JT
Posts:
1,254
Registered:
4/7/12


Re: 7!= 2*5! * 2*3! *7
Posted:
Jun 11, 2013 1:09 PM


On 11 Juni, 18:39, JT <jonas.thornv...@gmail.com> wrote: > On 11 Juni, 18:32, JT <jonas.thornv...@gmail.com> wrote: > > > > > > > > > > > On 11 Juni, 18:12, Brian Chandler <imaginator...@despammed.com> wrote: > > > > JT wrote: > > > > Ooops got it wrong there, *2 of course > > > > 13!=2*11!*2*9!*2*7!*2*5!*2*3!*13 > > > > > Try me, how can you use this? > > > > Well, you can divide both sides by 13 getting: > > > > 12! = 2*11!* 2*9!*2*7!*2*5!*2*3! > > > > And if this isn't enough: divide by 12: > > > > 11! = 2*11!* 2*9!*2*7!*2*5! > > > > Then 1 = 2*2*9!*2*7!*2*5! > > > > So even allowing for experimental error, we get 1 > 2^18 > > > > Which means 2 is a lot smaller than previously thought. > > > > Brian Chandler > > > I think that 2 somehow sneaked in when i was thinking about the > > permutation algorithm, bloppers. But i was correct second time. > > > 7!=5!*3!*7 > > 9!=7!*5!*3!*9 > > 11!=9!*7!*5!*3!*11 > > 13!=11!*9!*7!*5!*3!*13 > > No i was not 9! = 362880
Well 5!*3!*7 = 5040 and 7!=5040, tomorrow i give it another try. And just by chance 9!=362880 and 7!*5!*3!=3628800 so there may be something, that can be used in the permutation algorithm. I just have to think about it.

