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Let $a_n$ be the number of $n\times n$ zero-one matrices such that in any row and any column there are exactly two 1.

What is the value of $a_n$ in terms of $n$ ?

A A
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    trying to find a recursion relation. – A A Jul 27 '13 at 19:09
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    The reason I am asking is, if you write down your steps, somebody may pick it up and continue from that, people will be more willing to help, you will get an answer much faster. BTW, I have tried and basically nothing worked would also be a perfectly acceptable answer. – Lord Soth Jul 27 '13 at 19:12
  • Are the matrices of size $n\times n$? – Alex R. Jul 27 '13 at 20:41
  • yes matrices are of size $n\times n$ – A A Jul 27 '13 at 21:03
  • Figure it out for the first eight or so values of $n$ and try the sequence at O.E.I.S.; it's almost certain to be there. – coffeemath Jul 27 '13 at 22:23

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This is OEIS A001499; the entry has numerous references. The first few non-zero values are $a_2=1$, $a_3=6$, $a_4=90$, $a_5=2040$, $a_6=67950$, and $a_7=3110940$. The entry gives the exact formula

$$a_n=\frac{n!(n-1)\Gamma\left(n-\frac12\right)}{\Gamma\left(\frac12\right)}\cdot{_1F_1}\left[2-n;\frac32-n;-\frac12\right]\;,$$

the asymptotic formula

$$a_n\sim 2\sqrt{\pi}\left(\frac{n}e\right)^{2n+\frac12}\;,$$

and the recurrence

$$a_n=\frac12n(n-1)^2\left((2n-3)a_{n-2}+(n-2)^2a_{n-3}\right)\;.$$

Brian M. Scott
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