My Math Forum Factorial formulae

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 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 p-1! 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
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Re: Factorial formulae

A proof could look like this:

$(3k+1)!= (3^k) \cdot (k!)* \prod_{i=1}^k ((3i)^2 -1)$
Quote:
 Originally Posted by bogauss (I'm sorry I don't know how to write it in Latex)
Code:
$(3k+1)!= (3^k) \cdot (k!)* \prod_{i=1}^k ((3i)^2 -1)$
$(3k+1)!=\prod_{i=1}^k (3^k) \cdot (k!)* \prod_{i=1}^k ((3i)^2 -1)$

$(3k+1)!= (3i+1)\cdot\prod _{i=1}^{3k}i$

$(3^k) \cdot (k!) \cdot \left(\prod_{i=1}^k ((3i)^2 -1)\right) =\left(\prod_{i=1}^{k} 3\right) \cdot \left(\prod_{i=1}^{k} i \right)\cdot \left(\prod_{i=1}^{k}(3i-1) \right)\cdot \left(\prod_{i=1}^{k}(3i+1)\right)=\prod_{i=1}^k (3i-1)(3i)(3i+1)=2 \cdot 3 \cdot \;\;\cdots\;\; \cdot 3k \cdot (3k+1)=(3k+1)!$

For all we used is =, and both sides of the initial equation can be expressed the same, we have:
$(3k+1)!= (3^k) \cdot (k!)* \prod_{i=1}^k ((3i)^2 -1)$

I'm curious to a different/faster proof. Maybe induction could work.

February 10th, 2011, 03:17 PM   #16
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Re: Factorial formulae

Quote:
 Originally Posted by Hoempa I'm curious to a different/faster proof. Maybe induction could work.
That should give a three-step proof, but I'm not sure it's really that much more elegant.

 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. $k \in \mathbb{N}$ $f(k)=(3k+1)!$ $g(k)=3^k \cdot k! \cdot \prod_{i=1}^{k} ((3^i)^2-1)$ $f(1)=g(1);$ It works for k=1; $4!=3 \cdot 1 \cdot (3^2-1)=24.$ $\frac{f(k+1)}{f(k)}=(3k+1)(3k)(3k-1)$ $\frac{g(k+1)}{g(k)}=3 \cdot k \cdot (3k-1)(3k+1)$ $f(1)=g(1)$ and $\frac{f(k+1)}{f(k)}=\frac{g(k+1)}{g(k)}$ Therefore, $f(k)=g(k)= (3k+1)!=3^k \cdot k! \cdot \prod_{i=1}^{k} ((3^i)^2-1)$ I don't think, the induction-proof 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!

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