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 FFα(p2)=FFα(p)F_{F_{\alpha(p^2)}} = F_{F_{\alpha(p)}} , where Fp2∣FFα(p) F_{p^2}|F_{F_{\alpha(p)}} and α(FFα(p2))=α(FFα(p)) \alpha(F_{F_{\alpha(p^2)}}) = \alpha(F_{F_{\alpha(p)}}) . Let n=p1e1p2e2⋯pkek=Fα(p)n=p_{1}^{e_{1}}p_{2}^{e_{2}} \cdots p_{k}^{e_{k}}=F_{\alpha(p)}. Then

(i)

If pk=p,ek≥2, then α(Fp1e1Fp2e2⋯Fpkek)=α(Fp1e1Fp2e2⋯Fpk), for n≠p1e1,n=p1e1p2e2⋯pkek, where 2∦n, or 2∤n.\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{.}

If p1=p,e1≥2, then α(Fp1e1Fp2e2⋯Fpkek)=α(Fp1Fp2e2⋯Fpkek), for n≠p1e1,n=p1e1p2e2⋯pkek, where 2∦n, or 2∤n.\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{.}

If pi=p,ei≥2, then α(Fp1e1⋯Fpiei⋯Fpkek)=α(Fp1e1⋯Fpi⋯Fpkek), for n≠p1e1,n=p1e1p2e2⋯pkek, where 2∦n, or 2∤n.\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)

If pk=p,ek≥2, then α(F2Fp2e2Fp3e3⋯Fpkek)=α(F2Fp2e2Fp3e3⋯Fpk), for n≠p1e1,n≠p1e1p2e2,n=2p2e2p3e3⋯pkek, where 2∥n.\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{.}

If p2=p,e2≥2, then α(F2Fp2e2Fp3e3⋯Fpkek)=α(F2Fp2Fp3e3⋯Fpkek), for n≠p1e1,n≠p1e1p2e2,n=2p2e2p3e3⋯pkek, where 2∥n.\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{.}

If pi=p,ei≥2, then α(F2Fp2e2⋯Fpiei⋯Fpkek)=α(F2Fp2e2⋯Fpi⋯Fpkek), for n≠p1e1,n≠p1e1p2e2,n=2p2e2p3e3⋯pkek, where 2∥n.\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)

If p2=p,e2≥2, then α(F2Fp2e2)=α(F2Fp2), for n=2p2e2, where 2∥n.\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)

If p1=p,e1≥2, then α(Fp1e1)=α(Fp1), for n=p1e1, where 2∦n, or 2∤n.\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:

If d∣n, then Fd∣Fn, for d≤n,n≥1.\text{If }d|n \text{, then }F_{d}|F_{n} \text{, for }d \le n, n \ge 1 \text{.}

If Fd∤Fn, then d∤n, for d≤n,n≥1.\text{If }F_{d} \nmid F_{n}, \text{ then }d \nmid n \text{, for }d \le n, n \ge 1 \text{.}

α(Fn)=n, for n≠2.\alpha(F_n) = n \text{, for }n\ne 2 \text{.}

α(FFn)=Fn, for n≠2,n≠3.\alpha(F_{F_n}) = F_n \text{, for }n \ne 2,n\ne 3 \text{.}

α(ϕn)=α(Fn), for n≠2.\alpha(\phi_{n} )=\alpha(F_n) \text{, for } n \ne 2 \text{.}

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

βn=∏d∣nFd−μ(n/d), for proper divisors d of n.\beta_{n}= \prod_{d \mid n} F_{d}^{-\mu(n/d)} \text{, for proper divisors } d \text{ of } n \text{.}

Fn=βnϕn.F_{n}= \beta_{n} \phi_{n} \text{.}

We can prove Fpiei∤FFα(pi), for λ≥ei≥2. Then piei∤Fα(pi), for λ≥ei≥2.\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⁡(Fp1e1,Fp2e2)=Fgcd⁡(p1e1,p2e2)=F1=1, for distinct primes p1,p2,e1≥1,e2≥1.\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][a,b] denote the least common multiple of aa and bb. Then α([a,b])=[α(a),α(b)]\alpha([a, b]) = [\alpha(a), \alpha(b)] . {D. W. Robinson 1963}

(i) α(Fp1e1Fp2e2⋯Fpkek)=α([Fp1e1,Fp2e2,...,Fpkek])=[α(Fp1e1),α(Fp2e2),...,α(Fpkek)]=[p1e1,p2e2,...,pkek]=(p1e1p2e2⋯pkek)=n, for n≠p1e1,n=p1e1p2e2⋯pkek, where 2∦n, or 2∤n.(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) α(F2Fp2e2Fp3e3⋯Fpkek)=α([F2,Fp2e2,Fp3e3,...,Fpkek])=[α(F2),α(Fp2e2),α(Fp3e3),...,α(Fpkek)]=[1,p2e2,p3e3,...,pkek]=(p2e2p3e3⋯pkek)=n2, for n≠p1e1,n≠p1e1p2e2,n=p1e1p2e2p3e3⋯pkek, where 2∥n.(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) α(F2Fp2e2)=α([F2,Fp2e2])=[α(F2),α(Fp2e2)]=[1,p2e2]=(p2e2)=n2, for n=2p2e2, where 2∥n.(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) α(Fp1e1)=α([Fp1e1])=[α(Fp1e1)]=[p1e1]=(p1e1)=n, for n=p1e1, where 2∦n, or 2∤n.(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{.}

If pλ∥Fα(p), then Fpλ∣FFα(p), for λ≥2. According to the 4 proof points above, either \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 }

α(Fp1e1Fp2e2⋯Fpkek)=α(Fp1e1Fp2e2⋯Fpk)=(p1e1p2e2⋯pkek)=(p1e1p2e2⋯pk), for pk=p,λ≥ek≥2, or \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 }

α(Fp1e1Fp2e2⋯Fpkek)=α(Fp1Fp2e2⋯Fpkek)=(p1e1p2e2⋯pkek)=(p1p2e2⋯pkek), for p1=p,λ≥e1≥2, 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 }

α(Fp1e1⋯Fpiei⋯Fpkek)=α(Fp1e1⋯Fpi⋯Fpkek)=(p1e1⋯piei⋯pkek)=(p1e1⋯pi⋯pkek), for pi=p,λ≥ei≥2, 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 }

α(F2Fp2e2Fp3e3⋯Fpkek)=α(F2Fp2e2Fp3e3⋯Fpk)=(p2e2p3e3⋯pkek)=(p2e2p3e3⋯pk), for pk=p,λ≥ek≥2, 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 }

α(F2Fp2e2Fp3e3⋯Fpkek)=α(F2Fp2Fp3e3⋯Fpkek)=(p2e2p3e3⋯pkek)=(p2p3e3⋯pkek), for p2=p,λ≥e2≥2, 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 }

α(F2Fp2e2⋯Fpiei⋯Fpkek)=α(F2Fp2e2⋯Fpi⋯Fpkek)=(p2e2⋯piei⋯pkek)=(p2e2⋯pi⋯pkek), for pi=p,λ≥ei≥2, 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 }

α(F2Fp2e2)=α(F2Fp2)=(p2e2)=(p2), for p2=p,λ≥e2≥2, 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 }

α(Fp1e1)=α(Fp1)=(p1e1)=(p1), for p1=p,λ≥e1≥2, which is absurd.\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?

#NumberTheory

Note by Shane Findley
3 years, 3 months ago

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