Finding a better approximation to a prime number relation
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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.
ADDITION NEEDED:
I apologize a lot if making another edit here again cause the website to spiral into chaotic unrest somehow, but the following lemma pertaining to the residues of the floor function of natural numbers raised to unit fraction exponents (as we see playing one of the most significant role in the original formula I found and stated in this discussion page), is very significant to the context as previously stated:
$$y in{{j in mathbb N : j=k^m+1 land m inmathbb Nbackslash {{1}} land k inmathbb N }}Rightarrow notexists, n in mathbb N backslash {{1}}:lfloor y^{frac{1}{m}}rfloor^{m}equiv y (operatorname{mod}n)$$
$$quadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquad(operatorname{Residue000})$$
Attempting to simplify the above:
$$exists,\ m in mathbb N backslash {{1}},,: (y-1)^{frac{1}{m}}equiv 0 (operatorname{mod}1)Rightarrow notexists, n in mathbb N backslash {{1}}:lfloor y^{frac{1}{m}}rfloor^{m}equiv y (operatorname{mod}n)$$
$$quadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquad(operatorname{Residue001})$$
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 the following equalities:
$$lfloor (n^k -lfloor n^{frac{1}{k}} rfloorgcd(lfloor n^{frac{1}{k}} rfloor,vartheta _{{n,k-1}}))^{frac{1}{k}} rfloor+1=nquad forall k in mathbb Nbackslash {{1}} $$
$$quadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquad(R8)$$
$$lfloor (n^k -gcd(lfloor n^{frac{1}{k}} rfloor,Bigllfloor frac{p_n^{k}}{n^{k}} Bigrrfloor))^{frac{1}{k}} rfloor+1=nquad forall k in mathbb Nbackslash {{1}} $$
$$quadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquad(R9)$$
$$lfloor (n^k -{lfloor n^{frac{1}{k}} rfloor}^{k-1}gcd({lfloor n^{frac{1}{k}} rfloor}^{k-1},Bigllfloor frac{p_n^{k-1}}{n^{k-1}} Bigrrfloor))^{frac{1}{k}} rfloor+1=nquad forall k in mathbb Nbackslash {{1}} $$
$$quadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquad(R10)$$
$$n^k-Biggllfloor frac{n^k}{gcdBigl(Bigllfloor frac{p_n^{k}}{n^k}Bigrrfloor,n^kBigr)} Biggrrfloor gcdBigl(Bigllfloorfrac{p_n^{k}}{n^k}Bigrrfloor,n^kBigr)=0quadquadforall n,k in mathbb N$$
$$quadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquad(R11)$$
prime-numbers
asked Jun 10 '18 at 8:21
AdamAdam
553114
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show 26 more comments
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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.
ADDITION NEEDED:
I apologize a lot if making another edit here again cause the website to spiral into chaotic unrest somehow, but the following lemma pertaining to the residues of the floor function of natural numbers raised to unit fraction exponents (as we see playing one of the most significant role in the original formula I found and stated in this discussion page), is very significant to the context as previously stated:
$$y in{{j in mathbb N : j=k^m+1 land m inmathbb Nbackslash {{1}} land k inmathbb N }}Rightarrow notexists, n in mathbb N backslash {{1}}:lfloor y^{frac{1}{m}}rfloor^{m}equiv y (operatorname{mod}n)$$
$$quadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquad(operatorname{Residue000})$$
Attempting to simplify the above:
$$exists,\ m in mathbb N backslash {{1}},,: (y-1)^{frac{1}{m}}equiv 0 (operatorname{mod}1)Rightarrow notexists, n in mathbb N backslash {{1}}:lfloor y^{frac{1}{m}}rfloor^{m}equiv y (operatorname{mod}n)$$
$$quadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquad(operatorname{Residue001})$$
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 the following equalities:
$$lfloor (n^k -lfloor n^{frac{1}{k}} rfloorgcd(lfloor n^{frac{1}{k}} rfloor,vartheta _{{n,k-1}}))^{frac{1}{k}} rfloor+1=nquad forall k in mathbb Nbackslash {{1}} $$
$$quadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquad(R8)$$
$$lfloor (n^k -gcd(lfloor n^{frac{1}{k}} rfloor,Bigllfloor frac{p_n^{k}}{n^{k}} Bigrrfloor))^{frac{1}{k}} rfloor+1=nquad forall k in mathbb Nbackslash {{1}} $$
$$quadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquad(R9)$$
$$lfloor (n^k -{lfloor n^{frac{1}{k}} rfloor}^{k-1}gcd({lfloor n^{frac{1}{k}} rfloor}^{k-1},Bigllfloor frac{p_n^{k-1}}{n^{k-1}} Bigrrfloor))^{frac{1}{k}} rfloor+1=nquad forall k in mathbb Nbackslash {{1}} $$
$$quadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquad(R10)$$
$$n^k-Biggllfloor frac{n^k}{gcdBigl(Bigllfloor frac{p_n^{k}}{n^k}Bigrrfloor,n^kBigr)} Biggrrfloor gcdBigl(Bigllfloorfrac{p_n^{k}}{n^k}Bigrrfloor,n^kBigr)=0quadquadforall n,k in mathbb N$$
$$quadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquad(R11)$$
prime-numbers
asked Jun 10 '18 at 8:21
AdamAdam
553114
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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 $$ncdot lfloor frac{p_n}{n} rfloor$$
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– Peter
Jun 10 '18 at 8:28
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Further we can conclude that the left side is either $n-1$ or $n-q$
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– Peter
Jun 10 '18 at 8:30
1
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en.wikipedia.org/wiki/Prime-counting_function shows, among other things, the (proven) bounds from Pierre Dusart.
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– Peter
Jun 10 '18 at 10:21
2
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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.
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– Peter
Jun 10 '18 at 15:52
2
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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".
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– Gerry Myerson
Nov 14 '18 at 3:42
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show 26 more comments
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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.
ADDITION NEEDED:
I apologize a lot if making another edit here again cause the website to spiral into chaotic unrest somehow, but the following lemma pertaining to the residues of the floor function of natural numbers raised to unit fraction exponents (as we see playing one of the most significant role in the original formula I found and stated in this discussion page), is very significant to the context as previously stated:
$$y in{{j in mathbb N : j=k^m+1 land m inmathbb Nbackslash {{1}} land k inmathbb N }}Rightarrow notexists, n in mathbb N backslash {{1}}:lfloor y^{frac{1}{m}}rfloor^{m}equiv y (operatorname{mod}n)$$
$$quadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquad(operatorname{Residue000})$$
Attempting to simplify the above:
$$exists,\ m in mathbb N backslash {{1}},,: (y-1)^{frac{1}{m}}equiv 0 (operatorname{mod}1)Rightarrow notexists, n in mathbb N backslash {{1}}:lfloor y^{frac{1}{m}}rfloor^{m}equiv y (operatorname{mod}n)$$
$$quadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquad(operatorname{Residue001})$$
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 the following equalities:
$$lfloor (n^k -lfloor n^{frac{1}{k}} rfloorgcd(lfloor n^{frac{1}{k}} rfloor,vartheta _{{n,k-1}}))^{frac{1}{k}} rfloor+1=nquad forall k in mathbb Nbackslash {{1}} $$
$$quadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquad(R8)$$
$$lfloor (n^k -gcd(lfloor n^{frac{1}{k}} rfloor,Bigllfloor frac{p_n^{k}}{n^{k}} Bigrrfloor))^{frac{1}{k}} rfloor+1=nquad forall k in mathbb Nbackslash {{1}} $$
$$quadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquad(R9)$$
$$lfloor (n^k -{lfloor n^{frac{1}{k}} rfloor}^{k-1}gcd({lfloor n^{frac{1}{k}} rfloor}^{k-1},Bigllfloor frac{p_n^{k-1}}{n^{k-1}} Bigrrfloor))^{frac{1}{k}} rfloor+1=nquad forall k in mathbb Nbackslash {{1}} $$
$$quadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquad(R10)$$
$$n^k-Biggllfloor frac{n^k}{gcdBigl(Bigllfloor frac{p_n^{k}}{n^k}Bigrrfloor,n^kBigr)} Biggrrfloor gcdBigl(Bigllfloorfrac{p_n^{k}}{n^k}Bigrrfloor,n^kBigr)=0quadquadforall n,k in mathbb N$$
$$quadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquad(R11)$$
prime-numbers
asked Jun 10 '18 at 8:21
AdamAdam
553114
$endgroup$
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.
ADDITION NEEDED:
I apologize a lot if making another edit here again cause the website to spiral into chaotic unrest somehow, but the following lemma pertaining to the residues of the floor function of natural numbers raised to unit fraction exponents (as we see playing one of the most significant role in the original formula I found and stated in this discussion page), is very significant to the context as previously stated:
$$y in{{j in mathbb N : j=k^m+1 land m inmathbb Nbackslash {{1}} land k inmathbb N }}Rightarrow notexists, n in mathbb N backslash {{1}}:lfloor y^{frac{1}{m}}rfloor^{m}equiv y (operatorname{mod}n)$$
$$quadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquad(operatorname{Residue000})$$
Attempting to simplify the above:
$$exists,\ m in mathbb N backslash {{1}},,: (y-1)^{frac{1}{m}}equiv 0 (operatorname{mod}1)Rightarrow notexists, n in mathbb N backslash {{1}}:lfloor y^{frac{1}{m}}rfloor^{m}equiv y (operatorname{mod}n)$$
$$quadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquad(operatorname{Residue001})$$
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 the following equalities:
$$lfloor (n^k -lfloor n^{frac{1}{k}} rfloorgcd(lfloor n^{frac{1}{k}} rfloor,vartheta _{{n,k-1}}))^{frac{1}{k}} rfloor+1=nquad forall k in mathbb Nbackslash {{1}} $$
$$quadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquad(R8)$$
$$lfloor (n^k -gcd(lfloor n^{frac{1}{k}} rfloor,Bigllfloor frac{p_n^{k}}{n^{k}} Bigrrfloor))^{frac{1}{k}} rfloor+1=nquad forall k in mathbb Nbackslash {{1}} $$
$$quadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquad(R9)$$
$$lfloor (n^k -{lfloor n^{frac{1}{k}} rfloor}^{k-1}gcd({lfloor n^{frac{1}{k}} rfloor}^{k-1},Bigllfloor frac{p_n^{k-1}}{n^{k-1}} Bigrrfloor))^{frac{1}{k}} rfloor+1=nquad forall k in mathbb Nbackslash {{1}} $$
$$quadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquad(R10)$$
$$n^k-Biggllfloor frac{n^k}{gcdBigl(Bigllfloor frac{p_n^{k}}{n^k}Bigrrfloor,n^kBigr)} Biggrrfloor gcdBigl(Bigllfloorfrac{p_n^{k}}{n^k}Bigrrfloor,n^kBigr)=0quadquadforall n,k in mathbb N$$
$$quadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquad(R11)$$
prime-numbers
prime-numbers
asked Jun 10 '18 at 8:21
AdamAdam
553114
asked Jun 10 '18 at 8:21
AdamAdam
553114
edited Mar 24 at 8:49
Adam
asked Jun 10 '18 at 8:21
AdamAdam
553114
asked Jun 10 '18 at 8:21
AdamAdam
553114
asked Jun 10 '18 at 8:21
AdamAdam
553114
553114
$begingroup$
We can slightly simplify the left side to $$n-gcd(lfloor sqrt{n} rfloor,q)$$ where $q$ is the smallest prime factor of $$ncdot lfloor frac{p_n}{n} rfloor$$
$endgroup$
– Peter
Jun 10 '18 at 8:28
$begingroup$
Further we can conclude that the left side is either $n-1$ or $n-q$
$endgroup$
– Peter
Jun 10 '18 at 8:30
1
$begingroup$
en.wikipedia.org/wiki/Prime-counting_function shows, among other things, the (proven) bounds from Pierre Dusart.
$endgroup$
– Peter
Jun 10 '18 at 10:21
2
$begingroup$
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.
$endgroup$
– Peter
Jun 10 '18 at 15:52
2
$begingroup$
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".
$endgroup$
– Gerry Myerson
Nov 14 '18 at 3:42
|
show 26 more comments
$begingroup$
We can slightly simplify the left side to $$n-gcd(lfloor sqrt{n} rfloor,q)$$ where $q$ is the smallest prime factor of $$ncdot lfloor frac{p_n}{n} rfloor$$
$endgroup$
– Peter
Jun 10 '18 at 8:28
$begingroup$
Further we can conclude that the left side is either $n-1$ or $n-q$
$endgroup$
– Peter
Jun 10 '18 at 8:30
1
$begingroup$
en.wikipedia.org/wiki/Prime-counting_function shows, among other things, the (proven) bounds from Pierre Dusart.
$endgroup$
– Peter
Jun 10 '18 at 10:21
2
$begingroup$
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.
$endgroup$
– Peter
Jun 10 '18 at 15:52
2
$begingroup$
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".
$endgroup$
– Gerry Myerson
Nov 14 '18 at 3:42
$begingroup$
We can slightly simplify the left side to $$n-gcd(lfloor sqrt{n} rfloor,q)$$ where $q$ is the smallest prime factor of $$ncdot lfloor frac{p_n}{n} rfloor$$
$endgroup$
– Peter
Jun 10 '18 at 8:28
$begingroup$
We can slightly simplify the left side to $$n-gcd(lfloor sqrt{n} rfloor,q)$$ where $q$ is the smallest prime factor of $$ncdot lfloor frac{p_n}{n} rfloor$$
$endgroup$
– Peter
Jun 10 '18 at 8:28
$begingroup$
Further we can conclude that the left side is either $n-1$ or $n-q$
$endgroup$
– Peter
Jun 10 '18 at 8:30
$begingroup$
Further we can conclude that the left side is either $n-1$ or $n-q$
$endgroup$
– Peter
Jun 10 '18 at 8:30
1
1
$begingroup$
en.wikipedia.org/wiki/Prime-counting_function shows, among other things, the (proven) bounds from Pierre Dusart.
$endgroup$
– Peter
Jun 10 '18 at 10:21
$begingroup$
en.wikipedia.org/wiki/Prime-counting_function shows, among other things, the (proven) bounds from Pierre Dusart.
$endgroup$
– Peter
Jun 10 '18 at 10:21
2
2
$begingroup$
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.
$endgroup$
– Peter
Jun 10 '18 at 15:52
$begingroup$
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.
$endgroup$
– Peter
Jun 10 '18 at 15:52
2
2
$begingroup$
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".
$endgroup$
– Gerry Myerson
Nov 14 '18 at 3:42
$begingroup$
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".
$endgroup$
– Gerry Myerson
Nov 14 '18 at 3:42
|
show 26 more comments
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$begingroup$
We can slightly simplify the left side to $$n-gcd(lfloor sqrt{n} rfloor,q)$$ where $q$ is the smallest prime factor of $$ncdot lfloor frac{p_n}{n} rfloor$$
$endgroup$
– Peter
Jun 10 '18 at 8:28
$begingroup$
Further we can conclude that the left side is either $n-1$ or $n-q$
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– Peter
Jun 10 '18 at 8:30
1
$begingroup$
en.wikipedia.org/wiki/Prime-counting_function shows, among other things, the (proven) bounds from Pierre Dusart.
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– Peter
Jun 10 '18 at 10:21
2
$begingroup$
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.
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– Peter
Jun 10 '18 at 15:52
2
$begingroup$
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".
$endgroup$
– Gerry Myerson
Nov 14 '18 at 3:42