For which primes P is the following true?

  • Context: Graduate 
  • Thread starter Thread starter Wiz14
  • Start date Start date
  • Tags Tags
    Primes
Click For Summary

Discussion Overview

The discussion revolves around identifying which prime numbers "P" satisfy the condition that the function f(x) = x(x - 1) + P yields a prime number for all x less than P. Participants explore various primes, share observations, and inquire about the underlying reasons for the behavior of this function.

Discussion Character

  • Exploratory
  • Technical explanation
  • Debate/contested

Main Points Raised

  • Some participants note that the function f(x) does not yield primes for certain values of P, such as P = 7.
  • There is a suggestion that lucky numbers greater than 3 must be primes of the form 6n - 1, as otherwise, they would be divisible by 3.
  • One participant mentions that the prime P must be the smaller of a pair of twin primes, which narrows down the candidates.
  • Another participant points out that while being the smaller twin prime is necessary, it is not sufficient, providing an example with P = 29.
  • Some participants express curiosity about whether there is a method to determine if a prime will yield lucky numbers without exhaustive checking.
  • There is a discussion about the relationship between Euler's Lucky Numbers and Heegner numbers, suggesting further exploration of these concepts.

Areas of Agreement / Disagreement

Participants express various hypotheses regarding the primes that satisfy the condition, but there is no consensus on a definitive method or set of primes. Multiple competing views and uncertainties remain regarding the properties of lucky numbers and their relation to twin primes.

Contextual Notes

Some limitations include the lack of a simple test for determining lucky numbers and the dependence on specific forms of primes. The discussion also highlights unresolved mathematical steps and the complexity of properties related to primes.

Wiz14
Messages
20
Reaction score
0
For which primes "P" is the following true?

the function f(x) = x(x - 1) + p gives you a prime number for all x < p
I've tried this with 5,11, and 41, but it doesn't work for 7 since 5(5-1) + 7 is not a prime.
Btw, this isn't homework or anything, just a curiosity.
 
Physics news on Phys.org


Wiz14 said:
the function f(x) = x(x - 1) + p gives you a prime number for all x < p
I've tried this with 5,11, and 41, but it doesn't work for 7 since 5(5-1) + 7 is not a prime.
Btw, this isn't homework or anything, just a curiosity.
See Euler's Lucky Numbers http://oeis.org/A014556
 


ramsey2879 said:
See Euler's Lucky Numbers http://oeis.org/A014556

I guess my question is, why does it work for the numbers that it works for? Is there a way to know whether it will work for a certain prime without checking for every value below it? Are the numbers in that link the only "Lucky Numbers?"
 


Wiz14 said:
I guess my question is, why does it work for the numbers that it works for? Is there a way to know whether it will work for a certain prime without checking for every value below it? Are the numbers in that link the only "Lucky Numbers?"
There is a link within the page for a more complete listing of "Lucky Numbers"; however, I doubt that it is a complete listing very much. As to why certain primes are or are not lucky numbers, there is not much to say other than the fact that lucky numbers greater than 3 must be primes of the form 6*n -1 since otherwise 3|(1*2 + P). That being noted, I doubt that there is a simple test other than checking all non-negative numbers < P. You will find that many properties of primes are not easily explained in a simple manner.
 
Last edited:


The prime p has to be the smaller of a pair of twin primes. That thins out the candidates considerably.
 


mathsman1963 said:
The prime p has to be the smaller of a pair of twin primes. That thins out the candidates considerably.

This is only necessary but not sufficient: for example take p=29 and x = 3: 29 + 6 = 35 not prime

More unseful information at http://mathworld.wolfram.com/LuckyNumberofEuler.html
 


mathsman1963 said:
The prime p has to be the smaller of a pair of twin primes. That thins out the candidates considerably.

All twin primes are of the form 6n - 1 and 6n + 1. The smaller is always 6n - 1, which is what ramsey said.
 


With ARIBAS it is very simple to get a function 'IsLuckyNumber(n)', returning
the number of primes in f(i) = i(i - 1) + n with i a natural number < n
Code:
==> IsLuckyNumber(41).
IsLuckyNumber: i = 1 ; is prime 41
IsLuckyNumber: i = 2 ; is prime 43
IsLuckyNumber: i = 3 ; is prime 47
IsLuckyNumber: i = 4 ; is prime 53
IsLuckyNumber: i = 5 ; is prime 61
IsLuckyNumber: i = 6 ; is prime 71
IsLuckyNumber: i = 7 ; is prime 83
IsLuckyNumber: i = 8 ; is prime 97
IsLuckyNumber: i = 9 ; is prime 113
IsLuckyNumber: i = 10 ; is prime 131
IsLuckyNumber: i = 11 ; is prime 151
IsLuckyNumber: i = 12 ; is prime 173
IsLuckyNumber: i = 13 ; is prime 197
IsLuckyNumber: i = 14 ; is prime 223
IsLuckyNumber: i = 15 ; is prime 251
IsLuckyNumber: i = 16 ; is prime 281
IsLuckyNumber: i = 17 ; is prime 313
IsLuckyNumber: i = 18 ; is prime 347
IsLuckyNumber: i = 19 ; is prime 383
IsLuckyNumber: i = 20 ; is prime 421
IsLuckyNumber: i = 21 ; is prime 461
IsLuckyNumber: i = 22 ; is prime 503
IsLuckyNumber: i = 23 ; is prime 547
IsLuckyNumber: i = 24 ; is prime 593
IsLuckyNumber: i = 25 ; is prime 641
IsLuckyNumber: i = 26 ; is prime 691
IsLuckyNumber: i = 27 ; is prime 743
IsLuckyNumber: i = 28 ; is prime 797
IsLuckyNumber: i = 29 ; is prime 853
IsLuckyNumber: i = 30 ; is prime 911
IsLuckyNumber: i = 31 ; is prime 971
IsLuckyNumber: i = 32 ; is prime 1033
IsLuckyNumber: i = 33 ; is prime 1097
IsLuckyNumber: i = 34 ; is prime 1163
IsLuckyNumber: i = 35 ; is prime 1231
IsLuckyNumber: i = 36 ; is prime 1301
IsLuckyNumber: i = 37 ; is prime 1373
IsLuckyNumber: i = 38 ; is prime 1447
IsLuckyNumber: i = 39 ; is prime 1523
IsLuckyNumber: i = 40 ; is prime 1601
-: 40

==>
 


Wiz14 said:
All twin primes are of the form 6n - 1 and 6n + 1. The smaller is always 6n - 1, which is what ramsey said.

3 and 5? 3 satisfies the OP's condition. For that matter so does 2.
 
  • #10


FYI, Euler's Lucky Numbers map to the last 6 Heegner #s by the rule 4p - 1. So, a good place to begin to learn more about them is by learning more about the Heegner #'s.

http://en.wikipedia.org/wiki/Heegner_number
 

Similar threads

  • · Replies 6 ·
Replies
6
Views
2K
  • · Replies 21 ·
Replies
21
Views
2K
  • · Replies 31 ·
2
Replies
31
Views
3K
  • · Replies 2 ·
Replies
2
Views
2K
  • · Replies 1 ·
Replies
1
Views
3K
Replies
16
Views
3K
  • · Replies 2 ·
Replies
2
Views
2K
Replies
5
Views
2K
  • · Replies 14 ·
Replies
14
Views
3K
  • · Replies 2 ·
Replies
2
Views
4K