Mixing
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.
$$lambda_0(n)=sum _{i=1}^{pi (n ) } sum _{j=1}^{ {bigllfloorfrac {ln ( n ) }{ln ( p_{{i}} ) }}bigrrfloor +1} Bigllfloor {frac {n}{{p_{{i}}}^{j}}} Bigrrfloor ln
( p_{{i}}) -sum _{i=1}^{pi(n)} sum _{j=1}^{pi ( {lfloorfrac {n}{i}} rfloor ) } Bigllfloorfrac{ ln ( {frac {n}{i}} ) }{ ln p_{{j}} } Bigrrfloor ln ( p_{{j}})quadquadquadquad,,,,quadquadquadquadquad(0) $$
$${Biggl{(2-delta(n,1))frac{n!}{operatorname{e}^{lambda_0(n)}}}Biggr}=0 Rightarrowlambda_0 left( n right) =ln left( alpha_0 left( n right)
right) land left{ {frac {n!}{alpha_0 left( n right) }} right} =0quadquadquadquad,,,,(0.1)
$$
$delta(x,y)$ 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:
$$lambda_1(n)=sum _{k=pi (n ) +1}^{pi (n+1 ) }Biggl(nu_{{p_{k}}}( ((n+1)^2)! ) -nu_{{p_{k}}} ( n^2!) Biggr)ln(p_{k})quadquadquadquad,quadquadquadquad,,(1)$$
$$lambda_1(n)=0 operatorname{iff} n+1 not in mathbb Pquadquadquadquad,,,,,,,,quadquadquad,,,quadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquad(1.1)$$
$${Biggl{frac{lambda_1(n)}{({prod^{pi(n)}_{j=1}}p_j)^2ln(p_{pi(n+1)})}}Biggr}=0quadquadquadquadquad,,,,,,,,quadquadquad,,,quadquadquadquadquadquadquadquadquadquad(1.2)$$
$lambda_2(n)$ 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:
$$lambda_2(n)=sum _{k=pi (n ) +1}^{pi (n+1 ) }Biggl(nu_{{p_{k}}}( (n+1)! ) -nu_{{p_{k}}} ( n!) Biggr)ln(p_{k})quadquadquadquad,quadquadquadquad,,(2)$$
$$lambda_2(n)=0 operatorname{iff} n+1 not in mathbb Pquadquadquadquad,,,,,,,,quadquadquad,,,quadquadquadquadquadquadquadquadquadquadquadquadquadquadquad(2.1)$$
$${Biggl{frac{lambda_2(n)}{({prod^{pi(n)}_{j=1}}p_j)ln(p_{pi(n+1)})}}Biggr}=0quadquadquadquadquad,,,,,,,,quadquadquad,,,quadquadquadquadquadquadquadquad(2.2)$$
$lambda_3(n)$ 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:
$$lambda_3(n)=sum _{k=pi (n ) +1}^{pi (n+1 ) }Biggl(nu_{{p_{k}}}( (n+1)^2 ) -nu_{{p_{k}}} ( n^2) Biggr)ln(p_{k})quadquadquadquad,quadquadquadquad,,(3)$$
$$lambda_3(n)=cases{3ln(p_{pi(n+1)})&$,n+1in mathbb P $cr 0& $n+1notin mathbb P $cr}$$
$$quadquadquad,,,,,,,,quadquadquad,,,quadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquad(3.1)$$
$lambda_4(n)$ 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:
$$lambda_4(n)=sum _{k=pi (n ) +1}^{pi (n+1 ) }Biggl(nu_{{p_{k}}}( (n+1) ) -nu_{{p_{k}}} ( n) Biggr)ln(p_{k})quadquadquadquad,quadquadquadquad,,(4)$$
$$lambda_4(n)=cases{ln(p_{pi(n+1)})&$,n+1in mathbb P $cr 0& $n+1notin mathbb P $cr}$$
$$quadquadquad,,,,,,,,quadquadquad,,,quadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquad(4.1)$$
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 $n$:
$$nu_{{p}} left( n right) =sum _{j=1}^{ lfloor {frac {
ln n }{ln left( p right) }} rfloor +1} Bigllfloor {frac {n}{{p}^{j}}} Bigrrfloorquadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquad(operatorname{i})$$
Natural logarithm Sum to product identity:
$$ln left( n! right) =sum _{j=1}^{n}ln left( j right)quadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquad(operatorname{ii}) $$
Unique prime factorization Definition of the factorial of $n$:
$$n!=prod _{k=1}^{pi left( n right) }{p_{{k}}}^{nu_{{p_{{k}}}}
left( n! right) } quadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquad,,,,,,,,,,,,,,,,,,,,,,,,,,,,(operatorname{iii}) $$
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):
$$ln ( [x] ! ) =sum _{k=1}^{pi ( lfloor x rfloor ) }psi Bigl( {frac {x}{k}}
Bigr)
quadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquad(operatorname{iv}) $$
$pi(x)$ 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:
$$sum _{i=1}^{pi (n ) } sum _{j=1}^{ {bigllfloorfrac {ln ( n ) }{ln ( p_{{i}} ) }}bigrrfloor +1} Bigllfloor {frac {n}{{p_{{i}}}^{j}}} Bigrrfloor ln ( p_{{i}}) -sum _{i=1}^{n-1} sum _{j=1}^{pi ( {lfloorfrac {n}{i}} rfloor ) } Bigllfloorfrac{ ln ( {frac {n}{i}} ) }{ ln
p_{{j}} } Bigrrfloor ln ( p_{{j}}
) =0quadquadquad(operatorname{i land ii land iiiland iv})$$
I then took the following approach to furthering the above conclusion:
Consider the following identities:
$$sum _{i=1}^{pi (n ) } sum _{j=1}^{ {bigllfloorfrac {ln ( n +1) }{ln ( p_{{i}} ) }}bigrrfloor +1} Bigllfloor {frac {n+1}{{p_{{i}}}^{j}}} Bigrrfloor ln
( p_{{i}}) -sum _{i=1}^{pi(n)} sum _{j=1}^{pi ( {lfloorfrac {n}{i}} rfloor ) } Bigllfloorfrac{ ln ( {frac {n}{i}} ) }{ ln p_{{j}} } Bigrrfloor ln ( p_{{j}}) =0quadquadquadquadquadquadquadquadquadquad,,,,,,,,(operatorname{v})$$
$$sum _{i=1}^{pi (n+1 ) } sum _{j=1}^{ {bigllfloorfrac {ln ( n +1) }{ln ( p_{{i}} ) }}bigrrfloor +1} Bigllfloor {frac {n+1}{{p_{{i}}}^{j}}} Bigrrfloor ln
( p_{{i}}) -sum _{i=1}^{pi(n)} sum _{j=1}^{pi ( {lfloorfrac {n}{i}} rfloor ) } Bigllfloorfrac{ ln ( {frac {n}{i}} ) }{ ln p_{{j}} } Bigrrfloor ln ( p_{{j}}
) =ln(n+1)quadquadquadquadquadquad,,,,,,(operatorname{vi})$$
$$sum _{i=1}^{pi (n+1 ) } sum _{j=1}^{ {bigllfloorfrac {ln ( n ) }{ln ( p_{{i}} ) }}bigrrfloor +1} Bigllfloor {frac {n}{{p_{{i}}}^{j}}} Bigrrfloor ln
( p_{{i}}) -sum _{i=1}^{pi (n ) } sum _{j=1}^{ {bigllfloorfrac {ln ( n ) }{ln ( p_{{i}} ) }}bigrrfloor +1} Bigllfloor {frac {n}{{p_{{i}}}^{j}}} Bigrrfloor ln
( p_{{i}}) =0quadquadquadquadquadquadquadquadquadquad,,,,,,,,(operatorname{vii})$$
$$sum _{i=1}^{pi (n ) } sum _{j=1}^{ {bigllfloorfrac {ln ( n ) }{ln ( p_{{i}} ) }}bigrrfloor +1} Bigllfloor {frac {n}{{p_{{i}}}^{j}}} Bigrrfloor ln
( p_{{i}})=ln(n!)quadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquad,,,,,,(operatorname{viii})$$
$$sum _{i=1}^{pi (n+1 ) } sum _{j=1}^{ {bigllfloorfrac {ln ( n+1 ) }{ln ( p_{{i}} ) }}bigrrfloor +1} Bigllfloor {frac {n+1}{{p_{{i}}}^{j}}} Bigrrfloor ln ( p_{{i}}) -sum _{i=1}^{pi (n ) } sum _{j=1}^{ {bigllfloorfrac {ln ( n ) }{ln ( p_{{i}} ) }}bigrrfloor +1} Bigllfloor {frac {n}{{p_{{i}}}^{j}}} Bigrrfloor ln
( p_{{i}}) =ln(n+1)quadquadquadquad,,,,,,(operatorname{ix})$$
$$sum _{i=1}^{pi (n+1 ) } sum _{j=1}^{ {bigllfloorfrac {ln ( n ) }{ln ( p_{{i}} ) }}bigrrfloor +1} Bigllfloor {frac {n}{{p_{{i}}}^{j}}} Bigrrfloor ln
( p_{{i}})=ln(n!)quadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquad,,,,,(operatorname{x}) $$
$$sum _{i=1}^{pi (n ) } sum _{j=1}^{ {bigllfloorfrac {ln ( n+1 ) }{ln ( p_{{i}} ) }}bigrrfloor +1} Bigllfloor {frac {n+1}{{p_{{i}}}^{j}}} Bigrrfloor ln
( p_{{i}})=ln((n+1)!)quadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquad,,,,,,(operatorname{xi}) $$
$$sum _{i=1}^{pi (n ) } sum _{j=1}^{ {bigllfloorfrac {ln ( n+1 ) }{ln ( p_{{i}} ) }}bigrrfloor +1} Bigllfloor {frac {n+1}{{p_{{i}}}^{j}}} Bigrrfloor ln
( p_{{i}})-sum _{i=1}^{n } sum _{j=1}^{ {bigllfloorfrac {ln ( n ) }{ln ( p_{{i}} ) }}bigrrfloor +1} Bigllfloor {frac {n}{{p_{{i}}}^{j}}} Bigrrfloor ln
( p_{{i}})=sum _{j=1}^{N}alpha_{g(n,j)}ln(p
_{f(n,j)})deltaBiggl(frac{n}{2},Bigllfloor frac{n}{2}BigrrfloorBiggr)$$
$${{g(n,j),f(n,j)}} subset mathbb Nquadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquad,,,,,,quadquadquadquadquadquadquadquad(operatorname{xii})$$
Collectively imply:
$$sum _{i=pi (n )+1}^{pi (n+1 ) } sum _{j=1}^{ {bigllfloorfrac {ln ( n +1) }{ln ( p_{{i}} ) }}bigrrfloor +1} Bigllfloor {frac {n+1}{{p_{{i}}}^{j}}} Bigrrfloor ln
( p_{{i}})=ln(n+1)delta(tau(n+1),2)quadquadquadquadquadquadquadquad(0.2) $$
$tau(k)$ is the total number of divisors of $k$.
$delta(x,y)$ is the Kronecker delta function.
number-theory p-adic-number-theory
asked Sep 28 '18 at 18:17
Adam
54114
|
show 6 more comments
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.
$$lambda_0(n)=sum _{i=1}^{pi (n ) } sum _{j=1}^{ {bigllfloorfrac {ln ( n ) }{ln ( p_{{i}} ) }}bigrrfloor +1} Bigllfloor {frac {n}{{p_{{i}}}^{j}}} Bigrrfloor ln
( p_{{i}}) -sum _{i=1}^{pi(n)} sum _{j=1}^{pi ( {lfloorfrac {n}{i}} rfloor ) } Bigllfloorfrac{ ln ( {frac {n}{i}} ) }{ ln p_{{j}} } Bigrrfloor ln ( p_{{j}})quadquadquadquad,,,,quadquadquadquadquad(0) $$
$${Biggl{(2-delta(n,1))frac{n!}{operatorname{e}^{lambda_0(n)}}}Biggr}=0 Rightarrowlambda_0 left( n right) =ln left( alpha_0 left( n right)
right) land left{ {frac {n!}{alpha_0 left( n right) }} right} =0quadquadquadquad,,,,(0.1)
$$
$delta(x,y)$ 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:
$$lambda_1(n)=sum _{k=pi (n ) +1}^{pi (n+1 ) }Biggl(nu_{{p_{k}}}( ((n+1)^2)! ) -nu_{{p_{k}}} ( n^2!) Biggr)ln(p_{k})quadquadquadquad,quadquadquadquad,,(1)$$
$$lambda_1(n)=0 operatorname{iff} n+1 not in mathbb Pquadquadquadquad,,,,,,,,quadquadquad,,,quadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquad(1.1)$$
$${Biggl{frac{lambda_1(n)}{({prod^{pi(n)}_{j=1}}p_j)^2ln(p_{pi(n+1)})}}Biggr}=0quadquadquadquadquad,,,,,,,,quadquadquad,,,quadquadquadquadquadquadquadquadquadquad(1.2)$$
$lambda_2(n)$ 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:
$$lambda_2(n)=sum _{k=pi (n ) +1}^{pi (n+1 ) }Biggl(nu_{{p_{k}}}( (n+1)! ) -nu_{{p_{k}}} ( n!) Biggr)ln(p_{k})quadquadquadquad,quadquadquadquad,,(2)$$
$$lambda_2(n)=0 operatorname{iff} n+1 not in mathbb Pquadquadquadquad,,,,,,,,quadquadquad,,,quadquadquadquadquadquadquadquadquadquadquadquadquadquadquad(2.1)$$
$${Biggl{frac{lambda_2(n)}{({prod^{pi(n)}_{j=1}}p_j)ln(p_{pi(n+1)})}}Biggr}=0quadquadquadquadquad,,,,,,,,quadquadquad,,,quadquadquadquadquadquadquadquad(2.2)$$
$lambda_3(n)$ 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:
$$lambda_3(n)=sum _{k=pi (n ) +1}^{pi (n+1 ) }Biggl(nu_{{p_{k}}}( (n+1)^2 ) -nu_{{p_{k}}} ( n^2) Biggr)ln(p_{k})quadquadquadquad,quadquadquadquad,,(3)$$
$$lambda_3(n)=cases{3ln(p_{pi(n+1)})&$,n+1in mathbb P $cr 0& $n+1notin mathbb P $cr}$$
$$quadquadquad,,,,,,,,quadquadquad,,,quadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquad(3.1)$$
$lambda_4(n)$ 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:
$$lambda_4(n)=sum _{k=pi (n ) +1}^{pi (n+1 ) }Biggl(nu_{{p_{k}}}( (n+1) ) -nu_{{p_{k}}} ( n) Biggr)ln(p_{k})quadquadquadquad,quadquadquadquad,,(4)$$
$$lambda_4(n)=cases{ln(p_{pi(n+1)})&$,n+1in mathbb P $cr 0& $n+1notin mathbb P $cr}$$
$$quadquadquad,,,,,,,,quadquadquad,,,quadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquad(4.1)$$
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 $n$:
$$nu_{{p}} left( n right) =sum _{j=1}^{ lfloor {frac {
ln n }{ln left( p right) }} rfloor +1} Bigllfloor {frac {n}{{p}^{j}}} Bigrrfloorquadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquad(operatorname{i})$$
Natural logarithm Sum to product identity:
$$ln left( n! right) =sum _{j=1}^{n}ln left( j right)quadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquad(operatorname{ii}) $$
Unique prime factorization Definition of the factorial of $n$:
$$n!=prod _{k=1}^{pi left( n right) }{p_{{k}}}^{nu_{{p_{{k}}}}
left( n! right) } quadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquad,,,,,,,,,,,,,,,,,,,,,,,,,,,,(operatorname{iii}) $$
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):
$$ln ( [x] ! ) =sum _{k=1}^{pi ( lfloor x rfloor ) }psi Bigl( {frac {x}{k}}
Bigr)
quadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquad(operatorname{iv}) $$
$pi(x)$ 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:
$$sum _{i=1}^{pi (n ) } sum _{j=1}^{ {bigllfloorfrac {ln ( n ) }{ln ( p_{{i}} ) }}bigrrfloor +1} Bigllfloor {frac {n}{{p_{{i}}}^{j}}} Bigrrfloor ln ( p_{{i}}) -sum _{i=1}^{n-1} sum _{j=1}^{pi ( {lfloorfrac {n}{i}} rfloor ) } Bigllfloorfrac{ ln ( {frac {n}{i}} ) }{ ln
p_{{j}} } Bigrrfloor ln ( p_{{j}}
) =0quadquadquad(operatorname{i land ii land iiiland iv})$$
I then took the following approach to furthering the above conclusion:
Consider the following identities:
$$sum _{i=1}^{pi (n ) } sum _{j=1}^{ {bigllfloorfrac {ln ( n +1) }{ln ( p_{{i}} ) }}bigrrfloor +1} Bigllfloor {frac {n+1}{{p_{{i}}}^{j}}} Bigrrfloor ln
( p_{{i}}) -sum _{i=1}^{pi(n)} sum _{j=1}^{pi ( {lfloorfrac {n}{i}} rfloor ) } Bigllfloorfrac{ ln ( {frac {n}{i}} ) }{ ln p_{{j}} } Bigrrfloor ln ( p_{{j}}) =0quadquadquadquadquadquadquadquadquadquad,,,,,,,,(operatorname{v})$$
$$sum _{i=1}^{pi (n+1 ) } sum _{j=1}^{ {bigllfloorfrac {ln ( n +1) }{ln ( p_{{i}} ) }}bigrrfloor +1} Bigllfloor {frac {n+1}{{p_{{i}}}^{j}}} Bigrrfloor ln
( p_{{i}}) -sum _{i=1}^{pi(n)} sum _{j=1}^{pi ( {lfloorfrac {n}{i}} rfloor ) } Bigllfloorfrac{ ln ( {frac {n}{i}} ) }{ ln p_{{j}} } Bigrrfloor ln ( p_{{j}}
) =ln(n+1)quadquadquadquadquadquad,,,,,,(operatorname{vi})$$
$$sum _{i=1}^{pi (n+1 ) } sum _{j=1}^{ {bigllfloorfrac {ln ( n ) }{ln ( p_{{i}} ) }}bigrrfloor +1} Bigllfloor {frac {n}{{p_{{i}}}^{j}}} Bigrrfloor ln
( p_{{i}}) -sum _{i=1}^{pi (n ) } sum _{j=1}^{ {bigllfloorfrac {ln ( n ) }{ln ( p_{{i}} ) }}bigrrfloor +1} Bigllfloor {frac {n}{{p_{{i}}}^{j}}} Bigrrfloor ln
( p_{{i}}) =0quadquadquadquadquadquadquadquadquadquad,,,,,,,,(operatorname{vii})$$
$$sum _{i=1}^{pi (n ) } sum _{j=1}^{ {bigllfloorfrac {ln ( n ) }{ln ( p_{{i}} ) }}bigrrfloor +1} Bigllfloor {frac {n}{{p_{{i}}}^{j}}} Bigrrfloor ln
( p_{{i}})=ln(n!)quadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquad,,,,,,(operatorname{viii})$$
$$sum _{i=1}^{pi (n+1 ) } sum _{j=1}^{ {bigllfloorfrac {ln ( n+1 ) }{ln ( p_{{i}} ) }}bigrrfloor +1} Bigllfloor {frac {n+1}{{p_{{i}}}^{j}}} Bigrrfloor ln ( p_{{i}}) -sum _{i=1}^{pi (n ) } sum _{j=1}^{ {bigllfloorfrac {ln ( n ) }{ln ( p_{{i}} ) }}bigrrfloor +1} Bigllfloor {frac {n}{{p_{{i}}}^{j}}} Bigrrfloor ln
( p_{{i}}) =ln(n+1)quadquadquadquad,,,,,,(operatorname{ix})$$
$$sum _{i=1}^{pi (n+1 ) } sum _{j=1}^{ {bigllfloorfrac {ln ( n ) }{ln ( p_{{i}} ) }}bigrrfloor +1} Bigllfloor {frac {n}{{p_{{i}}}^{j}}} Bigrrfloor ln
( p_{{i}})=ln(n!)quadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquad,,,,,(operatorname{x}) $$
$$sum _{i=1}^{pi (n ) } sum _{j=1}^{ {bigllfloorfrac {ln ( n+1 ) }{ln ( p_{{i}} ) }}bigrrfloor +1} Bigllfloor {frac {n+1}{{p_{{i}}}^{j}}} Bigrrfloor ln
( p_{{i}})=ln((n+1)!)quadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquad,,,,,,(operatorname{xi}) $$
$$sum _{i=1}^{pi (n ) } sum _{j=1}^{ {bigllfloorfrac {ln ( n+1 ) }{ln ( p_{{i}} ) }}bigrrfloor +1} Bigllfloor {frac {n+1}{{p_{{i}}}^{j}}} Bigrrfloor ln
( p_{{i}})-sum _{i=1}^{n } sum _{j=1}^{ {bigllfloorfrac {ln ( n ) }{ln ( p_{{i}} ) }}bigrrfloor +1} Bigllfloor {frac {n}{{p_{{i}}}^{j}}} Bigrrfloor ln
( p_{{i}})=sum _{j=1}^{N}alpha_{g(n,j)}ln(p
_{f(n,j)})deltaBiggl(frac{n}{2},Bigllfloor frac{n}{2}BigrrfloorBiggr)$$
$${{g(n,j),f(n,j)}} subset mathbb Nquadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquad,,,,,,quadquadquadquadquadquadquadquad(operatorname{xii})$$
Collectively imply:
$$sum _{i=pi (n )+1}^{pi (n+1 ) } sum _{j=1}^{ {bigllfloorfrac {ln ( n +1) }{ln ( p_{{i}} ) }}bigrrfloor +1} Bigllfloor {frac {n+1}{{p_{{i}}}^{j}}} Bigrrfloor ln
( p_{{i}})=ln(n+1)delta(tau(n+1),2)quadquadquadquadquadquadquadquad(0.2) $$
$tau(k)$ is the total number of divisors of $k$.
$delta(x,y)$ is the Kronecker delta function.
number-theory p-adic-number-theory
asked Sep 28 '18 at 18:17
Adam
54114
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
Editing a question bumps it, which is a problem when the number of edits is that large. I recommend you use the sandbox at: math.meta.stackexchange.com/questions/4666/…
– Michael Greinecker♦
Oct 5 '18 at 20:31
|
show 6 more comments
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.
$$lambda_0(n)=sum _{i=1}^{pi (n ) } sum _{j=1}^{ {bigllfloorfrac {ln ( n ) }{ln ( p_{{i}} ) }}bigrrfloor +1} Bigllfloor {frac {n}{{p_{{i}}}^{j}}} Bigrrfloor ln
( p_{{i}}) -sum _{i=1}^{pi(n)} sum _{j=1}^{pi ( {lfloorfrac {n}{i}} rfloor ) } Bigllfloorfrac{ ln ( {frac {n}{i}} ) }{ ln p_{{j}} } Bigrrfloor ln ( p_{{j}})quadquadquadquad,,,,quadquadquadquadquad(0) $$
$${Biggl{(2-delta(n,1))frac{n!}{operatorname{e}^{lambda_0(n)}}}Biggr}=0 Rightarrowlambda_0 left( n right) =ln left( alpha_0 left( n right)
right) land left{ {frac {n!}{alpha_0 left( n right) }} right} =0quadquadquadquad,,,,(0.1)
$$
$delta(x,y)$ 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:
$$lambda_1(n)=sum _{k=pi (n ) +1}^{pi (n+1 ) }Biggl(nu_{{p_{k}}}( ((n+1)^2)! ) -nu_{{p_{k}}} ( n^2!) Biggr)ln(p_{k})quadquadquadquad,quadquadquadquad,,(1)$$
$$lambda_1(n)=0 operatorname{iff} n+1 not in mathbb Pquadquadquadquad,,,,,,,,quadquadquad,,,quadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquad(1.1)$$
$${Biggl{frac{lambda_1(n)}{({prod^{pi(n)}_{j=1}}p_j)^2ln(p_{pi(n+1)})}}Biggr}=0quadquadquadquadquad,,,,,,,,quadquadquad,,,quadquadquadquadquadquadquadquadquadquad(1.2)$$
$lambda_2(n)$ 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:
$$lambda_2(n)=sum _{k=pi (n ) +1}^{pi (n+1 ) }Biggl(nu_{{p_{k}}}( (n+1)! ) -nu_{{p_{k}}} ( n!) Biggr)ln(p_{k})quadquadquadquad,quadquadquadquad,,(2)$$
$$lambda_2(n)=0 operatorname{iff} n+1 not in mathbb Pquadquadquadquad,,,,,,,,quadquadquad,,,quadquadquadquadquadquadquadquadquadquadquadquadquadquadquad(2.1)$$
$${Biggl{frac{lambda_2(n)}{({prod^{pi(n)}_{j=1}}p_j)ln(p_{pi(n+1)})}}Biggr}=0quadquadquadquadquad,,,,,,,,quadquadquad,,,quadquadquadquadquadquadquadquad(2.2)$$
$lambda_3(n)$ 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:
$$lambda_3(n)=sum _{k=pi (n ) +1}^{pi (n+1 ) }Biggl(nu_{{p_{k}}}( (n+1)^2 ) -nu_{{p_{k}}} ( n^2) Biggr)ln(p_{k})quadquadquadquad,quadquadquadquad,,(3)$$
$$lambda_3(n)=cases{3ln(p_{pi(n+1)})&$,n+1in mathbb P $cr 0& $n+1notin mathbb P $cr}$$
$$quadquadquad,,,,,,,,quadquadquad,,,quadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquad(3.1)$$
$lambda_4(n)$ 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:
$$lambda_4(n)=sum _{k=pi (n ) +1}^{pi (n+1 ) }Biggl(nu_{{p_{k}}}( (n+1) ) -nu_{{p_{k}}} ( n) Biggr)ln(p_{k})quadquadquadquad,quadquadquadquad,,(4)$$
$$lambda_4(n)=cases{ln(p_{pi(n+1)})&$,n+1in mathbb P $cr 0& $n+1notin mathbb P $cr}$$
$$quadquadquad,,,,,,,,quadquadquad,,,quadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquad(4.1)$$
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 $n$:
$$nu_{{p}} left( n right) =sum _{j=1}^{ lfloor {frac {
ln n }{ln left( p right) }} rfloor +1} Bigllfloor {frac {n}{{p}^{j}}} Bigrrfloorquadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquad(operatorname{i})$$
Natural logarithm Sum to product identity:
$$ln left( n! right) =sum _{j=1}^{n}ln left( j right)quadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquad(operatorname{ii}) $$
Unique prime factorization Definition of the factorial of $n$:
$$n!=prod _{k=1}^{pi left( n right) }{p_{{k}}}^{nu_{{p_{{k}}}}
left( n! right) } quadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquad,,,,,,,,,,,,,,,,,,,,,,,,,,,,(operatorname{iii}) $$
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):
$$ln ( [x] ! ) =sum _{k=1}^{pi ( lfloor x rfloor ) }psi Bigl( {frac {x}{k}}
Bigr)
quadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquad(operatorname{iv}) $$
$pi(x)$ 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:
$$sum _{i=1}^{pi (n ) } sum _{j=1}^{ {bigllfloorfrac {ln ( n ) }{ln ( p_{{i}} ) }}bigrrfloor +1} Bigllfloor {frac {n}{{p_{{i}}}^{j}}} Bigrrfloor ln ( p_{{i}}) -sum _{i=1}^{n-1} sum _{j=1}^{pi ( {lfloorfrac {n}{i}} rfloor ) } Bigllfloorfrac{ ln ( {frac {n}{i}} ) }{ ln
p_{{j}} } Bigrrfloor ln ( p_{{j}}
) =0quadquadquad(operatorname{i land ii land iiiland iv})$$
I then took the following approach to furthering the above conclusion:
Consider the following identities:
$$sum _{i=1}^{pi (n ) } sum _{j=1}^{ {bigllfloorfrac {ln ( n +1) }{ln ( p_{{i}} ) }}bigrrfloor +1} Bigllfloor {frac {n+1}{{p_{{i}}}^{j}}} Bigrrfloor ln
( p_{{i}}) -sum _{i=1}^{pi(n)} sum _{j=1}^{pi ( {lfloorfrac {n}{i}} rfloor ) } Bigllfloorfrac{ ln ( {frac {n}{i}} ) }{ ln p_{{j}} } Bigrrfloor ln ( p_{{j}}) =0quadquadquadquadquadquadquadquadquadquad,,,,,,,,(operatorname{v})$$
$$sum _{i=1}^{pi (n+1 ) } sum _{j=1}^{ {bigllfloorfrac {ln ( n +1) }{ln ( p_{{i}} ) }}bigrrfloor +1} Bigllfloor {frac {n+1}{{p_{{i}}}^{j}}} Bigrrfloor ln
( p_{{i}}) -sum _{i=1}^{pi(n)} sum _{j=1}^{pi ( {lfloorfrac {n}{i}} rfloor ) } Bigllfloorfrac{ ln ( {frac {n}{i}} ) }{ ln p_{{j}} } Bigrrfloor ln ( p_{{j}}
) =ln(n+1)quadquadquadquadquadquad,,,,,,(operatorname{vi})$$
$$sum _{i=1}^{pi (n+1 ) } sum _{j=1}^{ {bigllfloorfrac {ln ( n ) }{ln ( p_{{i}} ) }}bigrrfloor +1} Bigllfloor {frac {n}{{p_{{i}}}^{j}}} Bigrrfloor ln
( p_{{i}}) -sum _{i=1}^{pi (n ) } sum _{j=1}^{ {bigllfloorfrac {ln ( n ) }{ln ( p_{{i}} ) }}bigrrfloor +1} Bigllfloor {frac {n}{{p_{{i}}}^{j}}} Bigrrfloor ln
( p_{{i}}) =0quadquadquadquadquadquadquadquadquadquad,,,,,,,,(operatorname{vii})$$
$$sum _{i=1}^{pi (n ) } sum _{j=1}^{ {bigllfloorfrac {ln ( n ) }{ln ( p_{{i}} ) }}bigrrfloor +1} Bigllfloor {frac {n}{{p_{{i}}}^{j}}} Bigrrfloor ln
( p_{{i}})=ln(n!)quadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquad,,,,,,(operatorname{viii})$$
$$sum _{i=1}^{pi (n+1 ) } sum _{j=1}^{ {bigllfloorfrac {ln ( n+1 ) }{ln ( p_{{i}} ) }}bigrrfloor +1} Bigllfloor {frac {n+1}{{p_{{i}}}^{j}}} Bigrrfloor ln ( p_{{i}}) -sum _{i=1}^{pi (n ) } sum _{j=1}^{ {bigllfloorfrac {ln ( n ) }{ln ( p_{{i}} ) }}bigrrfloor +1} Bigllfloor {frac {n}{{p_{{i}}}^{j}}} Bigrrfloor ln
( p_{{i}}) =ln(n+1)quadquadquadquad,,,,,,(operatorname{ix})$$
$$sum _{i=1}^{pi (n+1 ) } sum _{j=1}^{ {bigllfloorfrac {ln ( n ) }{ln ( p_{{i}} ) }}bigrrfloor +1} Bigllfloor {frac {n}{{p_{{i}}}^{j}}} Bigrrfloor ln
( p_{{i}})=ln(n!)quadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquad,,,,,(operatorname{x}) $$
$$sum _{i=1}^{pi (n ) } sum _{j=1}^{ {bigllfloorfrac {ln ( n+1 ) }{ln ( p_{{i}} ) }}bigrrfloor +1} Bigllfloor {frac {n+1}{{p_{{i}}}^{j}}} Bigrrfloor ln
( p_{{i}})=ln((n+1)!)quadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquad,,,,,,(operatorname{xi}) $$
$$sum _{i=1}^{pi (n ) } sum _{j=1}^{ {bigllfloorfrac {ln ( n+1 ) }{ln ( p_{{i}} ) }}bigrrfloor +1} Bigllfloor {frac {n+1}{{p_{{i}}}^{j}}} Bigrrfloor ln
( p_{{i}})-sum _{i=1}^{n } sum _{j=1}^{ {bigllfloorfrac {ln ( n ) }{ln ( p_{{i}} ) }}bigrrfloor +1} Bigllfloor {frac {n}{{p_{{i}}}^{j}}} Bigrrfloor ln
( p_{{i}})=sum _{j=1}^{N}alpha_{g(n,j)}ln(p
_{f(n,j)})deltaBiggl(frac{n}{2},Bigllfloor frac{n}{2}BigrrfloorBiggr)$$
$${{g(n,j),f(n,j)}} subset mathbb Nquadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquad,,,,,,quadquadquadquadquadquadquadquad(operatorname{xii})$$
Collectively imply:
$$sum _{i=pi (n )+1}^{pi (n+1 ) } sum _{j=1}^{ {bigllfloorfrac {ln ( n +1) }{ln ( p_{{i}} ) }}bigrrfloor +1} Bigllfloor {frac {n+1}{{p_{{i}}}^{j}}} Bigrrfloor ln
( p_{{i}})=ln(n+1)delta(tau(n+1),2)quadquadquadquadquadquadquadquad(0.2) $$
$tau(k)$ is the total number of divisors of $k$.
$delta(x,y)$ is the Kronecker delta function.
number-theory p-adic-number-theory
asked Sep 28 '18 at 18:17
Adam
54114
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.
$$lambda_0(n)=sum _{i=1}^{pi (n ) } sum _{j=1}^{ {bigllfloorfrac {ln ( n ) }{ln ( p_{{i}} ) }}bigrrfloor +1} Bigllfloor {frac {n}{{p_{{i}}}^{j}}} Bigrrfloor ln
( p_{{i}}) -sum _{i=1}^{pi(n)} sum _{j=1}^{pi ( {lfloorfrac {n}{i}} rfloor ) } Bigllfloorfrac{ ln ( {frac {n}{i}} ) }{ ln p_{{j}} } Bigrrfloor ln ( p_{{j}})quadquadquadquad,,,,quadquadquadquadquad(0) $$
$${Biggl{(2-delta(n,1))frac{n!}{operatorname{e}^{lambda_0(n)}}}Biggr}=0 Rightarrowlambda_0 left( n right) =ln left( alpha_0 left( n right)
right) land left{ {frac {n!}{alpha_0 left( n right) }} right} =0quadquadquadquad,,,,(0.1)
$$
$delta(x,y)$ 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:
$$lambda_1(n)=sum _{k=pi (n ) +1}^{pi (n+1 ) }Biggl(nu_{{p_{k}}}( ((n+1)^2)! ) -nu_{{p_{k}}} ( n^2!) Biggr)ln(p_{k})quadquadquadquad,quadquadquadquad,,(1)$$
$$lambda_1(n)=0 operatorname{iff} n+1 not in mathbb Pquadquadquadquad,,,,,,,,quadquadquad,,,quadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquad(1.1)$$
$${Biggl{frac{lambda_1(n)}{({prod^{pi(n)}_{j=1}}p_j)^2ln(p_{pi(n+1)})}}Biggr}=0quadquadquadquadquad,,,,,,,,quadquadquad,,,quadquadquadquadquadquadquadquadquadquad(1.2)$$
$lambda_2(n)$ 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:
$$lambda_2(n)=sum _{k=pi (n ) +1}^{pi (n+1 ) }Biggl(nu_{{p_{k}}}( (n+1)! ) -nu_{{p_{k}}} ( n!) Biggr)ln(p_{k})quadquadquadquad,quadquadquadquad,,(2)$$
$$lambda_2(n)=0 operatorname{iff} n+1 not in mathbb Pquadquadquadquad,,,,,,,,quadquadquad,,,quadquadquadquadquadquadquadquadquadquadquadquadquadquadquad(2.1)$$
$${Biggl{frac{lambda_2(n)}{({prod^{pi(n)}_{j=1}}p_j)ln(p_{pi(n+1)})}}Biggr}=0quadquadquadquadquad,,,,,,,,quadquadquad,,,quadquadquadquadquadquadquadquad(2.2)$$
$lambda_3(n)$ 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:
$$lambda_3(n)=sum _{k=pi (n ) +1}^{pi (n+1 ) }Biggl(nu_{{p_{k}}}( (n+1)^2 ) -nu_{{p_{k}}} ( n^2) Biggr)ln(p_{k})quadquadquadquad,quadquadquadquad,,(3)$$
$$lambda_3(n)=cases{3ln(p_{pi(n+1)})&$,n+1in mathbb P $cr 0& $n+1notin mathbb P $cr}$$
$$quadquadquad,,,,,,,,quadquadquad,,,quadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquad(3.1)$$
$lambda_4(n)$ 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:
$$lambda_4(n)=sum _{k=pi (n ) +1}^{pi (n+1 ) }Biggl(nu_{{p_{k}}}( (n+1) ) -nu_{{p_{k}}} ( n) Biggr)ln(p_{k})quadquadquadquad,quadquadquadquad,,(4)$$
$$lambda_4(n)=cases{ln(p_{pi(n+1)})&$,n+1in mathbb P $cr 0& $n+1notin mathbb P $cr}$$
$$quadquadquad,,,,,,,,quadquadquad,,,quadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquad(4.1)$$
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 $n$:
$$nu_{{p}} left( n right) =sum _{j=1}^{ lfloor {frac {
ln n }{ln left( p right) }} rfloor +1} Bigllfloor {frac {n}{{p}^{j}}} Bigrrfloorquadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquad(operatorname{i})$$
Natural logarithm Sum to product identity:
$$ln left( n! right) =sum _{j=1}^{n}ln left( j right)quadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquad(operatorname{ii}) $$
Unique prime factorization Definition of the factorial of $n$:
$$n!=prod _{k=1}^{pi left( n right) }{p_{{k}}}^{nu_{{p_{{k}}}}
left( n! right) } quadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquad,,,,,,,,,,,,,,,,,,,,,,,,,,,,(operatorname{iii}) $$
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):
$$ln ( [x] ! ) =sum _{k=1}^{pi ( lfloor x rfloor ) }psi Bigl( {frac {x}{k}}
Bigr)
quadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquad(operatorname{iv}) $$
$pi(x)$ 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:
$$sum _{i=1}^{pi (n ) } sum _{j=1}^{ {bigllfloorfrac {ln ( n ) }{ln ( p_{{i}} ) }}bigrrfloor +1} Bigllfloor {frac {n}{{p_{{i}}}^{j}}} Bigrrfloor ln ( p_{{i}}) -sum _{i=1}^{n-1} sum _{j=1}^{pi ( {lfloorfrac {n}{i}} rfloor ) } Bigllfloorfrac{ ln ( {frac {n}{i}} ) }{ ln
p_{{j}} } Bigrrfloor ln ( p_{{j}}
) =0quadquadquad(operatorname{i land ii land iiiland iv})$$
I then took the following approach to furthering the above conclusion:
Consider the following identities:
$$sum _{i=1}^{pi (n ) } sum _{j=1}^{ {bigllfloorfrac {ln ( n +1) }{ln ( p_{{i}} ) }}bigrrfloor +1} Bigllfloor {frac {n+1}{{p_{{i}}}^{j}}} Bigrrfloor ln
( p_{{i}}) -sum _{i=1}^{pi(n)} sum _{j=1}^{pi ( {lfloorfrac {n}{i}} rfloor ) } Bigllfloorfrac{ ln ( {frac {n}{i}} ) }{ ln p_{{j}} } Bigrrfloor ln ( p_{{j}}) =0quadquadquadquadquadquadquadquadquadquad,,,,,,,,(operatorname{v})$$
$$sum _{i=1}^{pi (n+1 ) } sum _{j=1}^{ {bigllfloorfrac {ln ( n +1) }{ln ( p_{{i}} ) }}bigrrfloor +1} Bigllfloor {frac {n+1}{{p_{{i}}}^{j}}} Bigrrfloor ln
( p_{{i}}) -sum _{i=1}^{pi(n)} sum _{j=1}^{pi ( {lfloorfrac {n}{i}} rfloor ) } Bigllfloorfrac{ ln ( {frac {n}{i}} ) }{ ln p_{{j}} } Bigrrfloor ln ( p_{{j}}
) =ln(n+1)quadquadquadquadquadquad,,,,,,(operatorname{vi})$$
$$sum _{i=1}^{pi (n+1 ) } sum _{j=1}^{ {bigllfloorfrac {ln ( n ) }{ln ( p_{{i}} ) }}bigrrfloor +1} Bigllfloor {frac {n}{{p_{{i}}}^{j}}} Bigrrfloor ln
( p_{{i}}) -sum _{i=1}^{pi (n ) } sum _{j=1}^{ {bigllfloorfrac {ln ( n ) }{ln ( p_{{i}} ) }}bigrrfloor +1} Bigllfloor {frac {n}{{p_{{i}}}^{j}}} Bigrrfloor ln
( p_{{i}}) =0quadquadquadquadquadquadquadquadquadquad,,,,,,,,(operatorname{vii})$$
$$sum _{i=1}^{pi (n ) } sum _{j=1}^{ {bigllfloorfrac {ln ( n ) }{ln ( p_{{i}} ) }}bigrrfloor +1} Bigllfloor {frac {n}{{p_{{i}}}^{j}}} Bigrrfloor ln
( p_{{i}})=ln(n!)quadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquad,,,,,,(operatorname{viii})$$
$$sum _{i=1}^{pi (n+1 ) } sum _{j=1}^{ {bigllfloorfrac {ln ( n+1 ) }{ln ( p_{{i}} ) }}bigrrfloor +1} Bigllfloor {frac {n+1}{{p_{{i}}}^{j}}} Bigrrfloor ln ( p_{{i}}) -sum _{i=1}^{pi (n ) } sum _{j=1}^{ {bigllfloorfrac {ln ( n ) }{ln ( p_{{i}} ) }}bigrrfloor +1} Bigllfloor {frac {n}{{p_{{i}}}^{j}}} Bigrrfloor ln
( p_{{i}}) =ln(n+1)quadquadquadquad,,,,,,(operatorname{ix})$$
$$sum _{i=1}^{pi (n+1 ) } sum _{j=1}^{ {bigllfloorfrac {ln ( n ) }{ln ( p_{{i}} ) }}bigrrfloor +1} Bigllfloor {frac {n}{{p_{{i}}}^{j}}} Bigrrfloor ln
( p_{{i}})=ln(n!)quadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquad,,,,,(operatorname{x}) $$
$$sum _{i=1}^{pi (n ) } sum _{j=1}^{ {bigllfloorfrac {ln ( n+1 ) }{ln ( p_{{i}} ) }}bigrrfloor +1} Bigllfloor {frac {n+1}{{p_{{i}}}^{j}}} Bigrrfloor ln
( p_{{i}})=ln((n+1)!)quadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquad,,,,,,(operatorname{xi}) $$
$$sum _{i=1}^{pi (n ) } sum _{j=1}^{ {bigllfloorfrac {ln ( n+1 ) }{ln ( p_{{i}} ) }}bigrrfloor +1} Bigllfloor {frac {n+1}{{p_{{i}}}^{j}}} Bigrrfloor ln
( p_{{i}})-sum _{i=1}^{n } sum _{j=1}^{ {bigllfloorfrac {ln ( n ) }{ln ( p_{{i}} ) }}bigrrfloor +1} Bigllfloor {frac {n}{{p_{{i}}}^{j}}} Bigrrfloor ln
( p_{{i}})=sum _{j=1}^{N}alpha_{g(n,j)}ln(p
_{f(n,j)})deltaBiggl(frac{n}{2},Bigllfloor frac{n}{2}BigrrfloorBiggr)$$
$${{g(n,j),f(n,j)}} subset mathbb Nquadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquad,,,,,,quadquadquadquadquadquadquadquad(operatorname{xii})$$
Collectively imply:
$$sum _{i=pi (n )+1}^{pi (n+1 ) } sum _{j=1}^{ {bigllfloorfrac {ln ( n +1) }{ln ( p_{{i}} ) }}bigrrfloor +1} Bigllfloor {frac {n+1}{{p_{{i}}}^{j}}} Bigrrfloor ln
( p_{{i}})=ln(n+1)delta(tau(n+1),2)quadquadquadquadquadquadquadquad(0.2) $$
$tau(k)$ is the total number of divisors of $k$.
$delta(x,y)$ is the Kronecker delta function.
number-theory p-adic-number-theory
number-theory p-adic-number-theory
asked Sep 28 '18 at 18:17
Adam
54114
asked Sep 28 '18 at 18:17
Adam
54114
edited Nov 21 '18 at 8:05
asked Sep 28 '18 at 18:17
Adam
54114
asked Sep 28 '18 at 18:17
Adam
54114
asked Sep 28 '18 at 18:17
Adam
54114
54114
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
Editing a question bumps it, which is a problem when the number of edits is that large. I recommend you use the sandbox at: math.meta.stackexchange.com/questions/4666/…
– Michael Greinecker♦
Oct 5 '18 at 20:31
|
show 6 more comments
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
Editing a question bumps it, which is a problem when the number of edits is that large. I recommend you use the sandbox at: math.meta.stackexchange.com/questions/4666/…
– Michael Greinecker♦
Oct 5 '18 at 20:31
is this a question?
– Alexander Gruber♦
Oct 4 '18 at 17:08
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
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
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
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
Editing a question bumps it, which is a problem when the number of edits is that large. I recommend you use the sandbox at: math.meta.stackexchange.com/questions/4666/…
– Michael Greinecker♦
Oct 5 '18 at 20:31
Editing a question bumps it, which is a problem when the number of edits is that large. I recommend you use the sandbox at: math.meta.stackexchange.com/questions/4666/…
– Michael Greinecker♦
Oct 5 '18 at 20:31
|
show 6 more comments
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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
Editing a question bumps it, which is a problem when the number of edits is that large. I recommend you use the sandbox at: math.meta.stackexchange.com/questions/4666/…
– Michael Greinecker♦
Oct 5 '18 at 20:31