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February 10th, 2010, 03:39 PM   #1
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An infinite number (?) of primes using the uple totient

Hi,

I discovered a way to generate (maybe) an infinite number of primes using the uple totient.(read this topic viewtopic.php?f=40&t=12181 )

Let the sequence U(n)={1,2,3,1,6,2,2,4,8,3,1,13,5...}

If we begin by the 2 numbers of the sequence and then we add each time the number following we will obtain a sequence of primes :

double totient ?(1,2)=5 (is prime)
3-uple totient ?(1,2,3)=13 (is prime)
4-uple totient ?(1,2,3,1)=23 (is prime)
5-ulpe totient ?(1,2,3,1,6)=47 (is prime)
6-uple totient ?(1,2,3,1,6,2)=61 (is prime)
and so on
?(1,2,3,1,6,2,2)=79
?(1,2,3,1,6,2,2,4)=107
?(1,2,3,1,6,2,2,4,=163
?(1,2,3,1,6,2,2,4,8,3)=199
?(1,2,3,1,6,2,2,4,8,3,1)=233
?(1,2,3,1,6,2,2,4,8,3,1,13)=373
?(1,2,3,1,6,2,2,4,8,3,1,13,5)=457

I have built the first 12 numbers.
Can someone continue the sequence.
If we have a huge sequence maybe some pattern or property will be discovered.

Thank you for any help.
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February 10th, 2010, 04:19 PM   #2
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Re: An infinite number (?) of primes using the uple totient

Here is the continuation of U(n)

1,2,3,1,6,2,2,4,8,3,1,13,5,3,29,14,7,3,29,13,3,13, 39,3,22,2,3,2....

The sequence of prime generated :

5
13
23
47
61
79
107
163
199
233
373
457
491
541
563
607
641
691
739
773
821
853
887
911
937
971
997

I think that the sequence of primes is as infinite as U(n).
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February 10th, 2010, 05:40 PM   #3
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Re: An infinite number (?) of primes using the uple totient

What is U(n)?
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February 10th, 2010, 05:49 PM   #4
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Re: An infinite number (?) of primes using the uple totient

Quote:
Originally Posted by CRGreathouse
What is U(n)?
U(n) is a sequence created on the basis of an algo very simple

We begin by ?(1,x)
We replace x by a number starting each time by 1
So we find 2 ---> ?(1,2) is equal to 5 (prime number)
Then we continue ?(1,2,x)
We find 3 ---> ?(1,2,3) = 13 (prime)
and so on ....

Once we have a big sequence U(n) then we can try to explain why to find some pattern and so on
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February 10th, 2010, 05:51 PM   #5
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Re: An infinite number (?) of primes using the uple totient

Here is the last one :

U(n)=1,2,3,1,6,2,2,4,8,3,1,13,5,3,29,14,7,3,29,13, 3,13,39,3,22,2,3,2,3,39,3,2,13,14,7....

If you look at it some numbers are repeated
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February 10th, 2010, 06:04 PM   #6
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Re: An infinite number (?) of primes using the uple totient

Here is a diagramme of the prime generated :

[attachment=0:1e4wh9yc]primegnertotient.GIF[/attachment:1e4wh9yc]

What do you think about it?
Attached Images
File Type: gif primegnertotient.GIF (6.8 KB, 374 views)
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February 10th, 2010, 06:18 PM   #7
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Re: An infinite number (?) of primes using the uple totient

Quote:
Originally Posted by momo
We begin by ?(1,x)
We replace x by a number starting each time by 1
So we find 2 ---> ?(1,2) is equal to 5 (prime number)
Then we continue ?(1,2,x)
We find 3 ---> ?(1,2,3) = 13 (prime)
and so on ....
Why 3? ?(1, 2, 2) = 11 is prime.
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February 10th, 2010, 06:28 PM   #8
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Re: An infinite number (?) of primes using the uple totient

Quote:
Originally Posted by CRGreathouse
Quote:
Originally Posted by momo
We begin by ?(1,x)
We replace x by a number starting each time by 1
So we find 2 ---> ?(1,2) is equal to 5 (prime number)
Then we continue ?(1,2,x)
We find 3 ---> ?(1,2,3) = 13 (prime)
and so on ....
Why 3? ?(1, 2, 2) = 11 is prime.
You are right.
When you work with Excel and you enter your data manually it happens.
Sorry for the mistake.
I have to recompute all the numbers!!!
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February 10th, 2010, 06:35 PM   #9
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Re: An infinite number (?) of primes using the uple totient

Can you please send me the right sequence (minimal sequence because we can have a lot of sequences).
Using Rapidshare will be very quick.

Thank you!
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February 10th, 2010, 06:38 PM   #10
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Re: An infinite number (?) of primes using the uple totient

I don't do Rapidshare, but here are the first 100:
1, 2, 2, 2, 2, 2, 3, 8, 3, 22, 2, 7, 3, 8, 10, 1, 4, 6, 4, 2, 1, 1, 7, 2, 16, 4, 1, 15, 12, 2, 10, 5, 3, 11, 3, 3, 3, 5, 5, 22, 3, 13, 6, 11, 33, 27, 16, 2, 2, 9, 5, 10, 1, 21, 5, 22, 7, 20, 12, 8, 12, 3, 10, 4, 24, 3, 8, 1, 4, 14, 13, 22, 7, 31, 24, 16, 9, 7, 6, 14, 4, 2, 6, 3, 14, 2, 9, 5, 18, 74, 23, 24, 8, 16, 3, 24, 5, 2, 4, 7

This comes from the following Pari code:
Code:
phiset(v)=my(s=sum(i=1,#v,v[i]));sum(i=1,s,sum(j=1,#v,gcd(v[j],i)==1))
find(v)=v=vector(#v+1,i,if(i<=#v,v[i],1));for(i=1,9e9,v[#v]=i;if(isprime(phiset(v)),return(i)))
v=[1];for(i=2,100,v=concat(v,find(v)));v
You can run this to larger values if you like, just replace the "100" with something larger.
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