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Slightly excessive numbers

  1. Jun 4, 2006 #1
    so...do they/ dont they exist????

    i dont think they do as they would have to be even with an even number of (odd*even) factor pairs.
    And if you assume this number is in the form of 2p where p is odd and has only two (odd*even) facots pairs, it must have the factors (2*psqaured) and (p * 2p)
    if is was slightly excessive;
    x + 1 = 1 + 2 + psqaured + p + 2p
    x = (p+1)(p+2
    so another factor pair....
     
  2. jcsd
  3. Jun 4, 2006 #2

    matt grime

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    Slightly excessive numbers, it turns out, are whta Singh calls quasiperfect numbers. n is quasi perfect if sigma(n)=2n+1, sigma of n means the sum of its divisors.

    Right, now, what I odn't get in your post is how you define p to be off and say it has two (odd*even) factors pairs. Whatever that means it seems to imply that and odd number has an even factor.
     
  4. Jun 5, 2006 #3
    no, i meant 2p has two (odd*even) pairs, not p
     
  5. Jun 5, 2006 #4

    shmoe

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    What makes you think a quasiperfect number would have to be even? The existence of quasiperfect numbers is still an open problem (though we can say a few things about them).

    I don't understand the rest of what you wrote. You seem to be claiming 2p^2 is a divisor of 2p? Maybe you could explain all that more clearly.
     
  6. Jun 5, 2006 #5

    matt grime

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    So, what is an (odd*even) pair?
     
  7. Jun 6, 2006 #6
    hang on shall i start from the beginning???!!!!!
    right. if the slightly excessive number was odd it would have only odd factors so when you add the factors together you would get 1 + odd + odd and as the odd numbers would always be in pairs ie there would be an even number of them, the total of all the facors would be odd (as you include 1). however if the odd numbers would slightly excessvie the total would be even.
    if the quasiperfect was even with only even factors it must be in the form of 2 to the power of n and these have already been proven slightly defective ( the factors add up to one less than original numbers)
    therefor the only chance of a number being quasiperfect is if it is even but has odd factors multiplied by an even number! to allow the total of factors + 1 to be odd there must be an even number of odd numbers multiplied by even numbers- this is what i mean by (o * e) pair!there may also be (e*e) pairs!
    i was saying..lets assume there is no (e*e) pairs and only two (o*e) pairs ie 18 (3*6) (9*2).
    one of the (o*e) pairs will be half the original number (call this z) and 2. the other odd factor must be the square root of half z (so it only has to factors) for example 30 has 15, but then also 3 and 5 but 18 has 9, and then only 3.
    so if the number is slightly excessive
    z + 1 = 1 + 2 + psquared + 2p + p
    z = psquared + 3p + 2
    however this total has other factors as it can be factorised
    (p+1)(p+2) this proofs that numbers like z cant be slightly excessive.
    i can also prove it for numbers like z but with (e*e) pairs but not for numbers with more than two (o*e) pairs!!!
     
  8. Jun 6, 2006 #7

    matt grime

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    15 is an odd number. Its factors are 1,3,5,15 the sum of these is even. So what does this do to your assertion that the sum of the factors of an odd number is odd? If you mean factors excluding the number itself then 9 is a counter example: its only proper factor is 3. So 1+3=4, again even. In either interpretation, you appear to have made a mistake.
     
  9. Jun 7, 2006 #8
    yeah, i did mean discluding the number itself and i realise my mistake now!!!
     
  10. Jun 7, 2006 #9
    i do hope you realise you have just broken my heart!!!!!!!! if i change it to diclude odd squares eg, 9 with the argument they would not be big enough? no that wouldnt work with numbers like 81 where their square root has factors in turn eg 81 = 1 + 9 + 3 + 27!aaaahhhhhh!ill keep working on it!:)!
     
  11. Jun 7, 2006 #10

    shmoe

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    If n is quasiperfect the sum of it's divisors is odd in any case.

    Two problems:

    If n is quasiperfect and odd, prove n is a perfect square. (easier)

    If n is quasiperfect, prove n is odd. (harder)
     
  12. Jun 10, 2006 #11
    answer to first one is that the factors and one must add up to be even. therefore you must have a factor which only counted once to allow the total to be even so therefore a sqaure
     
  13. Jun 10, 2006 #12
    answer to first one is that the factors and one must add up to be even. therefore you must have a factor which only counted once to allow the total to be even so therefore a sqaure
     
  14. Jun 10, 2006 #13

    shmoe

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    Why even again? If n is quasiperfect, the sum of it's divisors is 2n+1, which is odd.

    All divisors are counted only once in any case.

    Try considering the prime factorization of n. You can find the sum of the divisors in terms of this factorization (alternatively, you can count the number of divisors it in terms of this factorization, all divisors are odd, so..).
     
  15. Jun 11, 2006 #14
    no thats wrong slightly excessive numbers are actuallly where all n's factors add up to one more than n- i dont know where 2n came from!!!!!!
    so if n is odd all the factors must add to one more...even!
     
  16. Jun 11, 2006 #15

    Hurkyl

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    n is a factor of n...
     
  17. Jun 11, 2006 #16

    shmoe

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    I was going by the referece that matt found that for some reason you didn't bother objecting to. By the definition you seem to mean here, "slightly excessive numbers" are exactly the primes, so I would wonder what the point is. Maybe you mean the sum of n's proper divisors add up to n+1 (this would be equivalent to the sum of divisors adding up to 2n+1)?
     
  18. Jun 11, 2006 #17

    HallsofIvy

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    No, that's right. A number is perfect if all factors less than itself add up to itself: example, 6 has factors 1, 2, 3, 6. The sum of all numbers less than itself is 1+ 2+ 3= 6. Of course, that means that the sum of all factors of 6 is 1+ 2+ 3+ 6= 2*6. The sum of all factors of a perfect number, n, add up to 2n. The sum of all factors of an "excessive number" add up to 2n+ 1, the same as saying that the sum of all factors less than n add up to n+1.
     
  19. Jun 12, 2006 #18
    no, you dont count the number itself...
    ...ahhh now i see where sigma(n) +1 = 2n+1... thats when you include the number itself as afactor
     
  20. Jun 12, 2006 #19

    matt grime

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    This is what happens if you don't distinguish between proper divisors and divisors, you see.
     
  21. Jun 12, 2006 #20

    HallsofIvy

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    Is there a law to that effect? :rolleyes: You can define "perfect numbers" or "slightly excessive" numbers either way.
     
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