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#1 2012-10-28 20:51:18
- 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 2012-10-28 21:12:14
- anonimnystefy
- Real Member
- From: Harlan's World
- Registered: 2011-05-23
- Posts: 16,049
Re: Multifactorial
It can be defined recursively:http://en.wikipedia.org/wiki/Factorial#Multifactorials
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#3 2012-10-28 23:00:35
- Harold
- Guest
Re: Multifactorial
Can't this be another definition-
#4 2012-10-29 00:26:57
- bobbym
- bumpkin
- From: Bumpkinland
- Registered: 2009-04-12
- Posts: 109,606
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.
If it ain't broke, fix it until it is.
Always satisfy the Prime Directive of getting the right answer above all else.
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#5 2012-10-29 00:48:16
- 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 2012-10-29 01:09:14
- bobbym
- bumpkin
- From: Bumpkinland
- Registered: 2009-04-12
- Posts: 109,606
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.
If it ain't broke, fix it until it is.
Always satisfy the Prime Directive of getting the right answer above all else.
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#7 2012-10-29 01:47:18
- 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 2012-10-29 01:57:27
- bobbym
- bumpkin
- From: Bumpkinland
- Registered: 2009-04-12
- Posts: 109,606
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.
If it ain't broke, fix it until it is.
Always satisfy the Prime Directive of getting the right answer above all else.
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#9 2012-10-29 02:40:30
- 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 2012-10-29 02:44:34
- bobbym
- bumpkin
- From: Bumpkinland
- Registered: 2009-04-12
- Posts: 109,606
Re: Multifactorial
2!!! = 2 not 1.
In mathematics, you don't understand things. You just get used to them.
If it ain't broke, fix it until it is.
Always satisfy the Prime Directive of getting the right answer above all else.
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