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Jun 19, 2022 at 21:44 comment added Vladimir Dotsenko @TheAmplitwist this was written 12 years ago so I am not too sure, but probably the two articles are doi.org/10.1007/s11853-008-0021-4 and doi.org/10.1007/s11040-006-9010-3
Jun 17, 2022 at 14:58 comment added The Amplitwist The links to springerlink.com are broken. Perhaps you could take a look, whenever possible…
Apr 16, 2011 at 8:34 comment added Anirbit @Vladimir Oh..sorry..I was not accounting for the "-1". The number you quote is the minimum number that is representable and hence the Frobenius number would be one less than that. Okay!
Apr 13, 2011 at 22:13 comment added Vladimir Dotsenko Huh? $ab-a-b=(a-1)(b-1)-1$, so these two statements match perfectly.
Apr 12, 2011 at 16:03 comment added Anirbit @Vladimir In the answer by Pete on another question on MathOverflow mathoverflow.net/questions/19587/… , there seems to be a claim that Sylvester had shown that for the two variable case the Frobenius number is precisely $ab-a-b$ (in your notation) Doesn't that contradict what you saying in the two variable case?
Jul 8, 2010 at 16:18 vote accept TGM
Jul 8, 2010 at 16:17 vote accept TGM
Jul 8, 2010 at 16:17
Jun 11, 2010 at 23:36 history answered Vladimir Dotsenko CC BY-SA 2.5