On the back of this question comes the natural question of whether the string $$1234567891011121314\!\cdots$$ is even a number at all. While that sort of question is vague, given the lack of generic definition for the word "number", I would feel comfortable answering this question in the affirmative if we knew that it were a $p$-adic number.

Now, my first impressions are that this number is $10$-adic integer (although it is not quite as easy to show this as I initially thought). However, it seems rather unlikely that it is $p$-adic for any prime $p$. Does anyone know how to show that it is or isn't $p$-adic, or if there are similar questions which have been answered— I mean, the $p$-adicity of strings like $2481632\!\cdots$ or $23571113\!\cdots$?

(Sidebar: I believe that if it is a $p$-adic number, it has to be a $p$-adic integer, but I admit that I could be mistaken here.)

EDIT: mixedmath's answer shows that if the question is interpreted with $1$ as the "leftmost digit" then the question makes no sense. However, KCd points out that this notation is frequently used when we intend $1$ to be the "rightmost digit". In this case, the question becomes formalizable (probably), and almost surely much more interesting.

Therefore, the question becomes whether or not the following $10$-adic integer is $p$-adic for some prime $p$:

$$\sum_{k=0}^\infty k\exp_{10}\left(k+\sum_{i=1}^k\lfloor\log(i)\rfloor\right)$$

where $\exp_{10}(x)$ is a [formal] power $10^x$. This probably still doesn't make perfect sense formally, but it is at least easy to imagine formalizing it. We might reasonably interpret it to be a statement about the existence of sequences $b_m\in\Bbb Z_p$ and $e_m,\, f_m,\, g_m\to\infty$ such that the finite sums agree up to a point:

$$\sum_{m=0}^{f_N} b_m p^m \equiv \sum_{k=0}^{g_N} k\exp_{10}\left(k+\textstyle\sum\lfloor\log(i)\rfloor\right) \qquad \text{mod} \exp_{10}(e_N)$$

for all $N\in\Bbb N$. Perhaps something regarding rearrangements would work a little bit better (for this formalism I worry about some inessential objections regarding instability of the ones digit).

If anyone would like to answer this other interpretation of the question, I would give a bounty for it.

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    $\begingroup$ A string of digits is not any kind of definite "number" at all since it is not clearly defined unless you explain what your notation means. For instance, if I wrote .23232323... with the periodic repetition of 2 and 3, what is this? Well, if it means a number in base 10 then it is 23/99. If it means a number in base 11 it equals 23/120 (usual decimal notation). If it means a number in base 47 it equals 1/96 (usual decimal notation). If you want your expression to be a $p$-adic integer then it can be for any $p > 9$. There is no universal meaning once and for all of that string of digits. $\endgroup$ – KCd Jan 3 '15 at 4:02
  • $\begingroup$ @KCd: If the string is a $10$-adic integer, then I find it hard to believe that there is ambiguity about what this string means (I admit I could be mistaken, but $\sum_{m\geq k} a_m 10^m$ seems pretty unambiguous to me). I posted the question under this assumption before realizing that I was not able to find an $a_m$ sequence that was obviously correct. So perhaps it is not $10$-adic, in which case I agree that the question doesn't make sense, and I would accept an answer along those lines. $\endgroup$ – Eric Stucky Jan 3 '15 at 4:05
  • $\begingroup$ aaand two minutes later I get an answer along those lines :) $\endgroup$ – Eric Stucky Jan 3 '15 at 4:12
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    $\begingroup$ If you meant for the string to be $1 + 2 \cdot 10 + 3\cdot 10^2 + \cdots + 9 \cdot 10^8 + 1 \cdot 10^9 + 0 \cdot 10^{10} + 1 \cdot 10^{11} + \cdots$, then yes it is a $10$-adic integer. But the same exact notation could mean an $11$-adic integer $1 + 2 \cdot 11 + 3 \cdot 11^2 + \cdots$ and so on. It is all a matter of context. In practice $10$-adic integers are very rarely used. $\endgroup$ – KCd Jan 3 '15 at 4:17

This is not an $n$-adic integer or a $p$-adic integer. A pivotal idea of a $p$-adic integer is that it can be well-represented in a sort-of-base-$p$, $$ x = \sum_{k \geq \ell}a_kp^k,$$ and in particular it makes sense to speak of the first coefficient, the coefficient of $1$. This amounts to understanding what the number is mod $p$, which is very dependent on knowing what the "last digit" is. But your string does not have a well-defined last digit.

It is for this reason that you might see $p$-adic integers written in the form $$\cdots a_3a_2a_1,$$ where the expansion is on the left. In such a way, we have well-defined final digits, and so we can find any of the finite $p$-adic expansions.

  • $\begingroup$ Thank you; that clarifies the issue nicely… or at least it works for me :) $\endgroup$ – Eric Stucky Jan 3 '15 at 4:09
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    $\begingroup$ I think it is a fair interpretation to take the first digit to be $1$, the next to be $2$, and so on. It is common to write $p$-adic integers left to right also (try to write them in a computer algebra package :)). I agree that being more explicit about the built-in powers of 10 in the expression would take away any ambiguity. $\endgroup$ – KCd Jan 3 '15 at 4:20
  • $\begingroup$ @KCd: ah, this is something I hadn't considered. Yes, if we consider the number as secretly being written left to right instead of right to left, then this is a $p$-adic number and integer, matching your comments to the OP above. $\endgroup$ – davidlowryduda Jan 3 '15 at 4:22
  • $\begingroup$ @KCd: For what it's worth, I was not aware of this convention and did not intend that. But really I should pass responsibility for intention down to the OP of the linked post, who probably passes it down to the OP of their linked post, who probably wouldn't care either way, but would probably prefer a definitive answer to a "the question is nonsense" answer (lots of speculation in this comment :D). In any case the question in that form seems more interesting, so I thank you for your input. $\endgroup$ – Eric Stucky Jan 3 '15 at 4:59
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    $\begingroup$ @KCd: Maybe someone in Israel wrote them, and forgot to set the alignment! :-) $\endgroup$ – Asaf Karagila Jan 3 '15 at 12:55

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