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Interesting note on 2*Pn

  1. Sep 17, 2010 #1
    Take a prime[n] and double it, i.e., e.g., Pn (n=7) so prime[7] is 17, doubled to 34.

    Take all unique arrangements of two primes from P2 (3) to P7 (17):

    {3,3}, {3,5}, {5,5}, ..., {13, 17}, {17,17}

    Add the pairs together, e.g., {3,3} = 3 + 3 = 6

    Now eliminate duplicates, i.e., {5,5} and {3,7} both add up to 10, so just have one 10.

    Then for 19 and 109 I find that all even numbers less than or equal to 19*2 and 109*2 respectively appear after you eliminate duplicates, i.e., all primes less than or equal to Pn generate all even numbers less than or equal to 2*Pn for 19 and 109.
     
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  3. Sep 17, 2010 #2

    Office_Shredder

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  4. Sep 17, 2010 #3
    More specific than GC I think.

    While 34 is 17*2 no combination of primes from 3 to 17 can generate 32.

    The primes from 3 to 19 can generate all evens less than 19*2 = 38.

    Are there more besides 19 and 109?
     
  5. Sep 19, 2010 #4

    CRGreathouse

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    Suppose the limit is P. Then 2P can of course be expressed, but 2P - 2 cannot be expressed unless P-2 is prime. 34/2 - 2 is composite, so 34 - 2 is not the sum of two primes in {3, 5, ..., 17}.
     
  6. Sep 19, 2010 #5
    Not sure I follow. 31 is prime as is 29 - neither can express 56 in {3..31}...?
     
  7. Sep 19, 2010 #6

    CRGreathouse

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    Yes, I can see you didn't follow. But let's address this first: what particular claim are you making about 19 and 109? If you can answer this precisely, I may be able to expand the list.
     
  8. Sep 20, 2010 #7
    For any given prime P double the prime, e.g., 17 * 2 = 34.

    Take all unique combinations of the primes from 3 to P, e.g., 3,5,7,11,13,17 and combine pairs in all possible ways, e.g., {3,3}, {3,5}, ..., {13,17}, {17,17}.

    Add the pairs to get the sums - this generates even numbers (excluding 2 and 4) - ignore duplicates.

    You will see that 17 generates all even numbers less than or equal to 34 except 32.

    19 generates all even numbers including 2 * 19 = 38 as does 109.

    In general for P combinations of this sort generate most evens, but not all, from 3..2*P.

    I hope this helps...
     
  9. Sep 20, 2010 #8

    Char. Limit

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    Just a thought: all positive even numbers. I doubt, for example, that I could get -1000 using this.
     
  10. Sep 20, 2010 #9

    CRGreathouse

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    I think, perhaps, you're looking for primes P such that for all 2 <= n <= P, there are p <= q <= P with p + q = 2n. Is that right?

    In that case I find no examples larger than 109. Any example larger than 19 must have
    • P = 4 (mod 15)
    • P - 2 prime
    • P - 6 prime
    • P - 8 prime
    • P - 20 prime
    • P - 12 or P - 18 prime
    • P - 26 or P - 32 prime
    • P - 26 or P - 38 prime
    which should speed any searches.


    I have verified by computer that there are no examples below a billion (10^9).
     
    Last edited: Sep 20, 2010
  11. Sep 20, 2010 #10
    Yes - 3 (not 2) <= n <= P with p <= q <= P and p + q = 2n. (1)

    I have developed the following conjecture relative to this:

    There is a P2 such that for large values of P P <= P2 < 1.015P such that (1) generates all positive (thanks to Char Limit) evens less than or equal to 2*P.
     
  12. Sep 20, 2010 #11

    CRGreathouse

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    Of course the 2 comes free.

    As written this doesn't make sense. (For any given P2, "for large enough P with P <= P2 < 1.015P, foo is true" is true regardless of foo, since for large enough P the inequality fails to hold.) Also, you don't explain how you modify (1) to use P2. My guess:

    There is a k, 1 <= k <= 1.015, such that for large enough primes P and all 3 <= n <= P, there are primes p <= q <= kP with p + q = 2n.

    This is of course equivalent to

    For large enough primes P and all 3 <= n <= P, there are primes p <= q < 1.015P with p + q = 2n.
     
    Last edited: Sep 20, 2010
  13. Sep 20, 2010 #12

    CRGreathouse

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    No examples (for the original version) up to 4 billion. My latest algorithm takes (empirical) time ~ x^1.7 to find the examples up to x.
     
  14. Sep 20, 2010 #13
    Yes. Sorry about my notation.

    Given k represents 1.015 then it appears to me that k bounds the choices of p and q in an interesting way...

    I do not believe there are any other primes besides 19 and 109 that work where k = 1.
     
  15. Sep 20, 2010 #14

    CRGreathouse

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    I suspect it's true for k = 1 + epsilon for any epsilon > 0. In fact it's probably true with p <= q <= P + f(P) for many slow-growing functions f.
     
  16. Sep 21, 2010 #15
    Since you have to advance some integral number of primes past P to include all evens below 2P I don't see how arbitrarily small epsilons work. I calculate the missing evens below 109 (I guess its true for 6, 10, 14 and 26 as well but since those are easily checked by hand I forgot about them) as

    {6, {}},
    {10, {}},
    {14, {}},
    {22, {20}},
    {26, {}},
    {34, {32}},
    {38, {}},
    {46, {44}},
    {58, {44, 50, 54, 56}},
    {62, {56}},
    {74, {64, 70, 72}},
    {82, {76, 80}},
    {86, {76}},
    {94, {92}},
    {106, {92, 98, 102, 104}},
    {118, {92, 98, 104, 108, 110, 114, 116}},
    {122, {110, 116}},
    {134, {116, 124, 130, 132}},
    {142, {116, 136, 140}},
    {146, {136}},
    {158, {136, 148, 154, 156}},
    {166, {148, 160, 164}},
    {178, {164, 170, 174, 176}},
    {194, {174, 182, 184, 188, 190, 192}},
    {202, {182, 188, 192, 196, 200}},
    {206, {188, 196}},
    {214, {188, 212}},
    {218, {}},
    ...

    I have developed a good model for this behavior.
     
  17. Sep 21, 2010 #16

    CRGreathouse

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    Let's say you choose epsilon = 0.000001. Then perhaps "large enough P" is 10^50 + 151, at which point you have 10^44 primes past P to use.
     
  18. Sep 21, 2010 #17
    I was thinking of epsilon in terms of just enough primes (m) beyond Pn to complete the evens.

    k = (1/Pn+m) - (1/Pn)/(1/Pn), Pn * (1 + k) = Pm, epsilon >= k
     
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