## 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.

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 Quote by Wiz14 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

 Quote by ramsey2879 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?"

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## For which primes "P" is the following true?

 Quote by Wiz14 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.

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

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 Quote by mathsman1963 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

 Quote by mathsman1963 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.

 Recognitions: Gold Member 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 ==>

 Quote by Wiz14 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.

 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