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For a given positive integer $ 0 < n < 512$ and $l \,(l < n)$. Is to possible to find set of $n$ prime number such that choosing any $l$ prime numbers without repetition their sums or any basic math operation will not be equal unless the selected prime numbers are exactly the same.

Ex: n = 6 and l = 3, then prime numbers selected can be 2, 3, 5, 7, 11, 13.

However, if we choose (2, 5, 13) and (2, 7, 11) both add up to 20 though both of the sets have different prime numbers. Making the initial pick of the set of random numbers invalid.

letsBeePolite
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  • Why do you require $n<512$? – Dietrich Burde Aug 27 '17 at 16:22
  • @DietrichBurde the problem I am solving has the property that $n$ will never exceed 512. Hence, it might constrain the problem thereby helping in coming with a relatively easier solution. – letsBeePolite Aug 27 '17 at 16:24
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    We can chose any number of primes successively, so that no two different primes sum up to the same number. Apply induction! For $n=4$ we can start with $2,17,23$. Now chose a prime accordingly in addition - finished (take $29$). – Dietrich Burde Aug 27 '17 at 16:26
  • @DietrichBurde understood. Thanks – letsBeePolite Aug 27 '17 at 16:53
  • What do you mean by "any basic math operation"? We can discount pure products as they are taken care of by the fundamental theorem of arithmetic. Is some sort of ordered repetition necessary or not? e.g. alternating series/sequences. or do you mean any free form mixture of integer products, divisions, sums and differences? – James Arathoon Aug 27 '17 at 17:52
  • One way to build up such a set inductively: start with a set containing one number, and at each stage add a number that is greater than twice the largest number already in the set. Of course, this leads to rather large numbers if you really need 512 numbers in the set... – Michael Seifert Aug 30 '17 at 21:26
  • @MichaelSeifert does the value of $l$ will have influence in the strategy of building set. Let's say, the choosen size of prime numbers set will never exceed $10$. For it also, we have to add the number to the set which is greater than twice the largest number already in the set – letsBeePolite Aug 30 '17 at 21:29
  • My method guarantees that no sets of any size will have the same sum; but it doesn't guarantee that this is the "optimal" set, if you define "optimal" as having the smallest numbers possible. If you only need subsets of size up to $l$ (or exactly $l$) to have distinct sums, I strongly suspect that my method won't be optimal. Also, my method only takes care of the sums of the subsets. To discuss the "other basic math operations", you'll have to define what you mean by them.. – Michael Seifert Aug 30 '17 at 21:40
  • @MichaelSeifert can you describe as that property holds, when you keep extending set by adding a value which is twice the largest existing element in the set. – letsBeePolite Sep 09 '17 at 16:23

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