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July 9th, 2015, 07:43 AM  #1 
Banned Camp Joined: Dec 2013 Posts: 1,117 Thanks: 41  Conjecture about factorials and arithmetic sequence
For any positive integer n > 0 it always exists at least one number r >=1 such as : n+n+r+n+2r+n+3r+....+n+kr=m! k>=2 n=1 r=1 1+2+3=6=3! n=2 r=6 2+8+14=24=4! n=3 r=5 3+8+13=24=4! n=4 r=10 4+14+24+....+44=120=5! 
July 9th, 2015, 09:46 AM  #2 
Member Joined: Oct 2013 Posts: 57 Thanks: 5 
A new way to calculate factorials ?! sounds interesting to me. n*(n+1)+sum(k=0 .. n,k*r)=(n+1)! solving for r: [n,r]=(n+1)! [1,0]=2! [2,0]=3! [3,2]=4! [4,10]=5! [5,46]=6! [6,238]=7! [7,1438]=8! [8,10078]=9! can we generalize the sequence for r ? 
July 9th, 2015, 10:00 AM  #3  
Banned Camp Joined: Dec 2013 Posts: 1,117 Thanks: 41  Quote:
otherwise it will be easy all the factorials are divisible by 2.  
July 9th, 2015, 11:11 AM  #4  
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  Quote:
Conjecture (mobel, 2015): For any positive integer n there exist integers r >= 1, k >= 2, and m >= 3 such that r*k*(k+1)/2 + k*n + n = m!. This is of course true, since you can take k = 2, r = m!/3  n for any m large enough that r >= 1.  
July 10th, 2015, 05:40 AM  #5  
Banned Camp Joined: Dec 2013 Posts: 1,117 Thanks: 41  Quote:
But my problem is to find for some given r all the factorials possible. r=1 was designed for the initiator numbers and we have seen that not all the numbers are initiators (true or not?). My final goal by writing factorials as sums is to express factorial n mod (n+1) as sums simplified. I have just found a solution this morning. 5!=(1+2+3+4+5)+(19+20+21+22+23) Factorials could be expressed as 2 sums of consecutive numbers : the first one starting from 1 to n and the second one finishing by (n1)!1 Reduced mod 6 5!=2*(1+2+3+4+5)=5*6 I`m trying to find more general formula. So I could compute any n! mod (n+1) quickly. 6!=(1+2+3+4+5+6)+(114+115+116+117+118+119) 6! mod 7 = (1+2+3+4+5+6)+(2+3+4+5+6+0) and so on With a general formula it will make it easier the computation. I need to do it at least for the first 100 factorials to discover a pattern. Last edited by mobel; July 10th, 2015 at 06:30 AM.  
July 10th, 2015, 07:58 AM  #6 
Banned Camp Joined: Dec 2013 Posts: 1,117 Thanks: 41 
More generally : n!=sum(1 to n, step +1) + sum((n1)!1 to (n1)!n, step 1) Reduced mod n+1 The first sum remain unchanged The second sum is to clarify If we can find a close formula for the second sum it will be done. We can do it! 
July 10th, 2015, 09:12 AM  #7 
Banned Camp Joined: Dec 2013 Posts: 1,117 Thanks: 41 
10!=(1+2+3+....+9+10)+(362870+....+362879) 10! mod 11 = (1+2+...+10)+(2+3+4+......+10+0) = (1+10)+(2+9)+......(9+2)+10 =10 By regrouping and removing it is easy to conclude then. Now we could rewrite the Wilson theorem without using "factorials". 
July 10th, 2015, 09:36 AM  #8 
Banned Camp Joined: Dec 2013 Posts: 1,117 Thanks: 41 
8!=(1+2+3+....+7++(40312+....+40319) 8! mod 9 = (1+2+3+...++(1+2+3+....+= 0 because 8*9 mod 9 =0 
July 10th, 2015, 12:49 PM  #9  
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  Quote:
https://oeis.org/A061006 where I submitted a program last year: Code: (PARI) a(n)=if(isprime(n), n1, if(n==4, 2, 0))  
May 29th, 2017, 03:09 AM  #10 
Banned Camp Joined: Dec 2013 Posts: 1,117 Thanks: 41 
Here is another one not deeply studied even if it has some interest.


Tags 
arithmetic, arithmetical, conjecture, factorials, sequence 
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