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You are not logged in. #1 20121029 19:51:18
MultifactorialHow do you define multifactorial?I know what it is,but how do you define it?I know how to define factorial. #2 20121029 20:12:14
Re: MultifactorialIt can be defined recursively:http://en.wikipedia.org/wiki/Factorial#Multifactorials The limit operator is just an excuse for doing something you know you can't. “It's the subject that nobody knows anything about that we can all talk about!” ― Richard Feynman “Taking a new step, uttering a new word, is what people fear most.” ― Fyodor Dostoyevsky, Crime and Punishment #4 20121029 23:26:57
Re: MultifactorialHi Harold; In mathematics, you don't understand things. You just get used to them. I have the result, but I do not yet know how to get it. All physicists, and a good many quite respectable mathematicians are contemptuous about proof. #5 20121029 23:48:16
Re: MultifactorialWell,if I try n=11,k=3,then11!!!=11*8*5*2(as 11 mod 3=2) =880,is it correct? #6 20121030 00:09:14
Re: MultifactorialHi; In mathematics, you don't understand things. You just get used to them. I have the result, but I do not yet know how to get it. All physicists, and a good many quite respectable mathematicians are contemptuous about proof. #7 20121030 00:47:18
Re: MultifactorialWell,thank you,I love solving problems,now,20!!!!,so n=20 and k=4 #8 20121030 00:57:27
Re: MultifactorialHi; In mathematics, you don't understand things. You just get used to them. I have the result, but I do not yet know how to get it. All physicists, and a good many quite respectable mathematicians are contemptuous about proof. #9 20121030 01:40:30
Re: MultifactorialBut,I am confused,if I do 11!!! by the method in wikipedia,I get 11!!!=11(8!!!)=11(8(5!!!))=11(8(5(2!!!)))=11(8(5(1))) [as 0<2<3] =440 