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Given the natural integers 1,2,3,....see below?
and given a specific set of odd primes Sp = {3,5,7,11...Pz} (note Pz could be 3) and the rule that if 2x+1 is divisible by a odd prime p belonging to Sp then the integers x and x+1 are discarded, how would one most effectively compute the next integer that would not be discarded?
For example let the set of primes Sp={3,5} then 1+2 = 3 and so 1 and 2 are discarded; 2+3 = 5 and so 2 and 3 are discarded; but 3+4 = 7 and 7 does not belong to Sp so 4 is not discarded. Any thoughts? Thanks in advance, and I will not let this question go to voting since the voting system is so flawed.
Dr Brisbane, thanks for the response but it misses the point completely. And of course 2 is a prime (the only even prime)
mathbear: (thanks for the response)
Let's put it this way if the sum of two consecutive integers is divisible by a prime of Sp then those consecutive integers are eliminated. Let's say Sp = {3} then every pair of consecutive integers whose sum is divisible by 3 will be eliminated and the others won't. So if Sp ={3} then clearly only those integers divisible by 3 would be left. Gone would be (1,2) (4,5) (7,8) (10,11) etc, while 3,6,9,12,15 etc are left. If Sp also included 5 the (2,3)(7,8)(12,13)(17,18)..etc are gone. The first two integers missed by Sp ={3,5} are 6 and 9. Depending on the primes of Sp the first integer that will be skipped by those primes is explicitly computable, though it is increasingly difficult to do so; hence my question.
One other note: obviously it depends on the distribution of primes. But the primes will now be represented by consecutive integers and we can use multiple simultaneous arithmetic progressions to compute the next missed integer, which will introduce two new arithmetic progessions which can be fed into next computation etc.
4 Answers
- 1 decade agoFavorite Answer
I think that I understand what you are trying to say and heres what I did. I wrote a for loop in Matlab that first found the first 50 values of the sentence y=2x+1 with x starting at 1. I then found the first 26 prime numbers starting at 1 and put them in an array called p. And then I compared the two. There is no set pattern for when a prime number will appear compared to the y sequence. The best way to find the next not discarded integer would be to write a comparison for loop that would compare the two sequences and display there differences. Good Luck!
Source(s): ;) - Anonymous1 decade ago
First of all: All primes are odd, because if they weren't, two could divide them and so they would not be prime. Secondly, if one has no idea what the next integer would be other than it could be prime, this question seems unanswerable. I.E.1765199373.2^107520 is a prime, so in essence your problem could be {3, 7, 5, 11, 1765199373.2^107520, 5, x} how can I possibly guess what X would be with an algorithm?
I think there is more to your problem than you are stating here.
Anyone claiming to know the solution to this problem, state your answer and I'll show you where you're wrong. This problem is impossible.
- 1 decade ago
I don't understand your example. You say 4 is not discarded when Sp={3,5}, and yet setting x=4, you find that 2x+1 is divisible by 3, which is in Sp, so you end up discarding x=4 because of that (i.e., you can end up discarding a number either because it appears as x OR as x+1).
Please clarify.
- bequalmingLv 51 decade ago
There is no known pattern to prime numbers. So, there can be no solution to your problem.
