2}}p_{3}^{e_{3}}\cdots p_{k}^{e_{k}})=\tfrac{n}{2} \text{, for } n \ne p_{1}^{e_{1}},n \ne p_{1}^{e_{1}}p_{2}^{e_{2}},n=p_{1}^{e_{1}}p_{2}^{e_{2}}p_{3}^{e_{3}} \cdots p_{k}^{e_{k}} \text{, where }2 \parallel n \text{.}$

$(iii)\text{ } \alpha(F_{2}F_{p_{2}^{e_{2}}}) = \alpha([F_{2},F_{p_{2}^{e_{2}}}]) = [\alpha(F_{2}), \alpha(F_{p_{2}^{e_{2}}})]=[1,p_{2}^{e_{2}}]=(p_{2}^{e_{2}})=\frac{n}{2} \text{, for } n=2p_{2}^{e_{2}} \text{, where }2 \parallel n \text{.}$

$(iv)\text{ } \alpha(F_{p_{1}^{e_{1}}}) = \alpha([F_{p_{1}^{e_{1}}}]) = [\alpha(F_{p_{1}^{e_{1}}})]=[p_{1}^{e_{1}}]=(p_{1}^{e_{1}})=n \text{, for } n=p_{1}^{e_{1}} \text{, where }2 \nparallel n \text{, or }2 \nmid n \text{.}$

$\text{If }p^{\lambda} \parallel F_{\alpha(p)} \text{, then } F_{p^\lambda} \mid F_{F_{\alpha(p)}} \text{, for } \lambda \ge 2 \text{. According to the 4 proof points above, either }$

$\alpha{(F_{p_{1}^{e_{1}}}F_{p_{2}^{e_{2}}} \cdots F_{p_{k}^{e_{k}}})} = \alpha{(F_{p_{1}^{e_{1}}}F_{p_{2}^{e_{2}}} \cdots F_{p_{k}})}=(p_{1}^{e_{1}}p_{2}^{e_{2}} \cdots p_{k}^{e_{k}})=(p_{1}^{e_{1}}p_{2}^{e_{2}} \cdots p_{k}) \text{, for } p_{k}=p,\lambda \ge e_{k} \ge 2 \text{, or }$

$\alpha{(F_{p_{1}^{e_{1}}}F_{p_{2}^{e_{2}}} \cdots F_{p_{k}^{e_{k}}})} = \alpha{(F_{p_{1}}F_{p_{2}^{e_{2}}} \cdots F_{p_{k}^{e_{k}}})}=(p_{1}^{e_{1}}p_{2}^{e_{2}} \cdots p_{k}^{e_{k}})=(p_{1}p_{2}^{e_{2}} \cdots p_{k}^{e_{k}}) \text{, for } p_{1}=p,\lambda \ge e_{1} \ge 2 \text{, or }$

$\alpha{(F_{p_{1}^{e_{1}}} \cdots F_{p_{i}^{e_{i}}} \cdots F_{p_{k}^{e_{k}}})} = \alpha{(F_{p_{1}^{e_{1}}} \cdots F_{p_{i}} \cdots F_{p_{k}^{e_{k}}})}=(p_{1}^{e_{1}} \cdots p_{i}^{e_{i}} \cdots p_{k}^{e_{k}})=(p_{1}^{e_{1}} \cdots p_{i} \cdots p_{k}^{e_{k}}) \text{, for } p_{i}=p,\lambda \ge e_{i} \ge 2 \text{, or }$

$\alpha{(F_{2}F_{p_{2}^{e_{2}}}F_{p_{3}^{e_{3}}} \cdots F_{p_{k}^{e_{k}}})} = \alpha{(F_{2}F_{p_{2}^{e_{2}}}F_{p_{3}^{e_{3}}} \cdots F_{p_{k}})}=(p_{2}^{e_{2}}p_{3}^{e_{3}} \cdots p_{k}^{e_{k}})=(p_{2}^{e_{2}}p_{3}^{e_{3}} \cdots p_{k}) \text{, for } p_{k}=p,\lambda \ge e_{k} \ge 2 \text{, or }$

$\alpha{(F_{2}F_{p_{2}^{e_{2}}}F_{p_{3}^{e_{3}}} \cdots F_{p_{k}^{e_{k}}})} = \alpha{(F_{2}F_{p_{2}}F_{p_{3}^{e_{3}}} \cdots F_{p_{k}^{e_{k}}})}=(p_{2}^{e_{2}}p_{3}^{e_{3}} \cdots p_{k}^{e_{k}})=(p_{2}p_{3}^{e_{3}} \cdots p_{k}^{e_{k}}) \text{, for } p_{2}=p,\lambda \ge e_{2} \ge 2 \text{, or }$

$\alpha{(F_{2}F_{p_{2}^{e_{2}}} \cdots F_{p_{i}^{e_{i}}} \cdots F_{p_{k}^{e_{k}}})} = \alpha{(F_{2}F_{p_{2}^{e_{2}}} \cdots F_{p_{i}} \cdots F_{p_{k}^{e_{k}}})}=(p_{2}^{e_{2}} \cdots p_{i}^{e_{i}} \cdots p_{k}^{e_{k}})=(p_{2}^{e_{2}} \cdots p_{i} \cdots p_{k}^{e_{k}}) \text{, for } p_{i}=p,\lambda \ge e_{i} \ge 2 \text{, or }$

$\alpha{(F_{2}F_{p_{2}^{e_{2}}})} = \alpha{(F_{2}F_{p_{2}})}=(p_{2}^{e_{2}})=(p_{2}) \text{, for } p_{2}=p,\lambda \ge e_{2} \ge 2 \text{, or }$

$\alpha{(F_{p_{1}^{e_{1}}})} = \alpha{(F_{p_{1}})=(p_{1}^{e_{1}})=(p_{1})} \text{, for } p_{1}=p, \lambda \ge e_{1} \ge 2 \text{, which is absurd.}$

Does anyone have any comments, questions, suggestions, or corrections?

Note by Shane Findley
2 years, 3 months ago

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