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Jan 28, 2013 at 20:33 comment added Felix Goldberg It turns out there is an even simpler way to show that $S(2,k,v)$ has no parallel blocks for $v \geq 8$: a $(v,k,1)$-design with $b>v$ must be quasi-symmetric with $x=0,y=1$. And for $S(2,k,v)$ we have $b>v$ iff $v \geq 8$. QED. Thanks again, Yuichiro, for the nice solution!
Dec 28, 2012 at 10:53 vote accept Felix Goldberg
Dec 27, 2012 at 22:16 history edited Yuichiro Fujiwara CC BY-SA 3.0
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Dec 27, 2012 at 20:24 history edited Yuichiro Fujiwara CC BY-SA 3.0
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Dec 27, 2012 at 18:34 history edited Yuichiro Fujiwara CC BY-SA 3.0
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Dec 27, 2012 at 18:25 history edited Yuichiro Fujiwara CC BY-SA 3.0
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Dec 27, 2012 at 17:59 history answered Yuichiro Fujiwara CC BY-SA 3.0