Primes of the form ((p^p)%p#)/p

[kijinSeija]

Have these kinds of prime numbers been studied ?

I found nothing on factordb and OEIS.

I use the % for the modulo operation and # for the primorial numbers.

I used PFGW and I found these primes and PRP :

(((3^3)%3#)/3) is Unity (1)
(((5^5)%5#)/5) is Unity (1)
(((7^7)%7#)/7) is trivially prime!: 19
(((11^11)%11#)/11) is trivially prime!: 151
(((13^13)%13#)/13) is trivially prime!: 631
(((29^29)%29#)/29) is trivially prime!: 21946681
Switching to Exponentiating using GMP
(((41^41)%41#)/41) is 3-PRP! (0.0000s+0.0129s)
(((43^43)%43#)/43) is 3-PRP! (0.0000s+0.0012s)
(((47^47)%47#)/47) is 3-PRP! (0.0000s+0.0006s)
(((71^71)%71#)/71) is 3-PRP! (0.0000s+0.0002s)
(((167^167)%167#)/167) is 3-PRP! (0.0000s+0.0001s)
(((241^241)%241#)/241) is 3-PRP! (0.0000s+0.0001s)
(((257^257)%257#)/257) is 3-PRP! (0.0001s+0.0003s)
(((367^367)%367#)/367) is 3-PRP! (0.0001s+0.0002s)
Switching to Exponentiating using Woltman FFT's
(((769^769)%769#)/769) is 3-PRP! (0.0015s+0.0001s)
(((1031^1031)%1031#)/1031) is 3-PRP!
(((1459^1459)%1459#)/1459) is 3-PRP!
(((3533^3533)%3533#)/3533) is 3-PRP!
(((4219^4219)%4219#)/4219) is 3-PRP!
(((8933^8933)%8933#)/8933) is 3-PRP!
(((14843^14843)%14843#)/14843) is 3-PRP!

[kijinSeija]

(((30809^30809)%30809#)/30809) is 3-PRP!
(((42017^42017)%42017#)/42017) is 3-PRP!

[a1call]

If I am not mistaking:
An equivalent format would be
p^(p-1) - n*p#
where n is an integer which is not divisible by p.
The result will always be divisible by p and never by any primes less than p. While results (after division by p) are less than p^2, they are guaranteed to be 1 or prime. Just above p^2 they have a good chance of being prime. As the results get exponentially larger than p^2 their likeliness of being prime approaches any other random integer of the same size range.
AFAIK, they are not generally of a format that can be easily proven to be prime.
Just my 2 cents && FWIW. :)

[Cybertronic]

Hello kijnSeija,
this kind of numbers is me unknown.
Problem is , a PRP number is hard to prove it. The only address for PRP numbers is


http://www.primenumbers.net/prptop/prptop.php


otherwise the numbers enqueue in hundreds of other formulas, but interesting !


best

[kijinSeija]

thanks for your help and for the address

[a1call]

You can improve the probability of the numbers being Prime by replacing p^p with p^m for m>1 (the exponent being equal to p does not serve any purpose here). This will have the same mechanics but will over all yield smaller numbers (for the right m) which will be closer to p^2. In other words choose m such that it gives the smallest reminder (greater than p). Some results may be negative (if you add the n multiplier to the primorial and use subtraction rather than %) which you would have to change their sign.

[Dr Sardonicus]

In taking the remainder p_k^m % p_k-1# , you want p_k^m > p_k-1#. Thus m > log(p_k-1#)/log(p).

For p > 3, p_k < p_k-1#, so you get a remainder of p for m = 1. The number of possible remainders of p_k^m modulo p_k-1# is the multiplicative order h of p_k modulo p_k-1# and this is the lcm of the multiplicative orders of p_k modulo p_k-1, p_k-2,... 2.

I wrote a simple script to find the remainders of p_k^m % p_k-1# from the smallest m for which p_k^m > p_k-1# up to m = p-1.

? lpr=log(2);pr=2;v=[];forprime(p=3,20,n=1+floor(lpr/log(p));v=vector(p-n,i,(p^(i+n-1))%pr);print(p" "pr" "v);pr*=p;lpr+=log(p);)
3 2 [1, 1]
5 6 [1, 5, 1]
7 30 [19, 13, 1, 7, 19]
11 210 [71, 151, 191, 1, 11, 121, 71, 151]
13 2310 [841, 1693, 1219, 1987, 421, 853, 1849, 937, 631]
17 30030 [23461, 8447, 23479, 8753, 28681, 7097, 529, 8993, 2731, 16397, 8479, 24023, 18001]
19 510510 [434059, 78961, 479239, 426871, 452899, 436921, 133339, 491401, 147439, 248791, 132439, 474301, 333049, 201811]
?