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#1 2012-10-29 19:51:18
- Harold
- Guest
Multifactorial
How do you define multifactorial?I know what it is,but how do you define it?I know how to define factorial.
#2 2012-10-29 20:12:14
- anonimnystefy
- Real Member
Offline
Re: Multifactorial
It 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
“A secret's worth depends on the people from whom it must be kept.” ― Carlos Ruiz Zafón
#4 2012-10-29 23:26:57
- bobbym
- Administrator
Online
Re: Multifactorial
Hi Harold;
Did you try one? Say for k = 3.
In mathematics, you don't understand things. You just get used to them.
90% of mathematicians do not understand 90% of currently published mathematics.
I am willing to wager that over 75% of the new words that appeared were nothing more than spelling errors that caught on.
#5 2012-10-29 23:48:16
- Harold
- Guest
Re: Multifactorial
Well,if I try n=11,k=3,then11!!!=11*8*5*2(as 11 mod 3=2) =880,is it correct?
#6 2012-10-30 00:09:14
- bobbym
- Administrator
Online
Re: Multifactorial
Hi;
That is correct. Now what do you get for 20!!!! ?
In mathematics, you don't understand things. You just get used to them.
90% of mathematicians do not understand 90% of currently published mathematics.
I am willing to wager that over 75% of the new words that appeared were nothing more than spelling errors that caught on.
#7 2012-10-30 00:47:18
- Harold
- Guest
Re: Multifactorial
Well,thank you,I love solving problems,now,20!!!!,so n=20 and k=4
20!!!!=20(20-4)(20-4*2)(20-4*3)...(4) [as 20 mod 4=0]=20*16*12*8*4=122880.
#8 2012-10-30 00:57:27
- bobbym
- Administrator
Online
Re: Multifactorial
Hi;
That is correct. Your formula looks fine to me. Although there is a better one on that page that anonimnystefy sent you.
In mathematics, you don't understand things. You just get used to them.
90% of mathematicians do not understand 90% of currently published mathematics.
I am willing to wager that over 75% of the new words that appeared were nothing more than spelling errors that caught on.
#9 2012-10-30 01:40:30
- Harold
- Guest
Re: Multifactorial
But,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
and by my method,I get 11!!!=11(11-3)(11-2*3)...(2)=11*8*5*2=880
which is correct ??
#10 2012-10-30 01:44:34
- bobbym
- Administrator
Online
Re: Multifactorial
2!!! = 2 not 1.
In mathematics, you don't understand things. You just get used to them.
90% of mathematicians do not understand 90% of currently published mathematics.
I am willing to wager that over 75% of the new words that appeared were nothing more than spelling errors that caught on.