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DRAFTING REGION

I need to establish a proof or minimal counter example for the following:

$$f(n)=Biggllfloor {frac { left( left( n-1 right) !
right) ^{n}+ left( -1 right) ^{n}}{n}} Biggrrfloor +Biggllfloor {frac { left( left( n-1 right) !
right) ^{n}- left( -1 right) ^{n}}{n}} Biggrrfloor$$

$$n notin mathbb P land n gt 4 Rightarrow {Bigl{d_{j,f(n),p}}Bigr}_{j=lfloorln_p(f(n))rfloor+1-mathcal L_{n,p}..lfloorln_p(f(n))rfloor+1}={{p-1}}$$
$$quadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquadoperatorname{(MAIN)}$$

where $d_{j,N,p}$ is the $j^{th}$ digit of a number $N$ in it's base $p$ representation

*also for the above use of curly brackets is in reference to the use of standard set builder notation.

Or simply stated, if $n$ is a composite number greater than four, we are assured that the final $mathcal L_{n,p}$ digits of $f(n)$ in base $p$ are all equal to $p-1$.

Here are some numerical evaluations to clarify my premise:

begin{align*}
f left( 8 right) =&104084078179918440038399999999\
f left( 9 right) =&62585279654387966547717097783295999999999\
f left( 10 right) =
&7918817322448503864877511415653369155260776447999999999\
f left( 14 right) =
&1882896245429872511932094597390061412789401819279245213ddots\
&5414507239528144216026955851728814590763658312910110719ddots\
&999999999999999999999999999
end{align*}

Taking the following lemma to be true lead me to making the above conjecture:

$${Biggl{frac{(n-1)!^n-(-1)^n}{n}}Biggr}+{Biggl{frac{(n-1)!^n+(-1)^n}{n}}Biggr}=1 operatorname{iff} n not in mathbb P$$

$$quadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquadoperatorname{(A0)}$$

$${Biggl{frac{(n-1)!^n-(-1)^n}{n}}Biggr}+{Biggl{frac{(n-1)!^n+(-1)^n}{n}}Biggr}=frac{n-2}{n} operatorname{iff} n in mathbb P$$

$$quadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquadoperatorname{(A1)}$$

where ${{x}}$ is the fractional part of $x$

$operatorname{(A0)}$ and $operatorname{(A1)}$, (Which both readily follow from Wilson's Theorem) then lead me to stating the lemma:

$frac{n}{2} left( Biggllfloor {frac { left( left( n-1 right) !
right) ^{n}+ left( -1 right) ^{n}}{n}} Biggrrfloor +Biggllfloor {frac { left( left( n-1 right) !
right) ^{n}- left( -1 right) ^{n}}{n}} Biggrrfloor +1-2,{frac { left( left( n-1 right) !
right) ^{n}}{n}} right) -frac{1}{2}delta left( n,1 right) =cases{1&$n in mathbb P$cr 0&otherwisecr}
$

$$quadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquadoperatorname{(B0)}$$

EDIT:29/10/2018

Let

$$f_1(n)=Biggllfloor {frac { left( left( n-1 right) !
right) ^{n}+ left( -1 right) ^{n}}{n}} Biggrrfloor$$

$$f_2(n)=Biggllfloor {frac { left( left( n-1 right) !
right) ^{n}- left( -1 right) ^{n}}{n}} Biggrrfloor$$

We have the following parity relations:
$$(f_1(n)-f_2(n))(-1)^{delta(frac{n}{2},lfloorfrac{n}{2}rfloor)}-1=0$$

$$quadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquadoperatorname{(C0)}$$

$$f_1(n)(-1)^{delta(frac{n}{2},lfloorfrac{n}{2}rfloor)}+f_2(n)(-1)^{delta(frac{n+1}{2},lfloorfrac{n+1}{2}rfloor)}+1=0$$

$$quadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquadoperatorname{(C1)}$$

Concluding Remarks:

I know that this proof must involve Kummer's Theorem somehow, and not only for the proof, in the case that the conjecture is true, derive an exact expression for $mathcal L_{n,p}$, which based on inductive reasoning alone seems to be $n-2$, however I think this would be a great opportunity for someone to teach me how to apply Kummer's Theorem in a proof for a specific problem involving p-adic integers.

My current intuitive view is directing me to focus on the factor of $frac{1}{2}$ we see in $operatorname{(B0)}$,since there is a constant factor in the denominator that is the minimum prime digit for $p=10$,($2$) which is non zero if and only if $n$ has primality $1$,conversely, we see the trailing digits in $f(n)$ are all equal to the maximum digit value in base $p=10$,$(9)$ if and only if $n$ has primality of $0$ in $operatorname{(MAIN)}$ .

The following lemma are of primary consideration for the proof of the main question of this post also:

Let $mathcal K_{N,m,0}$ be the cardinality of the finite set :

$$S_0=
{Biggl{frac { left( (ncdot p_k)! right) ^{m}pm left( -1 right) ^{ncdot m+1}}{p_k}-Biggllfloor frac { left( (ncdot p_k)! right) ^{m}pm left( -1 right) ^{ncdot m+1}}{p_k}Biggrrfloor:n leq N land k leq N}Biggr}
$$

$$mathcal K_{N,m,0}=delta left( frac{m}{2},Bigllfloor frac{m}{2} Bigrrfloor right) left( -1 right) ^{delta left( frac{m}{2},Bigllfloor frac{m}{2} Bigrrfloor right) }p_{{pi left( N right) }}
+ Bigl( 2-delta left( frac{m}{2},Bigllfloor frac{m}{2} Bigrrfloor right) Bigr) left( N-1 right) quadquadquadquadquadquadquad(operatorname{WILSON003)}$$

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

$pi(x)$ is the prime counting function.

Let $mathcal K_{N,m,1}$ be the cardinality of the finite set :

$$S_1=
{Biggl{frac { left( (ncdot p_k)! right) ^{m}pm left( -1 right) ^{n}}{p_k}-Biggllfloor frac { left( (ncdot p_k)! right) ^{m}pm left( -1 right) ^{n}}{p_k}Biggrrfloor:n leq N land k leq N}Biggr}
$$

we have:

$mathcal K_{N,m,1}=2N-1quadquadquadquadquadquadquad(operatorname{WILSON004)}$

Linked Questions relevant:

First

Second

Third

Fourth

Fifth

Many thanks in advance.