Csdrhrt

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:

$$Biggl(frac{n^k -{lfloor n^{frac{1}{k}} rfloor}^{k-1}gcd({lfloor n^{frac{1}{k}} rfloor}^{k-1},Bigllfloor frac{p_n^{k-1}}{n^{k-1}} Bigrrfloor)}{n^k -{lfloor n^{frac{1}{k}} rfloor}gcd({lfloor n^{frac{1}{k}} rfloor},Bigllfloor frac{p_n^{k}}{n^{k}} Bigrrfloor)}Biggr)^{frac{1}{k}}
sim1quadforall n,k in mathbb Nbackslash {{1}}$$

$$quadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquad(A0)$$

$$frac{Bigllfloor bigl(n^k -{lfloor n^{frac{1}{k}} rfloor}^{k-1}gcd({lfloor n^{frac{1}{k}} rfloor}^{k-1},Bigllfloor frac{p_n^{k-1}}{n^{k-1}} Bigrrfloor)bigr)^{frac{1}{k}}Bigrrfloor
}{Bigllfloor bigl(n^k -{lfloor n^{frac{1}{k}} rfloor}gcd({lfloor n^{frac{1}{k}} rfloor},Bigllfloor frac{p_n^{k}}{n^{k}} Bigrrfloor)bigr)^{frac{1}{k}}Bigrrfloor} =1quadforall n,k in mathbb Nbackslash {{1}}$$

$$quadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquad(A1)$$

$${gcdBigl({lfloor n^{frac{1}{k}} rfloor},Bigllfloor frac{p_n^k}{n^k} BigrrfloorBigr)}quad Biggl|quad{{lfloor n^{frac{1}{k}} rfloor}^{k-1}gcdBigl({lfloor n^{frac{1}{k}} rfloor}^{k-1},Bigllfloor frac{p_n^{k-1}}{n^{k-1}} BigrrfloorBigr)}
$$

$$quadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquad(A2)$$

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 $i=j$:
$$varsigma_{n,k,j}= lfloor n^{frac{1}{k}} rfloor^{j-1}$$

$m^k leq n lt (m+1)^k Rightarrow varsigma_{n,k,j}=m^{j-1}$

$$quadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquad(A6)$$

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) that will be rigorous and indisputable?

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