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July 9th, 2015, 08:43 AM   #1
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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!
Thanks from Martin Hopf
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July 9th, 2015, 10:46 AM   #2
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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 ?
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July 9th, 2015, 11:00 AM   #3
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Quote:
Originally Posted by Martin Hopf View Post
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 ?
r must be >=1
otherwise it will be easy all the factorials are divisible by 2.
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July 9th, 2015, 12:11 PM   #4
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Quote:
Originally Posted by mobel View Post
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
Translation:

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.
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July 10th, 2015, 06:40 AM   #5
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Quote:
Originally Posted by CRGreathouse View Post
Translation:

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.
You are right.
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 (n-1)!-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 07:30 AM.
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July 10th, 2015, 08:58 AM   #6
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More generally :

n!=sum(1 to n, step +1) + sum((n-1)!-1 to (n-1)!-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!
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July 10th, 2015, 10:12 AM   #7
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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".
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July 10th, 2015, 10:36 AM   #8
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8!=(1+2+3+....+7++(40312+....+40319)

8! mod 9 = (1+2+3+...++(1+2+3+....+= 0

because 8*9 mod 9 =0
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July 10th, 2015, 01:49 PM   #9
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Quote:
Originally Posted by mobel View Post
I`m trying to find more general formula.
So I could compute any n! mod (n+1) quickly.
If you look this up in the OEIS you'll find
https://oeis.org/A061006
where I submitted a program last year:

Code:
(PARI) a(n)=if(isprime(n), n-1, if(n==4, 2, 0))
This is pretty easy to read, but just in case: if n is prime n! mod (n+1) is n-1, if n is 4 it is 2, and otherwise it is 0.
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May 29th, 2017, 04:09 AM   #10
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Here is another one not deeply studied even if it has some interest.
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