Re: More on triangle numbers and primes!




Dan wrote:
> Dan wrote:
> >> ( * ) used for spacers only.
> >> 1
> >> 3***2
> >> 6***3***5*|
> >> 10**4***7*| 9
> >> 15**5***9*|12**14
> >> 21**6**11*|15**18**20
> >> 28**7**13*|18**22**25**27
> >> 36**8**15*|21**26**30**33**35
> >> 45**9**17*|24**30**35**39**42**44
> >> 55*10**19*|27**34**40**45**49**52**54
> >> 66*11**21*|30**38**45**51**56**60**63*65
> >> 78*12**23*|33**42**50**57**63**68**72*75*77
> >> 91*13**25*|36**46**55**63**70**76**81*85*88*90
> >> etc.. ---->oo
> >
> >> Correction here in last entry of 8th column.
> >> Changed from 69 to 70.
> >> |
> >> V
> >> For all statements below --
> >> As far as even composites right of the column
> >> separator (|).
> >> I also believe this to be true, there are no
> >> even perfect numbers. 6,28,496,8128,... OEIS
> >> A000396.
>
> >Every even perfect number is of the form 2^(p-1) *
> >(2^p - 1), where
> >2^p - 1 is prime. It shouldn't be too hard to solve the >(Diophantine)
> >equation
>
> >n (n-1) / 2 - m (m - 1) / 2 = 2^(p-1) * (2^p - 1),
>
> >once you factor the left-hand side (like I did >elsewhere in this
> >thread), and then throw out the solutions where n - m =
> >1 or 2.
>
> So does this prove there are no perfect numbers
> to the right of the separator (|)?

I submitted a proof of this before I saw this reply. So, yes.

> >> Also no powers of 2.
>
> >Ditto, but for
>
> >n (n-1) / 2 - m (m - 1) / 2 = 2^M.
>
> But, isn't M a power of 2?

No, 2^M is the power of 2, so M is allowed to be any positive integer.
(The powers of 2 are 2^0, 2^1, 2^2, 2^3, ..., not 2^(2^0), 2^(2^1),
2^(2^2), etc.)

> The idea is no power of 2 is to the right of the
> separator(|)in the table above?

No power of 2 appears to the right of (|). (Again, check the post I
just made.)

> >> Maybe someone can come up with a proof here!
> >> I am not sure what else is missing in the
> >> even composites right of the (|) but
> >> I believe that is it. (ALL OF THE EVEN INTEGERS I MENTIONED)
> ------------------------------------------------------
> >> If some odd integer > 9 other than the primes
> >>is not found, then you have discovered the
> >> first odd perfect number. ;-)
>
> >I don't see how this follows.
>
> A new statement that is why I added "---".
>
> I have read somewhere that an odd perfect
> number does not exist but they cannot prove
> that it doesn't.

Where did you hear this?

> So if all the even perfect
> numbers do not exist right of the (|) then
> the first odd composite =>9 not found and not being
> prime might just be the 1st odd perfect number.

No odd primes occur to the right of the (|); that can be proven; see
one of my old posts:

] Thus, for any odd number N, if N is prime, there are exactly two
] solutions [to T(n) - T(m) = N]:
]
] ((N - 1) / 2, (N + 3) / 2) and (N, N + 1).

Note that in the first one, n - m = 2, so N appears in the column to
the left of the (|), and in the second one, n - m = 1, so N appears in
the column to the left of that one. N appears nowhere else.

> Because outside of the primes all odd integers
> =>9,I believe are represented to the right of
> the separator (|) in the above table --->oo.
> So if 1 odd composite is not found, then it would
> follow as an odd perfect number?

Every odd composite integer appears to the right of (|), so your
statement is of the form FALSE -> P, so it's "vacuuously" true. Another
example would be: If N is an integer such that N^2 = -1, then N is a
perfect number. The statement is true, because there's no way to have
the first part happen.

The proof when N is an odd composite number was another part of a post
I made to this thread:

] Now assume N is not prime, so that N = A * B (where A and B are odd
and
] thus >= 3). Then let
]
] m = B - A/2 + 1/2 and n = A/2 + B + 1/2.
]
] m and n are integers, and n * (n-1) / 2 - m * (m-1) / 2 = N.
]
] Note that n - m is A >= 3.

The condition that n - m >= 3 is equivalent to N appearing to the right
of the vertical bar (|).

> Unless of course it could be prime! ;-)

No, no prime numbers are perfect. (1 =/= p.)

--- Christopher Heckman

.



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