February 1st, 2011, 07:08 AM  #11 
Senior Member Joined: Jan 2011 Posts: 560 Thanks: 1  Re: Factorial formulae
Once I expand and contract the general formula for : N=((3^n)*k + ((3^n)1)/2) The general formula for ((3^n)*k + ((3^n)1)/2)! will be formed with 3 parts : 1. 3^m (where is some value depending on k) 2. k! 3. The product of the products powered (P(1to k))^n(1) * (P(k+1 to 3k+1))^n(2) ....... I have to study separately the part 3 to make it more compact. I use a pencil to do it but it is not finished yet. My big headache are miscalulations. Just wish me good luck or try to help me at least by writing down the general formula in Latex. Thank you. 
February 1st, 2011, 07:13 AM  #12 
Global Moderator Joined: Nov 2006 From: UTC 5 Posts: 16,046 Thanks: 938 Math Focus: Number theory, computational mathematics, combinatorics, FOM, symbolic logic, TCS, algorithms  Re: Factorial formulae
For different sizes it's sensible to split the number differently. That's why it's more sensible to write this as an algorithm than a formula.

February 1st, 2011, 09:05 AM  #13 
Senior Member Joined: Jan 2011 Posts: 560 Thanks: 1  Re: Factorial formulae
Once you have a formula for n=3k+1 it is easy to use the same formula for any n n=3k+2 then (3k+2)! = (3k+2)*((3k+1)!) n=3k + 3 = 3(k+1)! then (3(k+1))!=(3k+3)*(3k+2)*((3k+1)!) So the size does not matter. The problem is the third part. How to rearrange it such as you reach (in terms of modulos) the best form. You could create for testing each big number p where p1! is of the some form f(k) (you have to build the form) a specific formula to solve it. 
February 4th, 2011, 06:33 AM  #14 
Senior Member Joined: Jan 2011 Posts: 560 Thanks: 1  Re: Factorial formulae
I have found a way to write n! depending on what we intent to solve. Not finished yet. 
February 10th, 2011, 03:08 PM  #15  
Math Team Joined: Apr 2010 Posts: 2,780 Thanks: 361  Re: Factorial formulae
A proof could look like this: Quote:
Code: For all we used is =, and both sides of the initial equation can be expressed the same, we have: I'm curious to a different/faster proof. Maybe induction could work.  
February 10th, 2011, 03:17 PM  #16  
Global Moderator Joined: Nov 2006 From: UTC 5 Posts: 16,046 Thanks: 938 Math Focus: Number theory, computational mathematics, combinatorics, FOM, symbolic logic, TCS, algorithms  Re: Factorial formulae Quote:
 
February 10th, 2011, 03:41 PM  #17 
Math Team Joined: Apr 2010 Posts: 2,780 Thanks: 361  Re: Factorial formulae
Thank you, I think I have proof by induction. It works for k=1; and Therefore, I don't think, the inductionproof is more elegant. 
February 11th, 2011, 07:19 AM  #18 
Senior Member Joined: Jan 2011 Posts: 560 Thanks: 1  Re: Factorial formulae
Try to replace each k by 3k'+1 and reloop it and you will obtain a general formula. I do not know how to write it in latex 
August 15th, 2016, 03:49 PM  #19 
Banned Camp Joined: Dec 2013 Posts: 1,117 Thanks: 41 
Another one!


Tags 
factorial, formulae 
Search tags for this page 
Click on a term to search for related topics.

Thread Tools  
Display Modes  

Similar Threads  
Thread  Thread Starter  Forum  Replies  Last Post 
Formulae Required  qamath09  Algebra  6  June 11th, 2013 10:18 PM 
Formulae for permutations  apvkt  Algebra  1  October 26th, 2012 09:42 PM 
RFormulae  chocolate  Algebra  4  June 8th, 2012 06:05 PM 
transposition of formulae  randomFire  Algebra  6  February 1st, 2011 12:17 AM 
Transposition of formulae  skylineview945r2  Algebra  5  January 9th, 2011 04:23 PM 