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

$$f(n)=\Biggl\lfloor {\frac { \left( \left( n-1 \right) ! \right) ^{n}+ \left( -1 \right) ^{n}}{n}} \Biggr\rfloor +\Biggl\lfloor {\frac { \left( \left( n-1 \right) ! \right) ^{n}- \left( -1 \right) ^{n}}{n}} \Biggr\rfloor$$

$$n \not\in \mathbb P \land n \gt 4 \Rightarrow {\Bigl\{d_{j,f(n),p}}\Bigr\}_{j=\lfloor\ln_p(f(n))\rfloor+1-\mathcal L_{n,p}..\lfloor\ln_p(f(n))\rfloor+1}={\{p-1}\}$$ $$\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\operatorname{(MAIN)}$$

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) = &1882896245429872511932094597390061412789401819279245213\ddots\\ &5414507239528144216026955851728814590763658312910110719\ddots\\ &999999999999999999999999999 \end{align*}

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

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


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


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:

$\frac{n}{2} \left( \Biggl\lfloor {\frac { \left( \left( n-1 \right) ! \right) ^{n}+ \left( -1 \right) ^{n}}{n}} \Biggr\rfloor +\Biggl\lfloor {\frac { \left( \left( n-1 \right) ! \right) ^{n}- \left( -1 \right) ^{n}}{n}} \Biggr\rfloor +1-2\,{\frac { \left( \left( n-1 \right) ! \right) ^{n}}{n}} \right) -\frac{1}{2}\delta \left( n,1 \right) =\cases{1&$n \in \mathbb P$\cr 0&otherwise\cr} $




$$f_1(n)=\Biggl\lfloor {\frac { \left( \left( n-1 \right) ! \right) ^{n}+ \left( -1 \right) ^{n}}{n}} \Biggr\rfloor$$

$$f_2(n)=\Biggl\lfloor {\frac { \left( \left( n-1 \right) ! \right) ^{n}- \left( -1 \right) ^{n}}{n}} \Biggr\rfloor$$

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




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 :

$$S_0= {\Biggl\{\frac { \left( (n\cdot p_k)! \right) ^{m}\pm \left( -1 \right) ^{n\cdot m+1}}{p_k}-\Biggl\lfloor \frac { \left( (n\cdot p_k)! \right) ^{m}\pm \left( -1 \right) ^{n\cdot m+1}}{p_k}\Biggr\rfloor:n \leq N \land k \leq N}\Biggr\} $$

$$\mathcal K_{N,m,0}=\delta \left( \frac{m}{2},\Bigl\lfloor \frac{m}{2} \Bigr\rfloor \right) \left( -1 \right) ^{\delta \left( \frac{m}{2},\Bigl\lfloor \frac{m}{2} \Bigr\rfloor \right) }p_{{\pi \left( N \right) }} + \Bigl( 2-\delta \left( \frac{m}{2},\Bigl\lfloor \frac{m}{2} \Bigr\rfloor \right) \Bigr) \left( N-1 \right) \quad\quad\quad\quad\quad\quad\quad(\operatorname{WILSON003)}$$

$\delta(x,y)$ 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 :

$$S_1= {\Biggl\{\frac { \left( (n\cdot p_k)! \right) ^{m}\pm \left( -1 \right) ^{n}}{p_k}-\Biggl\lfloor \frac { \left( (n\cdot p_k)! \right) ^{m}\pm \left( -1 \right) ^{n}}{p_k}\Biggr\rfloor:n \leq N \land k \leq N}\Biggr\} $$

we have:

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

Linked Questions relevant:






Many thanks in advance.

  • $\begingroup$ Is $\mathcal L_{n,p}$ a number which is defined somewhere else and I don't see the definition, or is it an unknown quantity and part of your question is how to compute it? $\endgroup$ – Torsten Schoeneberg Oct 25 '18 at 22:50
  • $\begingroup$ yes the later is correct sir I thought I had mentioned this somewhere but I will try and clarify it, it appears to be $n-2$ but I have not been able to establish a derivation for it, only that it does appear to have proportion to the total number of digits of of $f(n)$. $\endgroup$ – Adam Oct 25 '18 at 23:15
  • $\begingroup$ And it's my current view that at least something similar to Kummer's theorem being applied would both establish a proof for the problem and allow for deriving an exact expression for $\mathcal L_{n,p}$ $\endgroup$ – Adam Oct 25 '18 at 23:16

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