Finding a better approximation to a prime number relation
FINAL EDIT AND SUMMARY.
The basis of this problem, and that which allows for the approximations to be made here, can be summarised in one approximation:
$$Biggl(frac{n^k -{lfloor n^{frac{1}{k}} rfloor}^{k-1}gcd({lfloor n^{frac{1}{k}} rfloor}^{k-1},Bigllfloor frac{p_n^{k-1}}{n^{k-1}} Bigrrfloor)}{n^k -{lfloor n^{frac{1}{k}} rfloor}gcd({lfloor n^{frac{1}{k}} rfloor},Bigllfloor frac{p_n^{k}}{n^{k}} Bigrrfloor)}Biggr)^{frac{1}{k}}
approx 1quadforall n,k in mathbb Nbackslash {{1}}$$
$$quadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquad(A0)$$
$$frac{Bigllfloor bigl(n^k -{lfloor n^{frac{1}{k}} rfloor}^{k-1}gcd({lfloor n^{frac{1}{k}} rfloor}^{k-1},Bigllfloor frac{p_n^{k-1}}{n^{k-1}} Bigrrfloor)bigr)^{frac{1}{k}}Bigrrfloor
}{Bigllfloor bigl(n^k -{lfloor n^{frac{1}{k}} rfloor}gcd({lfloor n^{frac{1}{k}} rfloor},Bigllfloor frac{p_n^{k}}{n^{k}} Bigrrfloor)bigr)^{frac{1}{k}}Bigrrfloor} =1quadforall n,k in mathbb Nbackslash {{1}}$$
$$quadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquad(A1)$$
$${gcdBigl({lfloor n^{frac{1}{k}} rfloor},Bigllfloor frac{p_n^k}{n^k} BigrrfloorBigr)}quad Biggl|quad gcdBigl({lfloor n^{frac{1}{k}} rfloor}^{k-1},Bigllfloor frac{p_n^{k-1}}{n^{k-1}} BigrrfloorBigr) $$
$$quadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquad(A2)$$
Defining the above ratio as varsigma:
$$varsigma_{n,k}= frac{{{lfloor n^{frac{1}{k}} rfloor}^{k-1}gcdBigl({lfloor n^{frac{1}{k}} rfloor}^{k-1},Bigllfloor frac{p_n^{k-1}}{n^{k-1}} BigrrfloorBigr)}
}{{gcdBigl({lfloor n^{frac{1}{k}} rfloor},Bigllfloor frac{p_n^k}{n^k} BigrrfloorBigr)}}$$
We have the following:
$n lt 2^k Rightarrow varsigma_{n,k}=1$
$$quadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquad(A3)$$
$varsigma_{n,k}$ is a perfect power $forall n,k in mathbb N, backslash ,{{1}}$
$$quadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquad(A4)$$
To generalize:
$$varsigma_{n,k,i,j}= frac{{{lfloor n^{frac{1}{k}} rfloor}^{j-1}gcdBigl({lfloor n^{frac{1}{k}} rfloor}^{j},Bigllfloor frac{p_n^{,,j}}{n^{,j}} BigrrfloorBigr)}
}{{gcdBigl({lfloor n^{frac{1}{k}} rfloor^{,i}},Bigllfloor frac{p_n^{,i}}{n^{,i}} BigrrfloorBigr)}}$$
provided that $j geq i$,we have:
$m^k leq n lt (m+1)^k Rightarrow varsigma_{n,k,i,j} in {{m^N:N in {{1,2,3,...,k}}}}$
$$quadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquad(A5)$$
reducing the above by setting $i=j$:
$$varsigma_{n,k,j}= lfloor n^{frac{1}{k}} rfloor^{j-1}$$
$m^k leq n lt (m+1)^k Rightarrow varsigma_{n,k,j}=m^{j-1}$
$$quadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquad(A6)$$
The maximum value of the integer remainder of the division of $n^k+m$ by $gcd(Bigllfloor frac{p_n^{, a}}{n^{,b}} Bigrrfloor,n^c)$ is equal to $m$, when $a gt 1$, $b gt 1$ and $1 leq c leq k$.
This stated in inequalities:
$$a gt 1land b gt 1 land 1 leq c leq k Rightarrow -m leq n^k-Biggllfloorfrac{n^k+m}{gcd(Bigllfloor frac{p_n^{, a}}{n^{,b}} Bigrrfloor,n^c)}BiggrrfloorgcdBiggl(Bigllfloor frac{p_n^{, a}}{n^{,b}} Bigrrfloor,n^cBiggr) leq 0$$
$$quadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquad(A7)$$
So the question I am now asking, for that fun person that wants to close this page, is how do I establish a proof for (A5) and (A7) that will be rigorous and indisputable?
note that for now, I will leave the lemma as the restriction
$${{a,b}} subset mathbb N land c in {{1,2,3,...,k}} Rightarrow n^k+m-Biggllfloorfrac{n^k+m}{gcd(Bigllfloor frac{p_n^{, a}}{n^{,b}} Bigrrfloor,n^c)}BiggrrfloorgcdBiggl(Bigllfloor frac{p_n^{, a}}{n^{,b}} Bigrrfloor,n^cBiggr) in {{0,1,2,...,m}}$$
But there most definitely exists congruence relations that have dependence in $(n,k)$ that allow us to reduce this condition to a specific subset of the least residue system modulo $m+1$ stated on the righthand side of the implicative arrow.
Yesterday I noticed quite a strong fit for the approximation:
$$vartheta _{{n}}=minBiggl(mathcal DBigl(ncdotBigllfloor frac{p_n}{n} BigrrfloorBigr) backslash {{1}}Biggr)$$
$$n-gcd(bigllfloor sqrt {n} bigrrfloor ,vartheta _{{n}})) approx A cdot (n-1) +B$$
where $A approx 1$ and $B approx -1/2$ and $mathcal D(n)$ denote the set of all divisors of $n$, $p_n$ is the $n^{th}$ prime.
$$sqrt{bigl( n^{2}-bigllfloor sqrt {n} bigrrfloorcdotgcd(bigllfloor sqrt {n} bigrrfloor ,vartheta _{{n}})bigr)}sim n $$
$$quadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquad(R1)$$
$$sqrt{bigl( n-gcd(n ,vartheta _{{n}})bigr)}+frac{1}{sqrt{n}}sim sqrt{n}$$
$$quadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquad(R2)$$
$$sqrt{{n}^{2}-max left( lfloor sqrt{n} rfloor ,n
right) min left( gcd left( lfloor sqrt{n} rfloor,vartheta_n right) ,gcd left( n,vartheta_n right) right)
}+1+delta_{{n}}sim n$$
Where $delta_n in {{-frac{1}{2},0,frac{1}{2}}}$ is a discrete function for which I am unable to determine as yet.
$$quadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquad(R3)$$
So I guess the best idea now would be for me to find either a value on $mathbb N$ that satisfies neither of the following equalities:
$$n- Biggl(Bigllfloorsqrt {{n}^{2}- lfloor sqrt{n}
rfloor cdot gcd left( lfloor sqrt{n} rfloor ,vartheta_n right) } Bigrrfloor+1Biggr) = 0 $$
$$quadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquad(R4)$$
$$n-Biggl(Bigllfloor sqrt{{n}^{2}-min left( lfloor
sqrt{n} rfloor ,n right)cdotminleft(gcd ( lfloor sqrt{n}rfloor,vartheta_n) ,gcd ( n,vartheta_n)
right) }Bigrrfloor +1Biggr) = 0$$
$$quadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquad(R5)$$
Figure 1:
Figure 2:
Defining a generalisation of vartheta:
$$vartheta _{{n,k}}=minBiggl(mathcal DBigl(n^{k}cdotBigllfloor frac{p_n^{k}}{n^{k}} BigrrfloorBigr), backslash, {{1}}Biggr)$$
$$quadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquad(R6)$$
Will allow for the following asymptotic relation as we would intuitively expect from the nature of the generalisation and the nature of $(R2)$:
$$(n^k -lfloor n^{frac{1}{k}} rfloorgcd(lfloor n^{frac{1}{k}} rfloor,vartheta _{{n,k-1}}))^{frac{1}{k}} sim n quad forall k in mathbb Nbackslash {{1}}$$
$$quadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquad(R7)$$
Which is based on an apparent equality:
$$lfloor (n^k -lfloor n^{frac{1}{k}} rfloorgcd(lfloor n^{frac{1}{k}} rfloor,vartheta _{{n,k-1}}))^{frac{1}{k}} rfloor+1=nquad forall k in mathbb Nbackslash {{1}} $$
$$quadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquad(R8)$$
$$lfloor (n^k -gcd(lfloor n^{frac{1}{k}} rfloor,Bigllfloor frac{p_n^{k}}{n^{k}} Bigrrfloor))^{frac{1}{k}} rfloor+1=nquad forall k in mathbb Nbackslash {{1}} $$
$$quadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquad(R9)$$
$$lfloor (n^k -{lfloor n^{frac{1}{k}} rfloor}^{k-1}gcd({lfloor n^{frac{1}{k}} rfloor}^{k-1},Bigllfloor frac{p_n^{k-1}}{n^{k-1}} Bigrrfloor))^{frac{1}{k}} rfloor+1=nquad forall k in mathbb Nbackslash {{1}} $$
$$quadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquad(R10)$$
$$lim _{krightarrow infty }((n^k -lfloor n^{frac{1}{k}} rfloorgcd(lfloor n^{frac{1}{k}} rfloor,vartheta _{{n,k-1}}))^{frac{1}{k}} )=n$$
$$quadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquad(R11)$$
$$quadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquad(R12)$$
Figure 5
prime-numbers
asked Jun 10 at 8:21
Adam
53414
|
show 26 more comments
FINAL EDIT AND SUMMARY.
The basis of this problem, and that which allows for the approximations to be made here, can be summarised in one approximation:
$$Biggl(frac{n^k -{lfloor n^{frac{1}{k}} rfloor}^{k-1}gcd({lfloor n^{frac{1}{k}} rfloor}^{k-1},Bigllfloor frac{p_n^{k-1}}{n^{k-1}} Bigrrfloor)}{n^k -{lfloor n^{frac{1}{k}} rfloor}gcd({lfloor n^{frac{1}{k}} rfloor},Bigllfloor frac{p_n^{k}}{n^{k}} Bigrrfloor)}Biggr)^{frac{1}{k}}
approx 1quadforall n,k in mathbb Nbackslash {{1}}$$
$$quadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquad(A0)$$
$$frac{Bigllfloor bigl(n^k -{lfloor n^{frac{1}{k}} rfloor}^{k-1}gcd({lfloor n^{frac{1}{k}} rfloor}^{k-1},Bigllfloor frac{p_n^{k-1}}{n^{k-1}} Bigrrfloor)bigr)^{frac{1}{k}}Bigrrfloor
}{Bigllfloor bigl(n^k -{lfloor n^{frac{1}{k}} rfloor}gcd({lfloor n^{frac{1}{k}} rfloor},Bigllfloor frac{p_n^{k}}{n^{k}} Bigrrfloor)bigr)^{frac{1}{k}}Bigrrfloor} =1quadforall n,k in mathbb Nbackslash {{1}}$$
$$quadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquad(A1)$$
$${gcdBigl({lfloor n^{frac{1}{k}} rfloor},Bigllfloor frac{p_n^k}{n^k} BigrrfloorBigr)}quad Biggl|quad gcdBigl({lfloor n^{frac{1}{k}} rfloor}^{k-1},Bigllfloor frac{p_n^{k-1}}{n^{k-1}} BigrrfloorBigr) $$
$$quadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquad(A2)$$
Defining the above ratio as varsigma:
$$varsigma_{n,k}= frac{{{lfloor n^{frac{1}{k}} rfloor}^{k-1}gcdBigl({lfloor n^{frac{1}{k}} rfloor}^{k-1},Bigllfloor frac{p_n^{k-1}}{n^{k-1}} BigrrfloorBigr)}
}{{gcdBigl({lfloor n^{frac{1}{k}} rfloor},Bigllfloor frac{p_n^k}{n^k} BigrrfloorBigr)}}$$
We have the following:
$n lt 2^k Rightarrow varsigma_{n,k}=1$
$$quadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquad(A3)$$
$varsigma_{n,k}$ is a perfect power $forall n,k in mathbb N, backslash ,{{1}}$
$$quadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquad(A4)$$
To generalize:
$$varsigma_{n,k,i,j}= frac{{{lfloor n^{frac{1}{k}} rfloor}^{j-1}gcdBigl({lfloor n^{frac{1}{k}} rfloor}^{j},Bigllfloor frac{p_n^{,,j}}{n^{,j}} BigrrfloorBigr)}
}{{gcdBigl({lfloor n^{frac{1}{k}} rfloor^{,i}},Bigllfloor frac{p_n^{,i}}{n^{,i}} BigrrfloorBigr)}}$$
provided that $j geq i$,we have:
$m^k leq n lt (m+1)^k Rightarrow varsigma_{n,k,i,j} in {{m^N:N in {{1,2,3,...,k}}}}$
$$quadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquad(A5)$$
reducing the above by setting $i=j$:
$$varsigma_{n,k,j}= lfloor n^{frac{1}{k}} rfloor^{j-1}$$
$m^k leq n lt (m+1)^k Rightarrow varsigma_{n,k,j}=m^{j-1}$
$$quadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquad(A6)$$
The maximum value of the integer remainder of the division of $n^k+m$ by $gcd(Bigllfloor frac{p_n^{, a}}{n^{,b}} Bigrrfloor,n^c)$ is equal to $m$, when $a gt 1$, $b gt 1$ and $1 leq c leq k$.
This stated in inequalities:
$$a gt 1land b gt 1 land 1 leq c leq k Rightarrow -m leq n^k-Biggllfloorfrac{n^k+m}{gcd(Bigllfloor frac{p_n^{, a}}{n^{,b}} Bigrrfloor,n^c)}BiggrrfloorgcdBiggl(Bigllfloor frac{p_n^{, a}}{n^{,b}} Bigrrfloor,n^cBiggr) leq 0$$
$$quadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquad(A7)$$
So the question I am now asking, for that fun person that wants to close this page, is how do I establish a proof for (A5) and (A7) that will be rigorous and indisputable?
note that for now, I will leave the lemma as the restriction
$${{a,b}} subset mathbb N land c in {{1,2,3,...,k}} Rightarrow n^k+m-Biggllfloorfrac{n^k+m}{gcd(Bigllfloor frac{p_n^{, a}}{n^{,b}} Bigrrfloor,n^c)}BiggrrfloorgcdBiggl(Bigllfloor frac{p_n^{, a}}{n^{,b}} Bigrrfloor,n^cBiggr) in {{0,1,2,...,m}}$$
But there most definitely exists congruence relations that have dependence in $(n,k)$ that allow us to reduce this condition to a specific subset of the least residue system modulo $m+1$ stated on the righthand side of the implicative arrow.
Yesterday I noticed quite a strong fit for the approximation:
$$vartheta _{{n}}=minBiggl(mathcal DBigl(ncdotBigllfloor frac{p_n}{n} BigrrfloorBigr) backslash {{1}}Biggr)$$
$$n-gcd(bigllfloor sqrt {n} bigrrfloor ,vartheta _{{n}})) approx A cdot (n-1) +B$$
where $A approx 1$ and $B approx -1/2$ and $mathcal D(n)$ denote the set of all divisors of $n$, $p_n$ is the $n^{th}$ prime.
$$sqrt{bigl( n^{2}-bigllfloor sqrt {n} bigrrfloorcdotgcd(bigllfloor sqrt {n} bigrrfloor ,vartheta _{{n}})bigr)}sim n $$
$$quadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquad(R1)$$
$$sqrt{bigl( n-gcd(n ,vartheta _{{n}})bigr)}+frac{1}{sqrt{n}}sim sqrt{n}$$
$$quadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquad(R2)$$
$$sqrt{{n}^{2}-max left( lfloor sqrt{n} rfloor ,n
right) min left( gcd left( lfloor sqrt{n} rfloor,vartheta_n right) ,gcd left( n,vartheta_n right) right)
}+1+delta_{{n}}sim n$$
Where $delta_n in {{-frac{1}{2},0,frac{1}{2}}}$ is a discrete function for which I am unable to determine as yet.
$$quadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquad(R3)$$
So I guess the best idea now would be for me to find either a value on $mathbb N$ that satisfies neither of the following equalities:
$$n- Biggl(Bigllfloorsqrt {{n}^{2}- lfloor sqrt{n}
rfloor cdot gcd left( lfloor sqrt{n} rfloor ,vartheta_n right) } Bigrrfloor+1Biggr) = 0 $$
$$quadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquad(R4)$$
$$n-Biggl(Bigllfloor sqrt{{n}^{2}-min left( lfloor
sqrt{n} rfloor ,n right)cdotminleft(gcd ( lfloor sqrt{n}rfloor,vartheta_n) ,gcd ( n,vartheta_n)
right) }Bigrrfloor +1Biggr) = 0$$
$$quadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquad(R5)$$
Figure 1:
Figure 2:
Defining a generalisation of vartheta:
$$vartheta _{{n,k}}=minBiggl(mathcal DBigl(n^{k}cdotBigllfloor frac{p_n^{k}}{n^{k}} BigrrfloorBigr), backslash, {{1}}Biggr)$$
$$quadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquad(R6)$$
Will allow for the following asymptotic relation as we would intuitively expect from the nature of the generalisation and the nature of $(R2)$:
$$(n^k -lfloor n^{frac{1}{k}} rfloorgcd(lfloor n^{frac{1}{k}} rfloor,vartheta _{{n,k-1}}))^{frac{1}{k}} sim n quad forall k in mathbb Nbackslash {{1}}$$
$$quadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquad(R7)$$
Which is based on an apparent equality:
$$lfloor (n^k -lfloor n^{frac{1}{k}} rfloorgcd(lfloor n^{frac{1}{k}} rfloor,vartheta _{{n,k-1}}))^{frac{1}{k}} rfloor+1=nquad forall k in mathbb Nbackslash {{1}} $$
$$quadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquad(R8)$$
$$lfloor (n^k -gcd(lfloor n^{frac{1}{k}} rfloor,Bigllfloor frac{p_n^{k}}{n^{k}} Bigrrfloor))^{frac{1}{k}} rfloor+1=nquad forall k in mathbb Nbackslash {{1}} $$
$$quadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquad(R9)$$
$$lfloor (n^k -{lfloor n^{frac{1}{k}} rfloor}^{k-1}gcd({lfloor n^{frac{1}{k}} rfloor}^{k-1},Bigllfloor frac{p_n^{k-1}}{n^{k-1}} Bigrrfloor))^{frac{1}{k}} rfloor+1=nquad forall k in mathbb Nbackslash {{1}} $$
$$quadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquad(R10)$$
$$lim _{krightarrow infty }((n^k -lfloor n^{frac{1}{k}} rfloorgcd(lfloor n^{frac{1}{k}} rfloor,vartheta _{{n,k-1}}))^{frac{1}{k}} )=n$$
$$quadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquad(R11)$$
$$quadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquad(R12)$$
Figure 5
prime-numbers
asked Jun 10 at 8:21
Adam
53414
We can slightly simplify the left side to $$n-gcd(lfloor sqrt{n} rfloor,q)$$ where $q$ is the smallest prime factor of $$ncdot lfloor frac{p_n}{n} rfloor$$
– Peter
Jun 10 at 8:28
Further we can conclude that the left side is either $n-1$ or $n-q$
– Peter
Jun 10 at 8:30
1
en.wikipedia.org/wiki/Prime-counting_function shows, among other things, the (proven) bounds from Pierre Dusart.
– Peter
Jun 10 at 10:21
2
Your goal is to construct a formula that avoids brute force calculation. If not, you do not need the formula and can calculate the values directly. "Predictible" means that a program can get the correct answer much faster and always correct and needs no cumbersome calculations.
– Peter
Jun 10 at 15:52
1
You have two chunks of text, both labeled "final edit and summary". I don't know which final edit and summary is the final final edit and summary. At the end of the top final edit and summary, you ask, "how do I establish a proof that will be rigorous and indisputable?" But it is not clear what you want to prove. There's nothing anywhere labeled "Theorem", or "Conjecture", just a multitude of equations and formulas. I'm very tempted to vote to close as "unclear what you are asking".
– Gerry Myerson
Nov 14 at 3:42
|
show 26 more comments
FINAL EDIT AND SUMMARY.
The basis of this problem, and that which allows for the approximations to be made here, can be summarised in one approximation:
$$Biggl(frac{n^k -{lfloor n^{frac{1}{k}} rfloor}^{k-1}gcd({lfloor n^{frac{1}{k}} rfloor}^{k-1},Bigllfloor frac{p_n^{k-1}}{n^{k-1}} Bigrrfloor)}{n^k -{lfloor n^{frac{1}{k}} rfloor}gcd({lfloor n^{frac{1}{k}} rfloor},Bigllfloor frac{p_n^{k}}{n^{k}} Bigrrfloor)}Biggr)^{frac{1}{k}}
approx 1quadforall n,k in mathbb Nbackslash {{1}}$$
$$quadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquad(A0)$$
$$frac{Bigllfloor bigl(n^k -{lfloor n^{frac{1}{k}} rfloor}^{k-1}gcd({lfloor n^{frac{1}{k}} rfloor}^{k-1},Bigllfloor frac{p_n^{k-1}}{n^{k-1}} Bigrrfloor)bigr)^{frac{1}{k}}Bigrrfloor
}{Bigllfloor bigl(n^k -{lfloor n^{frac{1}{k}} rfloor}gcd({lfloor n^{frac{1}{k}} rfloor},Bigllfloor frac{p_n^{k}}{n^{k}} Bigrrfloor)bigr)^{frac{1}{k}}Bigrrfloor} =1quadforall n,k in mathbb Nbackslash {{1}}$$
$$quadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquad(A1)$$
$${gcdBigl({lfloor n^{frac{1}{k}} rfloor},Bigllfloor frac{p_n^k}{n^k} BigrrfloorBigr)}quad Biggl|quad gcdBigl({lfloor n^{frac{1}{k}} rfloor}^{k-1},Bigllfloor frac{p_n^{k-1}}{n^{k-1}} BigrrfloorBigr) $$
$$quadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquad(A2)$$
Defining the above ratio as varsigma:
$$varsigma_{n,k}= frac{{{lfloor n^{frac{1}{k}} rfloor}^{k-1}gcdBigl({lfloor n^{frac{1}{k}} rfloor}^{k-1},Bigllfloor frac{p_n^{k-1}}{n^{k-1}} BigrrfloorBigr)}
}{{gcdBigl({lfloor n^{frac{1}{k}} rfloor},Bigllfloor frac{p_n^k}{n^k} BigrrfloorBigr)}}$$
We have the following:
$n lt 2^k Rightarrow varsigma_{n,k}=1$
$$quadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquad(A3)$$
$varsigma_{n,k}$ is a perfect power $forall n,k in mathbb N, backslash ,{{1}}$
$$quadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquad(A4)$$
To generalize:
$$varsigma_{n,k,i,j}= frac{{{lfloor n^{frac{1}{k}} rfloor}^{j-1}gcdBigl({lfloor n^{frac{1}{k}} rfloor}^{j},Bigllfloor frac{p_n^{,,j}}{n^{,j}} BigrrfloorBigr)}
}{{gcdBigl({lfloor n^{frac{1}{k}} rfloor^{,i}},Bigllfloor frac{p_n^{,i}}{n^{,i}} BigrrfloorBigr)}}$$
provided that $j geq i$,we have:
$m^k leq n lt (m+1)^k Rightarrow varsigma_{n,k,i,j} in {{m^N:N in {{1,2,3,...,k}}}}$
$$quadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquad(A5)$$
reducing the above by setting $i=j$:
$$varsigma_{n,k,j}= lfloor n^{frac{1}{k}} rfloor^{j-1}$$
$m^k leq n lt (m+1)^k Rightarrow varsigma_{n,k,j}=m^{j-1}$
$$quadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquad(A6)$$
The maximum value of the integer remainder of the division of $n^k+m$ by $gcd(Bigllfloor frac{p_n^{, a}}{n^{,b}} Bigrrfloor,n^c)$ is equal to $m$, when $a gt 1$, $b gt 1$ and $1 leq c leq k$.
This stated in inequalities:
$$a gt 1land b gt 1 land 1 leq c leq k Rightarrow -m leq n^k-Biggllfloorfrac{n^k+m}{gcd(Bigllfloor frac{p_n^{, a}}{n^{,b}} Bigrrfloor,n^c)}BiggrrfloorgcdBiggl(Bigllfloor frac{p_n^{, a}}{n^{,b}} Bigrrfloor,n^cBiggr) leq 0$$
$$quadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquad(A7)$$
So the question I am now asking, for that fun person that wants to close this page, is how do I establish a proof for (A5) and (A7) that will be rigorous and indisputable?
note that for now, I will leave the lemma as the restriction
$${{a,b}} subset mathbb N land c in {{1,2,3,...,k}} Rightarrow n^k+m-Biggllfloorfrac{n^k+m}{gcd(Bigllfloor frac{p_n^{, a}}{n^{,b}} Bigrrfloor,n^c)}BiggrrfloorgcdBiggl(Bigllfloor frac{p_n^{, a}}{n^{,b}} Bigrrfloor,n^cBiggr) in {{0,1,2,...,m}}$$
But there most definitely exists congruence relations that have dependence in $(n,k)$ that allow us to reduce this condition to a specific subset of the least residue system modulo $m+1$ stated on the righthand side of the implicative arrow.
Yesterday I noticed quite a strong fit for the approximation:
$$vartheta _{{n}}=minBiggl(mathcal DBigl(ncdotBigllfloor frac{p_n}{n} BigrrfloorBigr) backslash {{1}}Biggr)$$
$$n-gcd(bigllfloor sqrt {n} bigrrfloor ,vartheta _{{n}})) approx A cdot (n-1) +B$$
where $A approx 1$ and $B approx -1/2$ and $mathcal D(n)$ denote the set of all divisors of $n$, $p_n$ is the $n^{th}$ prime.
$$sqrt{bigl( n^{2}-bigllfloor sqrt {n} bigrrfloorcdotgcd(bigllfloor sqrt {n} bigrrfloor ,vartheta _{{n}})bigr)}sim n $$
$$quadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquad(R1)$$
$$sqrt{bigl( n-gcd(n ,vartheta _{{n}})bigr)}+frac{1}{sqrt{n}}sim sqrt{n}$$
$$quadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquad(R2)$$
$$sqrt{{n}^{2}-max left( lfloor sqrt{n} rfloor ,n
right) min left( gcd left( lfloor sqrt{n} rfloor,vartheta_n right) ,gcd left( n,vartheta_n right) right)
}+1+delta_{{n}}sim n$$
Where $delta_n in {{-frac{1}{2},0,frac{1}{2}}}$ is a discrete function for which I am unable to determine as yet.
$$quadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquad(R3)$$
So I guess the best idea now would be for me to find either a value on $mathbb N$ that satisfies neither of the following equalities:
$$n- Biggl(Bigllfloorsqrt {{n}^{2}- lfloor sqrt{n}
rfloor cdot gcd left( lfloor sqrt{n} rfloor ,vartheta_n right) } Bigrrfloor+1Biggr) = 0 $$
$$quadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquad(R4)$$
$$n-Biggl(Bigllfloor sqrt{{n}^{2}-min left( lfloor
sqrt{n} rfloor ,n right)cdotminleft(gcd ( lfloor sqrt{n}rfloor,vartheta_n) ,gcd ( n,vartheta_n)
right) }Bigrrfloor +1Biggr) = 0$$
$$quadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquad(R5)$$
Figure 1:
Figure 2:
Defining a generalisation of vartheta:
$$vartheta _{{n,k}}=minBiggl(mathcal DBigl(n^{k}cdotBigllfloor frac{p_n^{k}}{n^{k}} BigrrfloorBigr), backslash, {{1}}Biggr)$$
$$quadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquad(R6)$$
Will allow for the following asymptotic relation as we would intuitively expect from the nature of the generalisation and the nature of $(R2)$:
$$(n^k -lfloor n^{frac{1}{k}} rfloorgcd(lfloor n^{frac{1}{k}} rfloor,vartheta _{{n,k-1}}))^{frac{1}{k}} sim n quad forall k in mathbb Nbackslash {{1}}$$
$$quadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquad(R7)$$
Which is based on an apparent equality:
$$lfloor (n^k -lfloor n^{frac{1}{k}} rfloorgcd(lfloor n^{frac{1}{k}} rfloor,vartheta _{{n,k-1}}))^{frac{1}{k}} rfloor+1=nquad forall k in mathbb Nbackslash {{1}} $$
$$quadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquad(R8)$$
$$lfloor (n^k -gcd(lfloor n^{frac{1}{k}} rfloor,Bigllfloor frac{p_n^{k}}{n^{k}} Bigrrfloor))^{frac{1}{k}} rfloor+1=nquad forall k in mathbb Nbackslash {{1}} $$
$$quadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquad(R9)$$
$$lfloor (n^k -{lfloor n^{frac{1}{k}} rfloor}^{k-1}gcd({lfloor n^{frac{1}{k}} rfloor}^{k-1},Bigllfloor frac{p_n^{k-1}}{n^{k-1}} Bigrrfloor))^{frac{1}{k}} rfloor+1=nquad forall k in mathbb Nbackslash {{1}} $$
$$quadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquad(R10)$$
$$lim _{krightarrow infty }((n^k -lfloor n^{frac{1}{k}} rfloorgcd(lfloor n^{frac{1}{k}} rfloor,vartheta _{{n,k-1}}))^{frac{1}{k}} )=n$$
$$quadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquad(R11)$$
$$quadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquad(R12)$$
Figure 5
prime-numbers
asked Jun 10 at 8:21
Adam
53414
FINAL EDIT AND SUMMARY.
The basis of this problem, and that which allows for the approximations to be made here, can be summarised in one approximation:
$$Biggl(frac{n^k -{lfloor n^{frac{1}{k}} rfloor}^{k-1}gcd({lfloor n^{frac{1}{k}} rfloor}^{k-1},Bigllfloor frac{p_n^{k-1}}{n^{k-1}} Bigrrfloor)}{n^k -{lfloor n^{frac{1}{k}} rfloor}gcd({lfloor n^{frac{1}{k}} rfloor},Bigllfloor frac{p_n^{k}}{n^{k}} Bigrrfloor)}Biggr)^{frac{1}{k}}
approx 1quadforall n,k in mathbb Nbackslash {{1}}$$
$$quadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquad(A0)$$
$$frac{Bigllfloor bigl(n^k -{lfloor n^{frac{1}{k}} rfloor}^{k-1}gcd({lfloor n^{frac{1}{k}} rfloor}^{k-1},Bigllfloor frac{p_n^{k-1}}{n^{k-1}} Bigrrfloor)bigr)^{frac{1}{k}}Bigrrfloor
}{Bigllfloor bigl(n^k -{lfloor n^{frac{1}{k}} rfloor}gcd({lfloor n^{frac{1}{k}} rfloor},Bigllfloor frac{p_n^{k}}{n^{k}} Bigrrfloor)bigr)^{frac{1}{k}}Bigrrfloor} =1quadforall n,k in mathbb Nbackslash {{1}}$$
$$quadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquad(A1)$$
$${gcdBigl({lfloor n^{frac{1}{k}} rfloor},Bigllfloor frac{p_n^k}{n^k} BigrrfloorBigr)}quad Biggl|quad gcdBigl({lfloor n^{frac{1}{k}} rfloor}^{k-1},Bigllfloor frac{p_n^{k-1}}{n^{k-1}} BigrrfloorBigr) $$
$$quadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquad(A2)$$
Defining the above ratio as varsigma:
$$varsigma_{n,k}= frac{{{lfloor n^{frac{1}{k}} rfloor}^{k-1}gcdBigl({lfloor n^{frac{1}{k}} rfloor}^{k-1},Bigllfloor frac{p_n^{k-1}}{n^{k-1}} BigrrfloorBigr)}
}{{gcdBigl({lfloor n^{frac{1}{k}} rfloor},Bigllfloor frac{p_n^k}{n^k} BigrrfloorBigr)}}$$
We have the following:
$n lt 2^k Rightarrow varsigma_{n,k}=1$
$$quadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquad(A3)$$
$varsigma_{n,k}$ is a perfect power $forall n,k in mathbb N, backslash ,{{1}}$
$$quadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquad(A4)$$
To generalize:
$$varsigma_{n,k,i,j}= frac{{{lfloor n^{frac{1}{k}} rfloor}^{j-1}gcdBigl({lfloor n^{frac{1}{k}} rfloor}^{j},Bigllfloor frac{p_n^{,,j}}{n^{,j}} BigrrfloorBigr)}
}{{gcdBigl({lfloor n^{frac{1}{k}} rfloor^{,i}},Bigllfloor frac{p_n^{,i}}{n^{,i}} BigrrfloorBigr)}}$$
provided that $j geq i$,we have:
$m^k leq n lt (m+1)^k Rightarrow varsigma_{n,k,i,j} in {{m^N:N in {{1,2,3,...,k}}}}$
$$quadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquad(A5)$$
reducing the above by setting $i=j$:
$$varsigma_{n,k,j}= lfloor n^{frac{1}{k}} rfloor^{j-1}$$
$m^k leq n lt (m+1)^k Rightarrow varsigma_{n,k,j}=m^{j-1}$
$$quadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquad(A6)$$
The maximum value of the integer remainder of the division of $n^k+m$ by $gcd(Bigllfloor frac{p_n^{, a}}{n^{,b}} Bigrrfloor,n^c)$ is equal to $m$, when $a gt 1$, $b gt 1$ and $1 leq c leq k$.
This stated in inequalities:
$$a gt 1land b gt 1 land 1 leq c leq k Rightarrow -m leq n^k-Biggllfloorfrac{n^k+m}{gcd(Bigllfloor frac{p_n^{, a}}{n^{,b}} Bigrrfloor,n^c)}BiggrrfloorgcdBiggl(Bigllfloor frac{p_n^{, a}}{n^{,b}} Bigrrfloor,n^cBiggr) leq 0$$
$$quadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquad(A7)$$
So the question I am now asking, for that fun person that wants to close this page, is how do I establish a proof for (A5) and (A7) that will be rigorous and indisputable?
note that for now, I will leave the lemma as the restriction
$${{a,b}} subset mathbb N land c in {{1,2,3,...,k}} Rightarrow n^k+m-Biggllfloorfrac{n^k+m}{gcd(Bigllfloor frac{p_n^{, a}}{n^{,b}} Bigrrfloor,n^c)}BiggrrfloorgcdBiggl(Bigllfloor frac{p_n^{, a}}{n^{,b}} Bigrrfloor,n^cBiggr) in {{0,1,2,...,m}}$$
But there most definitely exists congruence relations that have dependence in $(n,k)$ that allow us to reduce this condition to a specific subset of the least residue system modulo $m+1$ stated on the righthand side of the implicative arrow.
Yesterday I noticed quite a strong fit for the approximation:
$$vartheta _{{n}}=minBiggl(mathcal DBigl(ncdotBigllfloor frac{p_n}{n} BigrrfloorBigr) backslash {{1}}Biggr)$$
$$n-gcd(bigllfloor sqrt {n} bigrrfloor ,vartheta _{{n}})) approx A cdot (n-1) +B$$
where $A approx 1$ and $B approx -1/2$ and $mathcal D(n)$ denote the set of all divisors of $n$, $p_n$ is the $n^{th}$ prime.
$$sqrt{bigl( n^{2}-bigllfloor sqrt {n} bigrrfloorcdotgcd(bigllfloor sqrt {n} bigrrfloor ,vartheta _{{n}})bigr)}sim n $$
$$quadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquad(R1)$$
$$sqrt{bigl( n-gcd(n ,vartheta _{{n}})bigr)}+frac{1}{sqrt{n}}sim sqrt{n}$$
$$quadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquad(R2)$$
$$sqrt{{n}^{2}-max left( lfloor sqrt{n} rfloor ,n
right) min left( gcd left( lfloor sqrt{n} rfloor,vartheta_n right) ,gcd left( n,vartheta_n right) right)
}+1+delta_{{n}}sim n$$
Where $delta_n in {{-frac{1}{2},0,frac{1}{2}}}$ is a discrete function for which I am unable to determine as yet.
$$quadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquad(R3)$$
So I guess the best idea now would be for me to find either a value on $mathbb N$ that satisfies neither of the following equalities:
$$n- Biggl(Bigllfloorsqrt {{n}^{2}- lfloor sqrt{n}
rfloor cdot gcd left( lfloor sqrt{n} rfloor ,vartheta_n right) } Bigrrfloor+1Biggr) = 0 $$
$$quadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquad(R4)$$
$$n-Biggl(Bigllfloor sqrt{{n}^{2}-min left( lfloor
sqrt{n} rfloor ,n right)cdotminleft(gcd ( lfloor sqrt{n}rfloor,vartheta_n) ,gcd ( n,vartheta_n)
right) }Bigrrfloor +1Biggr) = 0$$
$$quadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquad(R5)$$
Figure 1:
Figure 2:
Defining a generalisation of vartheta:
$$vartheta _{{n,k}}=minBiggl(mathcal DBigl(n^{k}cdotBigllfloor frac{p_n^{k}}{n^{k}} BigrrfloorBigr), backslash, {{1}}Biggr)$$
$$quadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquad(R6)$$
Will allow for the following asymptotic relation as we would intuitively expect from the nature of the generalisation and the nature of $(R2)$:
$$(n^k -lfloor n^{frac{1}{k}} rfloorgcd(lfloor n^{frac{1}{k}} rfloor,vartheta _{{n,k-1}}))^{frac{1}{k}} sim n quad forall k in mathbb Nbackslash {{1}}$$
$$quadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquad(R7)$$
Which is based on an apparent equality:
$$lfloor (n^k -lfloor n^{frac{1}{k}} rfloorgcd(lfloor n^{frac{1}{k}} rfloor,vartheta _{{n,k-1}}))^{frac{1}{k}} rfloor+1=nquad forall k in mathbb Nbackslash {{1}} $$
$$quadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquad(R8)$$
$$lfloor (n^k -gcd(lfloor n^{frac{1}{k}} rfloor,Bigllfloor frac{p_n^{k}}{n^{k}} Bigrrfloor))^{frac{1}{k}} rfloor+1=nquad forall k in mathbb Nbackslash {{1}} $$
$$quadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquad(R9)$$
$$lfloor (n^k -{lfloor n^{frac{1}{k}} rfloor}^{k-1}gcd({lfloor n^{frac{1}{k}} rfloor}^{k-1},Bigllfloor frac{p_n^{k-1}}{n^{k-1}} Bigrrfloor))^{frac{1}{k}} rfloor+1=nquad forall k in mathbb Nbackslash {{1}} $$
$$quadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquad(R10)$$
$$lim _{krightarrow infty }((n^k -lfloor n^{frac{1}{k}} rfloorgcd(lfloor n^{frac{1}{k}} rfloor,vartheta _{{n,k-1}}))^{frac{1}{k}} )=n$$
$$quadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquad(R11)$$
$$quadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquad(R12)$$
Figure 5
prime-numbers
prime-numbers
asked Jun 10 at 8:21
Adam
53414
asked Jun 10 at 8:21
Adam
53414
edited Dec 3 at 19:09
asked Jun 10 at 8:21
Adam
53414
asked Jun 10 at 8:21
Adam
53414
asked Jun 10 at 8:21
Adam
53414
53414
We can slightly simplify the left side to $$n-gcd(lfloor sqrt{n} rfloor,q)$$ where $q$ is the smallest prime factor of $$ncdot lfloor frac{p_n}{n} rfloor$$
– Peter
Jun 10 at 8:28
Further we can conclude that the left side is either $n-1$ or $n-q$
– Peter
Jun 10 at 8:30
1
en.wikipedia.org/wiki/Prime-counting_function shows, among other things, the (proven) bounds from Pierre Dusart.
– Peter
Jun 10 at 10:21
2
Your goal is to construct a formula that avoids brute force calculation. If not, you do not need the formula and can calculate the values directly. "Predictible" means that a program can get the correct answer much faster and always correct and needs no cumbersome calculations.
– Peter
Jun 10 at 15:52
1
You have two chunks of text, both labeled "final edit and summary". I don't know which final edit and summary is the final final edit and summary. At the end of the top final edit and summary, you ask, "how do I establish a proof that will be rigorous and indisputable?" But it is not clear what you want to prove. There's nothing anywhere labeled "Theorem", or "Conjecture", just a multitude of equations and formulas. I'm very tempted to vote to close as "unclear what you are asking".
– Gerry Myerson
Nov 14 at 3:42
|
show 26 more comments
We can slightly simplify the left side to $$n-gcd(lfloor sqrt{n} rfloor,q)$$ where $q$ is the smallest prime factor of $$ncdot lfloor frac{p_n}{n} rfloor$$
– Peter
Jun 10 at 8:28
Further we can conclude that the left side is either $n-1$ or $n-q$
– Peter
Jun 10 at 8:30
1
en.wikipedia.org/wiki/Prime-counting_function shows, among other things, the (proven) bounds from Pierre Dusart.
– Peter
Jun 10 at 10:21
2
Your goal is to construct a formula that avoids brute force calculation. If not, you do not need the formula and can calculate the values directly. "Predictible" means that a program can get the correct answer much faster and always correct and needs no cumbersome calculations.
– Peter
Jun 10 at 15:52
1
You have two chunks of text, both labeled "final edit and summary". I don't know which final edit and summary is the final final edit and summary. At the end of the top final edit and summary, you ask, "how do I establish a proof that will be rigorous and indisputable?" But it is not clear what you want to prove. There's nothing anywhere labeled "Theorem", or "Conjecture", just a multitude of equations and formulas. I'm very tempted to vote to close as "unclear what you are asking".
– Gerry Myerson
Nov 14 at 3:42
We can slightly simplify the left side to $$n-gcd(lfloor sqrt{n} rfloor,q)$$ where $q$ is the smallest prime factor of $$ncdot lfloor frac{p_n}{n} rfloor$$
– Peter
Jun 10 at 8:28
We can slightly simplify the left side to $$n-gcd(lfloor sqrt{n} rfloor,q)$$ where $q$ is the smallest prime factor of $$ncdot lfloor frac{p_n}{n} rfloor$$
– Peter
Jun 10 at 8:28
Further we can conclude that the left side is either $n-1$ or $n-q$
– Peter
Jun 10 at 8:30
Further we can conclude that the left side is either $n-1$ or $n-q$
– Peter
Jun 10 at 8:30
1
1
en.wikipedia.org/wiki/Prime-counting_function shows, among other things, the (proven) bounds from Pierre Dusart.
– Peter
Jun 10 at 10:21
en.wikipedia.org/wiki/Prime-counting_function shows, among other things, the (proven) bounds from Pierre Dusart.
– Peter
Jun 10 at 10:21
2
2
Your goal is to construct a formula that avoids brute force calculation. If not, you do not need the formula and can calculate the values directly. "Predictible" means that a program can get the correct answer much faster and always correct and needs no cumbersome calculations.
– Peter
Jun 10 at 15:52
Your goal is to construct a formula that avoids brute force calculation. If not, you do not need the formula and can calculate the values directly. "Predictible" means that a program can get the correct answer much faster and always correct and needs no cumbersome calculations.
– Peter
Jun 10 at 15:52
1
1
You have two chunks of text, both labeled "final edit and summary". I don't know which final edit and summary is the final final edit and summary. At the end of the top final edit and summary, you ask, "how do I establish a proof that will be rigorous and indisputable?" But it is not clear what you want to prove. There's nothing anywhere labeled "Theorem", or "Conjecture", just a multitude of equations and formulas. I'm very tempted to vote to close as "unclear what you are asking".
– Gerry Myerson
Nov 14 at 3:42
You have two chunks of text, both labeled "final edit and summary". I don't know which final edit and summary is the final final edit and summary. At the end of the top final edit and summary, you ask, "how do I establish a proof that will be rigorous and indisputable?" But it is not clear what you want to prove. There's nothing anywhere labeled "Theorem", or "Conjecture", just a multitude of equations and formulas. I'm very tempted to vote to close as "unclear what you are asking".
– Gerry Myerson
Nov 14 at 3:42
|
show 26 more comments
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We can slightly simplify the left side to $$n-gcd(lfloor sqrt{n} rfloor,q)$$ where $q$ is the smallest prime factor of $$ncdot lfloor frac{p_n}{n} rfloor$$
– Peter
Jun 10 at 8:28
Further we can conclude that the left side is either $n-1$ or $n-q$
– Peter
Jun 10 at 8:30
1
en.wikipedia.org/wiki/Prime-counting_function shows, among other things, the (proven) bounds from Pierre Dusart.
– Peter
Jun 10 at 10:21
2
Your goal is to construct a formula that avoids brute force calculation. If not, you do not need the formula and can calculate the values directly. "Predictible" means that a program can get the correct answer much faster and always correct and needs no cumbersome calculations.
– Peter
Jun 10 at 15:52
1
You have two chunks of text, both labeled "final edit and summary". I don't know which final edit and summary is the final final edit and summary. At the end of the top final edit and summary, you ask, "how do I establish a proof that will be rigorous and indisputable?" But it is not clear what you want to prove. There's nothing anywhere labeled "Theorem", or "Conjecture", just a multitude of equations and formulas. I'm very tempted to vote to close as "unclear what you are asking".
– Gerry Myerson
Nov 14 at 3:42