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The hypergeometric function $_2F_1(-n;-r;1;2)$ appears in many different situations. For instance, it counts the number of integer points within a sphere in the $l_1$ norm, i.e.,

$$_2F_1(-n;-r;1;2) = \mbox{#} \lbrace x \in \mathbb{Z}^n : |x_1| + \cdots + |x_n| \leq r \rbrace $$

and another formula for $_2F_1(-n;-r;1;2)$ is given by

$$_2F_1(-n;-r;1;2) = \sum_{i=1}^{\min\lbrace{n,r}\rbrace}{n \choose i}{r \choose i}2^i$$

Does anyone know an asymptotic formula for this function when $n$ is large?

I know there are some closed formulas for the Gaussian hypergeometric (i.e. $_2F_1(a,b;2,1)$) and in some other cases, but haven't find any clue in this case.

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The formula counts integers in the $l_2$ norm, not $l_1$. Presumably $l_1$ is what you meant, i.e. $|x_1|+\cdots+|x_n| \leq r$, not $|x_1|^2+\cdots+|x_n|^2 \leq r$. –  Noam D. Elkies Oct 9 '12 at 14:26
    
For fixed $r$, it is a polynomial in $n$ of degree $r$, so take leading term: ${}_2F_1(-n,-r;1;2) \sim {2^r n^r}/{r!}$ as $n \to \infty$. –  Gerald Edgar Oct 9 '12 at 15:25
    
These numbers are sometimes called Delannoy numbers; see mathworld.wolfram.com/DelannoyNumber.html. –  Ira Gessel Oct 10 '12 at 0:24
    
Thank you Noam. The mistake was corrected. Gerald, the approximation $2^r n^r/r!$ (volume of ball in the $l_1$ norm) is good for fixed $r$ and (very) large $n$. However, when $r$ also increases (for example, if r = O(n)), what should be the behavior? –  Campello Oct 11 '12 at 21:48
2  
The summands are all positive, and logarithmically concave down, so it's just a matter of locating the peak and estimating its width. If I did it right, when $r,n$ are large and comparable, the largest summand is around $i = n + r - \sqrt{n^2+r^2}$. Stirling's formula for $N!$ will let you estimate this maximal coefficient, and then a Gaussian-integral approximation around that peak will tell you the multiple of $\sqrt n$ to use for the peak width. The resulting formula won't be pretty but it should at least be correct. –  Noam D. Elkies Oct 11 '12 at 23:17
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