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My basic question that will relate to what I write here after is this: Is there any incorrect statements I have made in my description of the mathematics shown?

We start again with the Kronecker delta:

$$\delta \left( x,y \right) =\cases{1&$x=y$\cr 0&$x\neq y$\cr}\quad\qquad\quad\qquad\quad\qquad\quad\qquad\quad\qquad\quad\qquad\quad\qquad\quad\quad\quad\quad\quad\quad\quad\text{ (1)}$$

Which allows us to express the digits of a number 'a' in base 'b' in a computable integer sequence, in that we already know the exact length of the sequence which is of course the number of digits in total. The expression for this computation is:

$$d_{{n}} \left( a,b \right) =\sum _{k=1}^{ \Bigl\lfloor {\frac { \ln \left( a \right) }{\ln \left( b \right) }}\Bigr\rfloor +1} \left( \delta \left( n,k \right) -b\delta \left( n,k+1 \right) \right) \Bigl\lfloor{a{b}^{k- {\Bigl\lfloor\frac {\ln \left( a\right) }{\ln \left( b \right) }\Bigr\rfloor} -1}} \Bigr\rfloor \quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad$$

$$\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\qquad\quad\quad\quad\quad\quad\quad\quad\quad\quad\text{ (2)}$$

For example, $a=12345$ in base $b=10:$ will, purely coincidentally of course, evaluate to the arithmetic progression with initial value of 1 and d=1 of length 5: $$\left\{ d_{{1}} \left( 12345,10 \right) ,d_{{2}} \left( 12345,10 \right) ,d_{{3}} \left( 12345,10 \right) ,d_{{4}} \left( 12345,10 \right) ,d_{{5}} \left( 12345,10 \right) \right\} = \left\{ 1,2,3,4, 5 \right\} $$

But this(2) will compute the $n^{th}$ digit for the number in any base $b>1$, and thus these values correspond to the coefficients of the b-adic expansion * of the number thus we have as follows:

$$\mathcal{P} \left( a,b \right) =\sum _{n=0}^{ \Bigl\lfloor { \frac {\ln \left( a \right) }{\ln \left( b \right) }} \Bigr\rfloor +1}d_{ {n}} \left( a,b \right) {b}^{n}\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\text{ (3)} $$ Additionally, we may also use (2) for the calculation of the p adic valuation of the factorial of a number N for a prime p as follows:

$${\sum _{k=1}^{ \Bigl\lfloor {\frac {\ln \left( N \right) }{\ln \left( p \right) }} \Bigr\rfloor +1} \Bigl\lfloor {\frac {N}{{p}^{k}} } \Bigr\rfloor =\frac {N}{p-1}}-\frac{\sum _{j=1}^{ \Bigl\lfloor {\frac {\ln \left( N \right) }{\ln \left( p \right) }} \Bigr\rfloor +1}d \left( N,p,j \right)}{ \left( p-1 \right) }\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\text{ (4)}$$

And the lemma below I include for the reader to see my justification for the result stated in (6): where $ \text{There exists}\quad \alpha{\in \Bbb N}\quad\text{and}\quad\beta{\in \Bbb N}\quad\text{such that:}\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad $ $k \in {\{0,1,2,3,...,\Bigl\lfloor { \frac {\ln \left( a \right) }{\ln \left( \beta \right) }} \Bigr\rfloor\}}$ and $\alpha\in {\{\beta}^{k}\}$ $$\alpha\,\frac{\mathcal{P} \left( {\frac {\mathcal{P} \left( a,\beta \right) }{\beta}},\beta \right)} {\beta}=a\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\text{(5)} $$

Because of this almost "periodic" nature of an iteration of (3) returning us the original number in it's original base representation, I believe this to be why when any value greater that 1 is taken for b up to infinity, this recurrence reduces to a finite set under the axiomatic requisite of unique elements of a set to only values that are elements of the least residue system modulo the upper bound of a (N) excluding 0 will occur, thus having been enclosed in a nested set the infinite inner set will reduce to N-1 distinct values which will be those of an arithmetic progression with initial value of 1 and d=1 with total length N-1:

$\text{where } R_N\text{ is the least residue system modulo N:}\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad$

$ \left\{ \left\{ \frac{\mathcal{P} \left( {\frac {\mathcal{P} \left( a,b \right) }{b}},b \right) }{b} \right\} _{{b={2\ldots \infty }}} \right\} _{{a={1\ldots N-1}}} = R_{{N}} \backslash \left\{ 0 \right\}\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\text{(6)} $

We can then make a declaration of a congruence of $\mathcal{P} \left( a,b \right) $ and 0 modulo the product of $a$ and the base $b$ of the number system for which it is represented in:

$$\mathcal{P} \left( {\frac {\mathcal{P} \left( a,b \right) }{b}},b \right)\equiv 0\pmod {ab}\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\text{(7)} $$

I do however want to note in light of some recent concerning trends on the internet propagated by an individual who I will not name, that this in no way should be misconstrued as the author having a finitist view, this is entirely a consequence of the nature of this particular computation, please do not ask me if this is proof that "numbers have an end".

I have included this enumeration below if the reader is unclear as to what I have attempted to state in (4), as I do not know the formal terminology for describing this observed pattern.

$$\quad\,\mathcal{P} \left( \frac{1}{2}\,\mathcal{P} \left( 12,2 \right) ,2 \right) =12 $$ $$\quad\frac{1}{2}\mathcal{P} \left( \frac{1}{2}\mathcal{P} \left( 123,2 \right) ,2 \right) =123$$ $$\quad\qquad\mathcal{P}\left( \frac{1}{2}\mathcal{P} \left( 1234,2 \right) ,2 \right) = 1234\quad$$ $$\quad\quad\frac{1}{2}\,\mathcal{P} \left( \frac{1}{2}\mathcal{P} \left( 12345,2 \right) ,2 \right) =12345\quad$$

$$32\,\mathcal{P} \left( \frac{1}{2}\mathcal{P} \left( 123456,2 \right) ,2 \right) =123456 $$

$$\frac{1}{2}\mathcal{P} \left( \frac{1}{2}\mathcal{P} \left( 1234567,2 \right) ,2 \right) =1234567 $$

An additional note to make would be that one can clearly see this approach can be used to compute the digital root of a number or more readily the sum it's digits in any base, which has an intimate relationship with the p-adic order of the number, which I encourage the reader's review of here

I will end at this point to await feedback, seeing If anything I have stated above is incorrect, any of the further content on this subject must be brought into question.

Although it follows directly from assuming lemma (4) to be true, for this result I cannot verfiy to sufficiently high values without finding a more efficient means of calculation or purchasing a more powerful computer, so I encourage the reader to attempt to disprove or prove it as I will be, (unless this has already been done):

$$ \left\{ \left\{ \frac{\ln \left( \frac{a}{ \left( \mathcal{P} \left( {\frac {\mathcal{P} \left( a, \beta \right) }{\beta}},\beta \right) \right)} \right)}{ \left( \ln \left( \beta \right) \right) }+1 \right\} _{{a={1\ldots \infty }}} \right\} _{{\beta={2 \ldots \infty }}}=R_{{9}}\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\text{(8)} $$

It quite possibly can be explained by the fact that this is the least residue system for the number of distinct digits used in our number system, but so far everything here has been derived under the author's assumption that theoretically the base of the chosen number system can be any fixed natural number greater than 1, so if this were to be calculated in a number system of a higher base, we would obtain a least residue system of a correspondingly higher order, that is to say:

Defining $\mbox {U}_{{b}} \left( N \right) $ as the intersection of all of the digit sets D of all divisors of a number N over all number bases less than or equal to it's number base b:

$U_b(N)=\bigcap^{b}_{\beta=2}\bigcap^{\tau(N)}_{j=1}D(p_{N,b,j}^{v_{N,b,j}},\beta)$

The following distributions show the variance in the quantity with respect to N (first two plots) and the variance between consecutive values of b (last three plots):

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Adam Ledger
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  • ...to be blunt: I don't think there's anything that's incorrect, but I also don't think that any of this is saying anything meaningful, and it's using a lot of essentially pointless 'symbolage' to do so. – Steven Stadnicki Apr 28 '18 at 06:44
  • That's ok there is nothing wrong being with blunt, but you have a few other things you need to add to your comment in order for the criticism to be valid or helpful,1) which 'symbolage' is pointless, and what notation would you prefer? 2) meaningful to whom or what? if that is just your opinion which, yes, indeed it is just that, then I respect your right to have one, but I think you need to remember there is no absolute definition of 'meaning' I think you will find. – Adam Ledger Apr 28 '18 at 07:36
  • There have been several questions on your website that have asked relevant questions, the methods for which people have used have been incorrect and voted up sometimes, and my method is in not in anyway inferior to those that ive seen used, I'm a little sceptical in that I'm not sure you are making an unbias and accurate assessment here. – Adam Ledger Apr 28 '18 at 07:44
  • But that's fine all I wanted to know is if there was anything incorrect, that other stuff you said is kind of meaningless tbh. – Adam Ledger Apr 28 '18 at 07:47
  • I should note that it's not 'my' website; it's the community's site. As to the usage of symbols, I think you're using far too many formulas for things that are perfectly well defined (and well defined mathematically) by description. On the simplest example, you keep writing $\frac{\ln a}{\ln b}$ everywhere, but why not just $\log_ba$? – Steven Stadnicki Apr 28 '18 at 17:47
  • More relevantly, you don't need your complicated logs-and-deltas expression for $d_n(a,b)$;just say '$d_n(a,b)$ is the $n$th digit (from the left) of $a$ in its base-$b$ representation'. Likewise,your $\mathcal{P}$ is just a digit-reversing operation, postpending zero by the way you've defined it; not only is your more complicated expression $P(P(a,b)/b,b)\equiv 0\mod ab$, from everything I can tell it's exactly equal to $ab$ by definition. Basically, your overuse of 'explicit' formulae is making it hard to understand what the point is, or what you think is novel or interesting here – Steven Stadnicki Apr 28 '18 at 17:49
  • ok so how would I express a digit as a function of it's position in a less complicated way that details it's computation from the original number? And OK, no problem, I can change the ratio of the natural logs to $log_b(a)$ for you. And can you please show computation details of a more simple digit reversal operation, for the reader not satisfied with just reading "let P be the digit reversal function" – Adam Ledger Apr 28 '18 at 18:57
  • If you do know a simpler way of providing the same amount of information in a more simplified way, then great, I will make the necessary changes, but again, you are this type that says "I don't find this very interesting" instead of anything helpful. – Adam Ledger Apr 28 '18 at 19:03
  • In short - and I will try to write this up into an answer of sorts at some point - why do you think what you've written is less complicated than 'the $n$th base-$b$ digit of the number $a$' or 'let $\mathcal{P}$ be the digit reversal function'? More to the point, what benefit are you finding in the explicit formula that you've provided? – Steven Stadnicki Apr 29 '18 at 01:41
  • Ok, look the only thing achieved by what I have written is the explicit computation of the $n^{th}$ digit of the number in it's base b representation, that is the only thing over and above simply stating it that is achieved, but I think that this is important if we are to provide a rigorous explanation. If this is not significant in your view, then I will then point out several questions that have been asked here that require it. But at this point, I can't see any part of your critique as being meaningful in any way. – Adam Ledger Apr 30 '18 at 09:34
  • Let's rephrase the question slightly. Do you believe the explicit computation of the multiplicity of a prime $p$ in the integer factorization of a number $N=n!$ to be beneficial? And if so, would you be interested in utilizing the fact that this can be readily achieved with a closed form expression for the sum of the coefficients of the p adic expansion of n, in other words, the sum of it's digits in it's base p representation? – Adam Ledger May 03 '18 at 05:09
  • I'd say it's beneficial but also fairly well-understood; the usual way of expressing the formula is $\left|N!\right|p = \sum{k=1}^\infty \lfloor\frac{N}{p^k}\rfloor$ (which is directly related to the base-$p$ expansion of $N$, as you've noticed), and this leads to some nifty results like Kummer's Theorem; but the applications sadly don't go too much further than that. – Steven Stadnicki May 03 '18 at 23:56
  • (Also, I'd note that that Wikipedia page is a good example of how to talk about things like digit-sum, digits of a base-$p$ expansion, etc. in a way that's rigorous without relying on an explicit formula for each digit; instead they're implicitly defined, but in a fashion that allows for rigorous computation.) – Steven Stadnicki May 03 '18 at 23:58
  • lol awesome bro ill check it out – Adam Ledger May 08 '18 at 06:18

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(Caution: much of the below is my opinion, but I feel comfortable in saying that it's generally the 'common opinion' of the mathematical community.)

I think there's a misconception at work here that the goal of mathematics is to provide explicit formulas for things. Formulas can be a useful means to an end, but they are not an end in and of themselves; the ultimate goals of mathematics are to be able to understand mathematical structures and to be able to communicate understanding of those structures; a formula is useful insomuch as it provides either understanding or a useful means of communication.

Perhaps the canonical (negative) example of this is the explicit formulas for the prime numbers. For instance, by Wilson's formula, we have $(n-1)!+1$ exactly divisible by $n$ iff $n$ is prime (or $1$, of course), so $\left\lfloor\cos^2\left(\pi\dfrac{(n-1)!+1}{n}\right)\right\rfloor$ will be equal to $1$ iff $n$ is either $1$ or prime; this means that we can write the prime-counting function $\pi(n)$ as $\displaystyle \pi(n)=\sum_{j=2}^n\left\lfloor\cos^2\left(\pi\dfrac{(j-1)!+1}{j}\right)\right\rfloor$. But no meaningful discovery about the primes has come out of this formula! There's no structure in it; the pieces don't really cohere, and they're not especially amenable to deeper analysis.

By contrast, consider the explicit formula for the Fibonacci numbers $F_n=\dfrac{\phi^n-\bar{\phi}^n}{\phi-\bar{\phi}}$ (where $\phi=\frac12(1+\sqrt{5})$ is the golden ratio and $\bar{\phi}=\frac12(1-\sqrt{5})=-\phi^{-1}$ is its 'conjugate'). On the surface, it's not even clear that this should be an integer — but there's a lot of structure in this formula. Not only does it tell us exactly how fast the sequence grows (exponentially, with order of growth $\phi$) but we can also use it to prove things like the divisibility property $F_n|F_{mn}$ for all $m,n$ (since $\dfrac{F_{mn}}{F_n}=\dfrac{\phi^{mn}-\bar{\phi}^{mn}}{\phi^n-\bar{\phi}^n} = \sum_i\phi^{in}\bar{\phi}^{(m-i)n}$ and we can show that the latter expression is an integer); what's more, we can extend these results to general linear relations of the form $P_n=dP_{n-1}+P_{n-2}, P_0=0, P_1=1$ (with some small constraints on $d$) because all of these relations satisfy a formula with a nearly identical structure.

As things stand, the formulas that you've presented seem very much more akin to the explicit prime formulas than to the explicit formula for the Fibonaccis. Having an explicit formula for the digits of a number isn't a bad thing, but it still leaves the question: what understanding, or what communication, does it grant? I would contend that the formulas you've come up with are no better at expressing the notion of '$n$th digit of a number' than that phrase itself, and more to the point, they're no more rigorous than the usual definition. You need to show what your formula can do that can't be done with the usual definitions.

  • Sure I can see your point much more clearly here, but I find it very hard to come to terms with the statement that an explicit formula for something is no more rigorous that the phrase for something. If no formula is required for something as you say, then how are we to enumerate results with actual numerical arguments? when I show by example for a=12345 that in base 10 the formula provides us the sequence 1,2,3,4,5 this admittedly seems to be trivial, but I can immediately determine the 4th digit of the same number's base 3 representation as equal to 2 using the same formula – Adam Ledger May 02 '18 at 07:29
  • And it was through this means that I came to understand the concept of representing a number in different bases, and if finding the formula furthered my understanding, I just find it strange that the consensus is that nobody else would benefit in the same manner from it. – Adam Ledger May 02 '18 at 07:31