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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 $-p$" . – pedja Apr 10 '12 at 13:24
agree with pedja. $0>a>-p$ and $0>b>-p$. Brackets to denote difference to $0>a$ and $b>-p$. – example Apr 10 '12 at 13:25
Does it make sense for $(a,b)$ to mean $\gcd(a,b)$? – Thomas Apr 10 '12 at 13:29
Can you provide us with some of the (relevant) sentences before and after this? – Doug Spoonwood Apr 10 '12 at 13:42
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! – Brian M. Scott Apr 10 '12 at 22:43
up vote 3 down vote accepted

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. – JeffE Apr 11 '12 at 7:51
@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. – Marc van Leeuwen Apr 11 '12 at 7:57

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