Please look at
https://mathoverflow.net/questions/175757/conjecture-related-to-distinct-unique-permutations-of-digits. In my view, it is clearly a specific research question - it is not a general mathematics question ... What about the portion of the question where I am asking for references with regards to the topic of my question ? If though the question "denied" is indeed a general trivial mathematical question - would it be possible at least out of sheer politeness to give the references to the sources where the particular topic is covered ?

  • 2
    $\begingroup$ I would just like to say that I think the question would be helped by using LaTeX; it's very hard to read the mathematics (for example, I have a hard time seeing how exponents are bracketed). $\endgroup$
    – Todd Trimble Mod
    Jul 12, 2014 at 19:17
  • 7
    $\begingroup$ I would have closed that question as "unclear what you're asking". Your terminology is very unclear, and after reading it several times I still am not sure what you're trying to ask. $\endgroup$ Jul 12, 2014 at 20:19
  • $\begingroup$ @Eric Wofsey - Eric, I (actually the program) only looked at pairs within the range of bases from 2 to 10 ... And the formula which I have provided for the value of the ratio of such pairs in the given base is empirical (based on the above said range of the bases). I was asking to provide theoretical proof (or disproof), which would cover all possible bases. $\endgroup$
    – Alex
    Jul 12, 2014 at 20:39
  • $\begingroup$ @Eric Wofsey (continuation) - In more detailed language I was asking to provide a theoretical proof (or disproof) of my statement (that in every base such pairs exist) and also I was asking to confirm (or otherwise) that the formula derived by me (for the ratio value of such pairs) covers all possible bases. $\endgroup$
    – Alex
    Jul 12, 2014 at 20:59
  • 7
    $\begingroup$ The MO question is incomprehensible. Please edit it into a form in which someone other than yourself will be able to understand it, and then ask for it to be reopened. Maybe get someone to help you. $\endgroup$ Jul 13, 2014 at 1:26
  • $\begingroup$ @Gerry Myerson - I rewrote my MO question. Could you read it and let me know whether it became comprehendible now? Thanks in advance ! $\endgroup$
    – Alex
    Jul 13, 2014 at 3:04
  • $\begingroup$ Since responses to my post here suggested that my MO question lacks clarity - I rewrote it and reposted under the following link mathoverflow.net/questions/176028/… $\endgroup$
    – Alex
    Jul 14, 2014 at 3:12
  • $\begingroup$ I also put "reference request" tag onto my question - since if the issue is trivial as it was suggested - then surely those experts should be able to refer me to the source where my question was already covered ...- that is so much more helpful then downvoting and practically suggesting reference takes the same amount of time as downvoting $\endgroup$
    – Alex
    Jul 14, 2014 at 3:26
  • 11
    $\begingroup$ The problem isn't that the question is "trivial", it's that no one knows what the words you're using mean. For instance, what is a "complete set of all distinct permutations in [a] numerical base"? I think you're asking something about taking a number $n$ and permuting its digits to get a number $n'$ and counting how many ways there are to get some particular ratio $n/n'$ that way, but I'm not entirely sure. If you want to get any sort of helpful answer, you need to clearly define the terminology you're using. $\endgroup$ Jul 14, 2014 at 8:09

2 Answers 2


So following Eric's thoughts (which reflect my own), I would suggest (as a start) improving the notation and giving some examples of the basic concept. (It was only after looking at a relevant MSE post that I began to have a little idea what you were talking about.)

Listen: we mathematicians are human and sometimes need a little help. My impression is that you were trying to say too much in your post without giving a proper introduction. I see that you've posted before here with something on convergents to $\pi$ that attracted people's interest, and so it's possible that you're onto something interesting here as well, only people are having a hard time seeing what it is.

Please don't accept this answer as yet; I want to have a crack at rewriting some of it a little later. Or, perhaps better yet, please open up another answer box and try this yourself. If people can understand it, then maybe we can consider reopening or asking anew.


Ah: the question was confusing to me because there was this mention of some curious coincidences involving ratios of certain "permutation pairs", and counting the number of such and relating this to the Euler $\phi$ function; this seems to me to be the real meat of the OP's concerns. But if the OP just wants confirmation of his empirical formula for one such special ratio [and that is indeed the concrete question I see], then if I understand the question correctly, the answer is very easy and elementary (and thus would not be appropriate for MO).

So: my current understanding is that the question is about a formula which computes the analogue in any base $r$ (the "radix") of the base 10 expression

$$\frac{9876543210}{1234567890} = 8.0000000729.$$

In other words, OP wants a closed form that calculates

$$\frac{\sum_{k=1}^{r-1} k r^k}{\sum_{k=1}^{r-1} k r^{r-k}}$$

As for references which deal with this, there must be many, but the techniques required can be found in any decent text which deals in generating functions as used in combinatorics and probability, such as Concrete Mathematics by Graham, Knuth, and Patashnik. The numerator can be evaluated as the value $f(r)$ of the function

$$f(x) = x(\sum_{k=1}^{r-1} x^k)' = x \cdot \left(\frac{x^r - 1}{x-1}\right)'.$$

The denominator can similarly be evaluated as the value $g(r)$ of the function

$$g(x) = x^{r-1} (\sum_{k=1}^{r-1} x^{-k})' = x^{r-1} \cdot \left(\frac{x^{-r} - 1}{x^{-1} - 1}\right)'.$$

So it just becomes a matter of a little tedious calculus at this point. I'll note that the author's $n$ here is $r-1$. His formula does check out for $n=9$, and I'm sure it's correct even without doing the tedious calculus.

Edit 2:

Probably the author does want more than the elementary derivation above. My guess is that he is interested in counting how many ratios of "pairs of permutations", where here "permutation" means an expression of the form

$$\sum_{k=0}^{r-1} \sigma(k) \cdot r^k$$

where $\sigma: \{0, 1, \ldots, r-1\} \to \{0, 1, \ldots, r-1\}$ is a bijection, are exactly equal to the special ratio whose closed form was computed above. He notes in the mathematics.se post (that he links to) that there are $7$ (yes, $7$) such distinct pairs in the base $r = 10$ case. This type of question really does seem curious and of potential number-theoretic interest, although I haven't thought very hard about it yet.

  • 3
    $\begingroup$ Here is the introduction I just have put in my MO question: Consider, for example, the set S10, containing all possible distinct permutations of all digits in base 10: 1,2,...,8,9,0 One could see that the one pair (P) in that set {9876543210,1234567890} yields the ratio 9876543210/1234567890=8.0000000729... Then one could find another pair in this set (S10), which gives the same ratio 7901234568/987654312=8.0000000729... Then the question arises whether there are other pairs (P)in this set S10 which have the same ratio ... $\endgroup$
    – Alex
    Jul 14, 2014 at 13:28
  • $\begingroup$ I wanted to continue my intro within the MO question ... but it is locked now ... so I will do it here ... Another question which arises - is there only one ratio in the set S10 for which multiple pairs exist ? And, finally, same questions could be extended to distinct permutations in other (than base 10) sets Sn ... $\endgroup$
    – Alex
    Jul 14, 2014 at 13:57
  • 1
    $\begingroup$ Yes, that's a good start. But I didn't understand what this expression with terms like $(n+1)^n$ and the like is supposed to count. Could you explain that as well? $\endgroup$
    – Todd Trimble Mod
    Jul 14, 2014 at 14:01
  • $\begingroup$ It appears empirically (based on computer program results for bases from 2 to 10) that the value of the ratio could be expressed as (n^2 (n+1)^n-(n+1)^n+1) / (-n^2+n (n+1)^n+(n+1)^n-n-1) for n=1 ... infinity where n = r - 1 For example, if you will take the base 10 (r = 10) then n = r - 1 = 9 . If you put n = 9 into above formula - you will get 8.0000000729 ... $\endgroup$
    – Alex
    Jul 14, 2014 at 14:21
  • $\begingroup$ I still can't read that formula. Could you please write that in LaTeX? $\endgroup$
    – Todd Trimble Mod
    Jul 14, 2014 at 14:26
  • $\begingroup$ I don't know LaTex I only could put $ $ around the formula to make it "MathJax"ed ... Will this suffice ? $(n^2 (n+1)^n-(n+1)^n+1) / (-n^2+n (n+1)^n+(n+1)^n-n-1)$ $\endgroup$
    – Alex
    Jul 14, 2014 at 14:29
  • $\begingroup$ @Alex I think that the issue lies not just in LaTeX but in absence of brackets. For instance, when you write (n+1)^n+1, do you mean (n+1)^(n+1)? Or, in LaTeX, $(n+1)^n+1$ versus $(n+1)^{n+1}$ $\endgroup$
    – Yemon Choi
    Jul 14, 2014 at 17:55
  • $\begingroup$ @Yemon Choi It follows from the precedence of operations - but if you want - I will use the brackets for your example, but they are really not necessary ((n + 1)^n) + 1 $\endgroup$
    – Alex
    Jul 14, 2014 at 18:05
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    $\begingroup$ @Alex Humans are not computers, so we often prefer some extra unnecessary brackets to make sure we understand everything correctly, especially since we can never be sure that there are not some other brackets missing due to a typo or similar. $\endgroup$ Jul 14, 2014 at 18:33
  • $\begingroup$ @TobiasKildetoft - acknowledged. Wouldn't it be nice if the same caution approach would be used by some quick "down-voters" ? $\endgroup$
    – Alex
    Jul 14, 2014 at 18:39
  • $\begingroup$ What is puzzling that even by now when I have beaten here the explanation of what I have meant in my MO question "to death", there are still no comments or answers pointing me to references, where my question is already covered. There are also no answers showing that my findings come out as corollaries of "this" and "that" ... $\endgroup$
    – Alex
    Jul 14, 2014 at 18:54
  • $\begingroup$ I think, Todd, that what Alex wants is not quite what you write. I think the question is, given $n$, find integers $a,b$ such that there are as many $k$ as possible such that both $ka$ and $kb$ use all $n$ "digits" exactly once when written to base $n$ (with a leading zero permitted). It happens that in base 10 (and maybe in the other bases already studied) the ratio $a/b$ is that special ratio you discuss; perhaps it is part of the question, whether that ratio maximizes the number of integers $k$ for all $n$. But, gosh, it's hard to figure out what's meant! $\endgroup$ Jul 15, 2014 at 0:49
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    $\begingroup$ Okay, glad that this is getting sorted out. The MO post will unlock in about 10 hours. You might want to try rewriting your post in an answer box below; Gerry's comment seems clear and could be useful to you here. If what you write then seems clear to readers, then I think you could copy it over to the MO post that is currently under lock after the lock releases, and we'll see what happens. $\endgroup$
    – Todd Trimble Mod
    Jul 15, 2014 at 3:14
  • 1
    $\begingroup$ @Todd Trimble - please do not unlock my MO question yet. As you have suggested - first I want to formulate my question to acceptable by you and Gerry context here in meta. Thanks for your patience. I appreciate your skills, which you have shown as a Moderator ! $\endgroup$
    – Alex
    Jul 15, 2014 at 13:15
  • 1
    $\begingroup$ The unlocking takes place automatically (actually it is unlocked now), but don't worry about it. Just post an answer here and let's have a look, and whenever you feel ready you can copy it over to MO. Glad I could help in a small way. :-) $\endgroup$
    – Todd Trimble Mod
    Jul 15, 2014 at 13:48

Some numerical experimentation (which could be considered recreational mathematics) has led to some curious observations which I don't properly understand but which might have number-theoretic import. I am hoping experts here can shed some light. If this is not considered appropriate for MathOverflow, then I apologize, and I ask for advice on a more suitable place to ask.

Among recreational enthusiasts, it has often been noted that the rational number


is very close to being an integer (the exact value is $8.0000000729$). What seems to be less well-known is that there are a number of such fractions, in which each digit appears exactly once in the numerator and denominator, with the same value as above, provided that we allow $0$ to be a leading digit:

$$\frac{7901234568}{0987654312}\;\; \frac{6913580247}{0864197523}\;\; \frac{4938271605}{0617283945}\;\; \frac{3950617284}{0493827156}\;\; \frac{1975308642}{0246913578}\;\; \frac{0987654321}{0123456789}$$

Each of these has the same value $8.0000000729$, or $\frac{109739369}{13717421}$ as a rational number in reduced form.

In more formal terms, there are thus $7$ pairs of permutations $(\phi, \psi)$ on the set $\{0, 1, 2, 3, 4, 5, 6, 7, 8, 9\}$ such that the base 10 expansions

$$\frac{\sum_{i=0}^9 \phi(i) \cdot 10^i}{\sum_{i=0}^9 \psi(i) \cdot 10^i}$$

all yield the same value $\frac{a}{b}$. Some extensive computer experimentation, courtesy of R. Cano and described in detail in a Mathematics Stackexchange thread here, suggests that if we define two permutation pairs $(\phi, \psi)$ (with $\phi \neq \psi$) to be equivalent if they yield the same rational value as displayed above, then $7$ is the size of the largest equivalence class, and this size is attained just for the value $\frac{a}{b} = \frac{109739369}{13717421}$ or its reciprocal.

My questions have to do with what happens in other bases besides base 10.

Question 1: For other bases $r$, consider permutations $\phi$ on the set $\{0, 1, \ldots, r-1\}$, and again define an equivalence relation on pairs of permutations $(\phi, \psi)$ (again with $\phi \neq \psi$) where two such are equivalent if they yield the same rational value $$\frac{\sum_{i=0}^{r-1} \phi(i)\cdot r^i}{\sum_{i=0}^{r-1} \psi(i)\cdot r^i}.$$ What is the size of the largest equivalence class? Calculations for the cases $r = 2, 3, \ldots, 10$ seem to yield, respectively, the sizes $2,2,3,3,5,3,7,5,7$. This looks suspiciously like one of the OEIS sequences here, all having to do with Euler's totient function.

Question 2: Which rational value $\frac{a}{b}$ represents this largest equivalence class? Putting $n = r-1$, my conjecture is that
$$\frac{(n^2 - 1) (n+1)^n + 1}{n(n+1)^n - n^2}$$ yields the maximum number, although there could be more than one such fraction. (Note that this is just a closed form expression for the base-$r$ expression $\frac{n\; n-1\; \ldots\; 1}{1\; 2\; \ldots\; n}$, analogous to $\frac{987654321}{123456789}$.)

There are other questions that could be asked; more can be found at the Mathematics StackExchange link given above. But this would be a good start for me.

(End of suggested edit. What appears below is an earlier version of the question.)

As an introduction, consider, for example, the set S10, containing all possible distinct permutations of all digits in the base 10: 1,2,...,8,9,0

One could see that the one pair (P) in that set {9876543210,1234567890} yields the ratio 9876543210/1234567890=8.0000000729...

Then one could find another pair in this set (S10), which gives the same ratio


Then the question arises whether there are other pairs (P) in this set (S10) ,which have the same ratio ...

Another question which arises - is there only one ratio in the set (S10) for which multiple pairs exist ?

And, finally, same questions could be extended to distinct permutations in other (than base 10) sets (Sn).

Also (as shown later) it appears empirically (based on computer program results for bases from 2 to 10) that the value of the ratio could be expressed as

(1)$$(n^2 (n+1)^n-(n+1)^n+1) / (-n^2+n (n+1)^n+(n+1)^n-n-1)$$ for n=1 ... infinity, where n = r - 1 and where r is the radix of the of the base.

For example, if one will take the base 10 (r = 10) then n = r - 1 = 9 If then one puts n = 9 into above formula - he will get 8.0000000729

So now I will try to generalize my questions as following:

According to an exhaustive computer program written for numerical bases 2-10 (which searched for the maximum number of pairs with the smallest, less than "n", but wherever is possible greater than 1 ratio(R) ), it was discovered that within an entire set of distinct permutations in the covered range, pairs (P) of permutations, which satisfy above stated conditions can be found, and that the number of such pairs is equal to: {2,2,3,3,5,3,7,5,7,...}. If one would consider the latter as an integer sequence - it might be (according to OEIS A039649, A039650, A214288) related to phi, which is the Euler totient function .

I have derived the above empirical formula (1) for the values of the ratio (R) for pairs in the given base (based on conditions as defined above): This formula (1) could be also expressed as A221740(n)/A221741(n),
where A221740 and A221741 are OEIS's integer sequences (submitted by me) to cover the values, generated by the numerator and denominator (correspondingly), which appear in the formula (1).

Gerry Myerson rephrased my question as:
given "n", find integers "a", "b" such that there exists "k" (greater than one, if possible, but less than "n"), so "ka" and "kb" use all n "digits" exactly once when written to base "n (with a leading zero permitted).

In Gerry's terms "n" is what I call radix "r", {ka, kb} is what I call to be a pair (P). What he defines as "k", I call ratio (R) - it could be expressed as k=i/l, where both "i" and "l" are integers. For each "n" the value of "k" is different. As empirical formula shows, both "i" and "l" (and therefore "k") are functions of "n". In each covered base "n" (from 2 to 10) several pairs, satisfying that "n" specific ratio ("k" or "R") were found - the number of pairs for given "n" is also a function of "n".

Based on obtained results two following conjectures are made:
1) Every complete set of all distinct permutations in any numerical base (radix) r contains some prime number of pairs (P) with the ratio (R) as defined above.

2)The ratio (R) could be calculated by the (empirical) formula (1).

If anyone has references already covering this particular topic - please provide them - I will appreciate such reference.

PS My analysis of results on this issue are described in
and in

Could someone provide analytical proof or disproof of my conjectures stated above?

  • $\begingroup$ Thanks, Alex. I'll try to respond to this a little later today (sometime in the next six hours, I hope). $\endgroup$
    – Todd Trimble Mod
    Jul 15, 2014 at 15:39
  • $\begingroup$ @Todd Trimble - I see that you are busy - meanwhile as I noticed - downvoting on my MO question, as it was formulated originally, is continuing ...So since you have said:"whenever you feel ready you can copy it over to MO" - I have followed your advice and copied the new version of my answer from here to MO. Thanks, Alex. $\endgroup$
    – Alex
    Jul 16, 2014 at 10:46
  • $\begingroup$ Sorry for the delay. Would you mind if I tried editing your answer above? I think it could be tightened and the formatting improved. For one I would use LaTeX, for another I would insert some links, and finally I would try to make it more to-the-point. (We may have to accept that this question already has a troubled history and just isn't going to fly at MO after all.) $\endgroup$
    – Todd Trimble Mod
    Jul 16, 2014 at 12:14
  • $\begingroup$ @Todd Trimble - surely, please do it ! $\endgroup$
    – Alex
    Jul 16, 2014 at 12:56
  • $\begingroup$ @Todd Trimble - irregardless of what will happen to my question - your mentioning of how "troubled history" may affect the outcome of accepting or rejecting someone's question on MO is interesting on its own. Perhaps, being a moderator, you could write some helpful memo advising those, who are new to MO, how it is important to workout the question prior to posing it on MO. $\endgroup$
    – Alex
    Jul 16, 2014 at 18:07
  • $\begingroup$ Okay, I have written a reformulation that I suspect would gain better traction. Please read to make sure I got it right. If it looks right, I am happy to unlock the question and post it, with an explanatory note at the beginning. $\endgroup$
    – Todd Trimble Mod
    Jul 16, 2014 at 18:22
  • $\begingroup$ @Todd Trimble Oops - I misread your message and posted your version in the form of the edit. I will add the note crediting your generous contribution in there. Thanks again! $\endgroup$
    – Alex
    Jul 16, 2014 at 18:41

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