primes?
 

Just a question. I am interested in looking for primes such that the following is true. Does any one know anything about their distribution.

Let such a prime be p.

Then take the first p prime P(1), P(2).... P(p).

Let set A= {P(1),....P(p)} (mod p)

Let set B= all numbers from 0 to p-1.

Then are there any primes for which the elements of set A and B are identical?

2 is the only such prime I know of .

Since P(1)=2
P(2)=3

Then A={0,1)
B={0,1}
Hence both the sets have the same elements.

Are there any more?

Some thoughts that I have:
The probability of finding k*p+n for fixed p is the same irrespective of n. Hence there should be more of these primes but I have not found any such examples upto 30 Million.

Citrix

Just thinking statistically, this is hopeless. You want the first p primes to fall into unique residue classes mod p. The probability of this occurring is something like (1/p)^p, which gets very small very quickly as p increases.

[QUOTE=PBMcL]Just thinking statistically, this is hopeless. You want the first p primes to fall into unique residue classes mod p. The probability of this occurring is something like (1/p)^p, which gets very small very quickly as p increases.[/QUOTE]

Why is the probability 1/p^p? Can you show how you get to it?

Some other ways to look at the problem.

Find p such that, where # is primorial

((P(p)#/p)+1)=0 (mod p)

Another way to look at the problem: Sum of first P(p) =0 (mod p)


Though the other 2 methods of looking at this problem will produce false positives, but only these numbers can be tested then.


Citrix
edit: I would say the probability to be 1 in p for the above 2 conditions.

If p≥3, 4∈B but obviously ∉ A as it is not prime. But this is kinda trivial... maybe I didn't understand your question.

Alex

Alex, what do the boxes mean?

The probability if primes were completely random would be p!/(p^p). This is also huge. But primes are not completely random and that should make a difference.

If primes were half the time random and 1/2 the times not then the probability would reduce down to the square root of p!/(p^p).

Citrix

[QUOTE=Citrix]Let such a prime be p.

Then take the first p prime P(1), P(2).... P(p).

Let set A= {P(1),....P(p)} (mod p)

Let set B= all numbers from 0 to p-1.[/QUOTE]
Alex,

I think what Citrix means is that

for p = 3:
A = {2, 3, 5} (mod 3) = (2,0,2)
B = (0,1,2)

for p = 5:
A = {2, 3, 5, 7, 11} (mod 5) = (2,3,0,2,1)
B = (0,1,2,3,4)

for p = 7:
A = {2, 3, 5, 7, 11, 13, 17} (mod 7) = (2,3,5,0,4,6,3)
B = (0,1,2,3,4,5,6)

So for p = 7, 4 ∈ A

[QUOTE=Citrix]Alex, what do the boxes mean?

The probability if primes were completely random would be p!/(p^p). This is also huge. But primes are not completely random and that should make a difference.

Citrix[/QUOTE]

You are right. I forgot the p! part. Using Stirling's formula p!/(p^p) is roughly
sqrt(2*pi*p)*exp(-p). But even considering only those primes exceeding p as "random", you end up approaching this formula as p gets large.

[QUOTE=Citrix]((P(p)#/p)+1)=0 (mod p)[/QUOTE]

Shouldn't it be ((P(p)#/p) - 1) [tex]\eq[/tex] 0 (mod p) ?

Or alternately P(p)#/p [tex]\eq[/tex]1 (mod p)?

Oops, I overlooked the (mod p) - sorry.

Alex

for p = 3:
A = {2, 3, 5} (mod 3) = {2,0,2} -- 2 is duplicated; 1 is missing
B = {0,1,2}

for p = 5:
A = {2, 3, 5, 7, 11} (mod 5) = {2,3,0,2,1} -- 2 is duplicated; 4 is missing
B = {0,1,2,3,4}

for p = 7:
A = {2, 3, 5, 7, 11, 13, 17} (mod 7) = {2,3,5,0,4,6,3} -- 3 is duplicated; 1 is missing
B = {0,1,2,3,4,5,6}

for p = 11:
A = {2, 3, 5, 7, 11, 13, 17, 19, 23, 29, 31} (mod 11) = {2,3,5,7,0,2,6,8,1,7,9} -- 2 and 7 are duplicated; 4 and 10 are missing
B = {0,1,2,3,4,5,6,7,8,9,10}

for p = 13:
A = {2, 3, 5, 7, 11, 13, 17, 19, 23, 29, 31, 37, 41} (mod 13) = {2,3,5,7,11,0,4,6,10,3,5,11,2} -- 2, 3, 5, 11 are duplicated; 1, 8, 9, 12 are missing
B = {0,1,2,3,4,5,6,7,8,9,10,11,12}

for p = 17:
A = {2, 3, 5, 7, 11, 13, 17, 19, 23, 29, 31, 37, 41, 43, 47, 53, 59} (mod 17) = {2,3,5,7,11,13,0,2,6,12,14,3,7,9,13,2,8} -- 2 is triplicated; 3, 7, 13 are duplicated; 1, 4, 10, 15, 16 are missing
B = {0,1,2,3,4,5,6,7,8,9,10,11,12,13,14,15,16}

Is there anything profound in the pattern of congruence classes missing from or duplicated in A? (... other than that, so far, the only missing elements are 1 or composites, and the only duplicates are primes -- are those just law-of-small-number coincidences?)

[QUOTE=axn1]Shouldn't it be ((P(p)#/p) - 1) [tex]\eq[/tex] 0 (mod p) ?

Or alternately P(p)#/p [tex]\eq[/tex]1 (mod p)?[/QUOTE]

See [url]http://mathworld.wolfram.com/WilsonsTheorem.html[/url]
the formula comes from wilson's theorem