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I am reading an article about counting hexagonal p-minos (the article is in a combinatorics book) and I saw a notation I don't understand:

$0>(a,b)>-p$ .

$a,b,p$ are integers and so "$>$" means 'bigger', but what can be the meaning of "$(a,b)$" ?

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    perhaps ,"both $a$ and $b$ are less than $0$ and greater than $-p$" .2012-04-10
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    agree with pedja. $0>a>-p$ and $0>b>-p$. Brackets to denote difference to $0>a$ and $b>-p$.2012-04-10
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    Does it make sense for $(a,b)$ to mean $\gcd(a,b)$?2012-04-10
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    Can you provide us with some of the (relevant) sentences before and after this?2012-04-10
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    @Thomas $\gcd(a,b) > 0$ for all integers $a$ and $b$...2012-04-10
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    @pedja Yes, that's right. But I was wondering if it might be a form of "signed" gcd. So for example $\gcd(-4,2) = -2$... and with the $p$ which might denote a prime, that's where my thoughts went... anyway, just a thought. I am guessing that it probably doesn't make sense in the context of the problem...2012-04-10
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    I think that tou are right about that both a and b...p isn't prime and gcd doesn't make sense...thanks for the help!2012-04-10
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    It clearly does not mean that $0>a>-p$ and $0>b>-p$, since the full statement is about ‘the triangular region $x\le 0,y\le 0,z\ge p$ (see Fig. 9) with $a=b=0,0\ge(a,b)\ge-p$, and $0\le c'\le c\le p$’. (I’m assuming that this refers to the Lunnon article.)2012-04-10
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    @BrianM.Scott - yes, I am reffering to Lunnon article (how did you know ?). if it doesn't mean that, what does it mean ?2012-04-10
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    I knew from your earlier question. On further examination of Fig. 9, my best guess is that it’s a typo: $(a,b)$ should be $(a',b')$, and the intent is to say that $0\ge a',b'\ge -p$. I’ve not waded through the information on his coordinate system, though, so I make no guarantees!2012-04-10
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    @BrianM.Scott - I will look at this and try to understand..his coordinate system looks weird to me and I didn't really understand it yet. I also believe I found two other mistakes in this article (mainly delta is wrong..), perhaps you understood the algorithm itself ? I am unsure about how new cells are added (i.e. how do we use the growth criterion to add cells) and what does the algorithm do with p-mino that he built but did not touch the boundaries..(and how does he even get to the position that he have such a p-mino...the growth criterion should of helped in getting only complete p-minos)2012-04-10
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    also, thanks for the help, it's much appreciated!2012-04-10
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    No, I didn’t try to figure out the algorithm; to be honest, I just wasn’t interested enough to spend the time, since it looks like pretty dense going.2012-04-10

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From the comments it seems pretty clear that it means $0>a>-p$ and $0>b>-p$ (and also that $a$ and $b$ should be $a'$ and $b'$ respectively). The notation is awful, but I think I know why the author doesn't write $0>a,b>-p$: that can be easily misread as $0>a$, $b>-p$, which is a much weaker condition (look closely, if like me you don't see any difference, right-click on the formulas to see the TeX source). I regularly have difficulty avoiding this kind of ambiguity when writing; one could promise to the reader to never to write two conditions separated by just a comma that means "and", but that is an annoying constraint as well, in situations where one needs a somewhat complicated set like $\{(x,y)\in\mathbb R^2\mid x\geq 1, 0\leq y\leq x^2\}$ (not all readers are used to "$\land$" meaning "and"; by the way the perversion of writing "$(a,b)$" instead of "$\gcd(a,b)$" also sometimes takes the alternative form of writing it "$a\land b$"; ah, the delights of laziness…).

It is true that with his private notation the author has managed to make clear that he does not mean $0>a$ and $b>-p$, but at the price of totally obfuscating what he does mean, and all that to save a few keystrokes.

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    I disagree that never using "," to mean "and" is "an annoying constraint"; I consider an ironclad rule. I would write your example as $\{(x,y)\in\mathbb{R^2} \mid x\ge 1 \text{ and } 0\le y\le x^2\}$. Yes, with "and" spelled out.2012-04-11
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    @JeffE: I totally agree with you in principle. Yet "annoying" is subjective, and it sometimes happens that one has four or more conditions to combine, and that writing out "and" each time causes problems fitting the formula within the narrow margins, and so forth. And also: for this rule to function as disambiguation you not only need do adhere to it, but also make sure the reader is aware. So you _have to_ explicitly make the promise to the reader.2012-04-11