# P-adic Valuation and Chebyshev Function Lemma Proof

all following content subject to prompt and drastic changes required due to author stupidity and or lazyness

I am trying to rigorously prove / disprove the following lemma, (as labelled $$((0.1),(1.1),(1.2))$$)

EDIT: 10/10/2018:

The two primary points of concern remaining are:

1) The complete set of factors for the denominator of the expression inside the fractional part brackets of $$(1.2)$$ and $$(2.2)$$

2) Most importantly, if any of what I have stated here is to be true, there must be a true statement regarding the number of prime numbers that exist between $$\pi(n)+1$$ and $$\pi(n+1)$$ that exists. I haven't been able to work out how to state this, but the assurance it must exist would be a requisite of any of the piece wise evaluations that assert the primality of $$n+1$$. It will probably be so elementary it's embarrassing.
$$\lambda_0(n)=\sum _{i=1}^{\pi (n ) } \sum _{j=1}^{ {\bigl\lfloor\frac {\ln ( n ) }{\ln ( p_{{i}} ) }}\bigr\rfloor +1} \Bigl\lfloor {\frac {n}{{p_{{i}}}^{j}}} \Bigr\rfloor \ln ( p_{{i}}) -\sum _{i=1}^{\pi(n)} \sum _{j=1}^{\pi ( {\lfloor\frac {n}{i}} \rfloor ) } \Bigl\lfloor\frac{ \ln ( {\frac {n}{i}} ) }{ \ln p_{{j}} } \Bigr\rfloor \ln ( p_{{j}})\quad\quad\quad\quad\,\,\,\,\quad\quad\quad\quad\quad(0)$$

$${\Biggl\{(2-\delta(n,1))\frac{n!}{\operatorname{e}^{\lambda_0(n)}}}\Biggr\}=0 \Rightarrow\lambda_0 \left( n \right) =\ln \left( \alpha_0 \left( n \right) \right) \land \left\{ {\frac {n!}{\alpha_0 \left( n \right) }} \right\} =0\quad\quad\quad\quad\,\,\,\,(0.1)$$

$$\delta(x,y)$$ is the Kronecker delta function.

$${\{x}\}$$ is the fractional part of $$x$$.

$$\lambda_1(n)$$ is the sum of the products of the natural logarithms of the $$k^{th}$$ prime numbers and the differences between the $$p_{k}$$-adic valuations of $$((n+1)^2)!)!$$ and $$((n)^2)!)!$$ for $$k$$ from $$\pi(n)+1$$ to $$\pi(n+1)$$ inclusively:

$$\lambda_1(n)=\sum _{k=\pi (n ) +1}^{\pi (n+1 ) }\Biggl(\nu_{{p_{k}}}( ((n+1)^2)! ) -\nu_{{p_{k}}} ( n^2!) \Biggr)\ln(p_{k})\quad\quad\quad\quad\,\quad\quad\quad\quad\,\,(1)$$

$$\lambda_1(n)=0 \operatorname{iff} n+1 \not \in \mathbb P\quad\quad\quad\quad\,\,\,\,\,\,\,\,\quad\quad\quad\,\,\,\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad(1.1)$$

$${\Biggl\{\frac{\lambda_1(n)}{({\prod^{\pi(n)}_{j=1}}p_j)^2\ln(p_{\pi(n+1)})}}\Biggr\}=0\quad\quad\quad\quad\quad\,\,\,\,\,\,\,\,\quad\quad\quad\,\,\,\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad(1.2)$$

$$\lambda_2(n)$$ is the sum of the products of the natural logarithms of the $$k^{th}$$ odd prime numbers and the differences between the $$p_{k}$$-adic valuations of $$(n+1)!!$$ and $$n!!$$ over all $$p_{k+1}$$ for $$k$$ from $$\pi(n)+1$$ to $$\pi(n+1)$$ inclusively:

$$\lambda_2(n)=\sum _{k=\pi (n ) +1}^{\pi (n+1 ) }\Biggl(\nu_{{p_{k}}}( (n+1)! ) -\nu_{{p_{k}}} ( n!) \Biggr)\ln(p_{k})\quad\quad\quad\quad\,\quad\quad\quad\quad\,\,(2)$$

$$\lambda_2(n)=0 \operatorname{iff} n+1 \not \in \mathbb P\quad\quad\quad\quad\,\,\,\,\,\,\,\,\quad\quad\quad\,\,\,\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad(2.1)$$

$${\Biggl\{\frac{\lambda_2(n)}{({\prod^{\pi(n)}_{j=1}}p_j)\ln(p_{\pi(n+1)})}}\Biggr\}=0\quad\quad\quad\quad\quad\,\,\,\,\,\,\,\,\quad\quad\quad\,\,\,\quad\quad\quad\quad\quad\quad\quad\quad(2.2)$$

$$\lambda_3(n)$$ is the sum of the products of the natural logarithms of the $$k^{th}$$ prime numbers $$p_{k}$$ and the differences between the $$p_{k}$$-adic valuations of $$(n+1)^2!$$ and $$n^2!$$ for $$k$$ from $$\pi(n)+1$$ to $$\pi(n+1)$$ inclusively:

$$\lambda_3(n)=\sum _{k=\pi (n ) +1}^{\pi (n+1 ) }\Biggl(\nu_{{p_{k}}}( (n+1)^2 ) -\nu_{{p_{k}}} ( n^2) \Biggr)\ln(p_{k})\quad\quad\quad\quad\,\quad\quad\quad\quad\,\,(3)$$

$$\lambda_3(n)=\cases{3\ln(p_{\pi(n+1)})&\,n+1\in \mathbb P \cr 0& n+1\not\in \mathbb P \cr}$$ $$\quad\quad\quad\,\,\,\,\,\,\,\,\quad\quad\quad\,\,\,\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad(3.1)$$

$$\lambda_4(n)$$ is the sum of the products of the natural logarithms of the $$k^{th}$$ odd prime numbers $$p_{k}$$ and the differences between the $$p_{k}$$-adic valuations of $$(n+1)!$$ and $$n!$$ for $$k$$ from $$\pi(n)+1$$ to $$\pi(n+1)$$ inclusively:

$$\lambda_4(n)=\sum _{k=\pi (n ) +1}^{\pi (n+1 ) }\Biggl(\nu_{{p_{k}}}( (n+1) ) -\nu_{{p_{k}}} ( n) \Biggr)\ln(p_{k})\quad\quad\quad\quad\,\quad\quad\quad\quad\,\,(4)$$

$$\lambda_4(n)=\cases{\ln(p_{\pi(n+1)})&\,n+1\in \mathbb P \cr 0& n+1\not\in \mathbb P \cr}$$ $$\quad\quad\quad\,\,\,\,\,\,\,\,\quad\quad\quad\,\,\,\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad(4.1)$$

So far the current draft of the considerations leading to (0.1),(1.1)&(1.2):

Legendre's formula for the p-adic valuation of the factorial of $$n$$:

$$\nu_{{p}} \left( n \right) =\sum _{j=1}^{ \lfloor {\frac { \ln n }{\ln \left( p \right) }} \rfloor +1} \Bigl\lfloor {\frac {n}{{p}^{j}}} \Bigr\rfloor\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad(\operatorname{i})$$

Natural logarithm Sum to product identity:

$$\ln \left( n! \right) =\sum _{j=1}^{n}\ln \left( j \right)\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad(\operatorname{ii})$$

Unique prime factorization Definition of the factorial of $$n$$:

$$n!=\prod _{k=1}^{\pi \left( n \right) }{p_{{k}}}^{\nu_{{p_{{k}}}} \left( n! \right) } \quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,(\operatorname{iii})$$

An identity for the natural logarithm of the factorial of the nearest integer to $$x$$ in terms of the first Chebyshev function (used in the first line for the proof of Bertrand's Postulate):

$$\ln ( [x] ! ) =\sum _{k=1}^{\pi ( \lfloor x \rfloor ) }\psi \Bigl( {\frac {x}{k}} \Bigr) \quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad(\operatorname{iv})$$

$$\pi(x)$$ is the number of prime numbers less than or equal to $$x$$.

$${\{x}\}$$ is the fractional part of $$x$$.

Collectively the four above lemmas imply the identity:

$$\sum _{i=1}^{\pi (n ) } \sum _{j=1}^{ {\bigl\lfloor\frac {\ln ( n ) }{\ln ( p_{{i}} ) }}\bigr\rfloor +1} \Bigl\lfloor {\frac {n}{{p_{{i}}}^{j}}} \Bigr\rfloor \ln ( p_{{i}}) -\sum _{i=1}^{n-1} \sum _{j=1}^{\pi ( {\lfloor\frac {n}{i}} \rfloor ) } \Bigl\lfloor\frac{ \ln ( {\frac {n}{i}} ) }{ \ln p_{{j}} } \Bigr\rfloor \ln ( p_{{j}} ) =0\quad\quad\quad(\operatorname{i \land ii \land iii\land iv})$$

I then took the following approach to furthering the above conclusion:

Consider the following identities:

$$\sum _{i=1}^{\pi (n ) } \sum _{j=1}^{ {\bigl\lfloor\frac {\ln ( n +1) }{\ln ( p_{{i}} ) }}\bigr\rfloor +1} \Bigl\lfloor {\frac {n+1}{{p_{{i}}}^{j}}} \Bigr\rfloor \ln ( p_{{i}}) -\sum _{i=1}^{\pi(n)} \sum _{j=1}^{\pi ( {\lfloor\frac {n}{i}} \rfloor ) } \Bigl\lfloor\frac{ \ln ( {\frac {n}{i}} ) }{ \ln p_{{j}} } \Bigr\rfloor \ln ( p_{{j}}) =0\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\,\,\,\,\,\,\,\,(\operatorname{v})$$

$$\sum _{i=1}^{\pi (n+1 ) } \sum _{j=1}^{ {\bigl\lfloor\frac {\ln ( n +1) }{\ln ( p_{{i}} ) }}\bigr\rfloor +1} \Bigl\lfloor {\frac {n+1}{{p_{{i}}}^{j}}} \Bigr\rfloor \ln ( p_{{i}}) -\sum _{i=1}^{\pi(n)} \sum _{j=1}^{\pi ( {\lfloor\frac {n}{i}} \rfloor ) } \Bigl\lfloor\frac{ \ln ( {\frac {n}{i}} ) }{ \ln p_{{j}} } \Bigr\rfloor \ln ( p_{{j}} ) =\ln(n+1)\quad\quad\quad\quad\quad\quad\,\,\,\,\,\,(\operatorname{vi})$$

$$\sum _{i=1}^{\pi (n+1 ) } \sum _{j=1}^{ {\bigl\lfloor\frac {\ln ( n ) }{\ln ( p_{{i}} ) }}\bigr\rfloor +1} \Bigl\lfloor {\frac {n}{{p_{{i}}}^{j}}} \Bigr\rfloor \ln ( p_{{i}}) -\sum _{i=1}^{\pi (n ) } \sum _{j=1}^{ {\bigl\lfloor\frac {\ln ( n ) }{\ln ( p_{{i}} ) }}\bigr\rfloor +1} \Bigl\lfloor {\frac {n}{{p_{{i}}}^{j}}} \Bigr\rfloor \ln ( p_{{i}}) =0\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\,\,\,\,\,\,\,\,(\operatorname{vii})$$ $$\sum _{i=1}^{\pi (n ) } \sum _{j=1}^{ {\bigl\lfloor\frac {\ln ( n ) }{\ln ( p_{{i}} ) }}\bigr\rfloor +1} \Bigl\lfloor {\frac {n}{{p_{{i}}}^{j}}} \Bigr\rfloor \ln ( p_{{i}})=\ln(n!)\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\,\,\,\,\,\,(\operatorname{viii})$$ $$\sum _{i=1}^{\pi (n+1 ) } \sum _{j=1}^{ {\bigl\lfloor\frac {\ln ( n+1 ) }{\ln ( p_{{i}} ) }}\bigr\rfloor +1} \Bigl\lfloor {\frac {n+1}{{p_{{i}}}^{j}}} \Bigr\rfloor \ln ( p_{{i}}) -\sum _{i=1}^{\pi (n ) } \sum _{j=1}^{ {\bigl\lfloor\frac {\ln ( n ) }{\ln ( p_{{i}} ) }}\bigr\rfloor +1} \Bigl\lfloor {\frac {n}{{p_{{i}}}^{j}}} \Bigr\rfloor \ln ( p_{{i}}) =\ln(n+1)\quad\quad\quad\quad\,\,\,\,\,\,(\operatorname{ix})$$

$$\sum _{i=1}^{\pi (n+1 ) } \sum _{j=1}^{ {\bigl\lfloor\frac {\ln ( n ) }{\ln ( p_{{i}} ) }}\bigr\rfloor +1} \Bigl\lfloor {\frac {n}{{p_{{i}}}^{j}}} \Bigr\rfloor \ln ( p_{{i}})=\ln(n!)\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\,\,\,\,\,(\operatorname{x})$$

$$\sum _{i=1}^{\pi (n ) } \sum _{j=1}^{ {\bigl\lfloor\frac {\ln ( n+1 ) }{\ln ( p_{{i}} ) }}\bigr\rfloor +1} \Bigl\lfloor {\frac {n+1}{{p_{{i}}}^{j}}} \Bigr\rfloor \ln ( p_{{i}})=\ln((n+1)!)\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\,\,\,\,\,\,(\operatorname{xi})$$

$$\sum _{i=1}^{\pi (n ) } \sum _{j=1}^{ {\bigl\lfloor\frac {\ln ( n+1 ) }{\ln ( p_{{i}} ) }}\bigr\rfloor +1} \Bigl\lfloor {\frac {n+1}{{p_{{i}}}^{j}}} \Bigr\rfloor \ln ( p_{{i}})-\sum _{i=1}^{n } \sum _{j=1}^{ {\bigl\lfloor\frac {\ln ( n ) }{\ln ( p_{{i}} ) }}\bigr\rfloor +1} \Bigl\lfloor {\frac {n}{{p_{{i}}}^{j}}} \Bigr\rfloor \ln ( p_{{i}})=\sum _{j=1}^{N}\alpha_{g(n,j)}\ln(p _{f(n,j)})\delta\Biggl(\frac{n}{2},\Bigl\lfloor \frac{n}{2}\Bigr\rfloor\Biggr)$$

$${\{g(n,j),f(n,j)}\} \subset \mathbb N\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\,\,\,\,\,\,\quad\quad\quad\quad\quad\quad\quad\quad(\operatorname{xii})$$

Collectively imply:

$$\sum _{i=\pi (n )+1}^{\pi (n+1 ) } \sum _{j=1}^{ {\bigl\lfloor\frac {\ln ( n +1) }{\ln ( p_{{i}} ) }}\bigr\rfloor +1} \Bigl\lfloor {\frac {n+1}{{p_{{i}}}^{j}}} \Bigr\rfloor \ln ( p_{{i}})=\ln(n+1)\delta(\tau(n+1),2)\quad\quad\quad\quad\quad\quad\quad\quad(0.2)$$

$$\tau(k)$$ is the total number of divisors of $$k$$.

$$\delta(x,y)$$ is the Kronecker delta function.

• is this a question? – Alexander Gruber Oct 4 '18 at 17:08
• Well yes of course it is, If I simply state what I am attempting, everybody complains that I need to demonstrate that I have made an attempt before posting, so I posted what I had done at the time I made the question – Adam Oct 5 '18 at 1:05
• I will admit that since then I have near on almost provided a rigorous enough proof, but there are still holes in it otherwise I would feel comfortable looking at it – Adam Oct 5 '18 at 1:06
• Ah for example, the denominators of $(1.2)$ and $(2.2)$ are not the maximal value that will adhere to the divisibility condition, they are only factors of it, there is a big hole right there that needs a plug – Adam Oct 5 '18 at 1:08
• @Adam one thing you may want to do is highlight the question part of your post by putting a > in the beginning of the line (see for example this question of mine from years ago). This type of formatting is in no way required, but it can be helpful for getting people to read your question, since it is easier to locate at a glance what the post is about. – Alexander Gruber Oct 8 '18 at 4:20