- Apr 14 8:09 PMOne more thing, the gap are added or subtracted from the number

below until there are to primes at the ends.

89 (-7) ((96)) (+1) 97 (+4) 101 (+2) 103

103-89 = 14

96 +/- 7

383 (-6) 389 (-8) 397 (-4) 401 (-1) ((402)) (+7) 409 (+10) 419 (+2)

421

421 - 383 = 38

402 +/- 19

509 (-7) ((516)) (+5) 521 (+2) 523

523 - 507 = 14

516 +/- 7

761 (-8) 769 (-4) 773 (-13) ((786)) (+1) 787 (+10) 797 (+12) 809

(+2) 811

811- 761 = 50

786 +/- 25

883 (-4) 887 (-19) ((906)) (+1) 907 (+4) 911 (+8) 919 (+10) 929

929 - 883 = 46

906 +/- 23

1109 (-7) ((1116)) (+1) 1117 (+6) 1123

1123 1109 = 14

1116 +/- 7

1129 (-17) ((1146)) (+5) 1151 (+2) 1153 (+10) 1163

1163 1129 = 34

1146 +/- 17

1249 (-10)1259 (-7) ((1266)) (+11) 1277 (+2) 1279 (+4) 1283

1283 -1249 = 34

1266 +/- 17

1303 (-4)1307 (-12) 1319 (-2) 1321 (-6) 1327 (-29) ((1356)) (+5)

1361 (+6) 1367 (+6) 1373 (+8) 1381 (+18) 1399 (+10) 1409

1406 1303 = 106

1356 +/- 53

3221 (-8) 3229 (-17) ((3246)) (+5) 3251 (+2) 3253 (+4) 3257 (+2)

3259 (+12) 3271

3271 3221 = 50

3246 +/- 25

4201 (-5) ((4206)) (+5) 4211

4211 4201 = 10

4206 +/- 5

John

--- In primenumbers@yahoogroups.com, "John W. Nicholson"

<reddwarf2956@y...> wrote:>

wrong,

> I have been looking at this paper of Harvey's. I think I have

> discovered a few things which are very interesting. First let me

> list the numbers which he found not to be middle numbers < 2*10^10:

>

> N = 96, 402, 516, 786, 906, 1116, 1146, 1266, 1356, 3246, and 4206.

>

> Harvey stated to me the correction of 4208 should be 4206.

>

> Note that all except the second one, 402, is == 1 (mod 5). Also as

> would be expected all are divisible by 6, but they are a complete

> residue set with (mod 4) == (mod 24) (I might be stating this

> so let me restate this when divided by 24 the remainder is one of

namely

> 0,1,2, or 3.) There is only one of these with == 0 (mod 24),

> 96. There is also a interesting thing with (mod 11) and N/6 ==

(mod

> 11) most have a factor of 2 except the first three. I did look at

state

> (mod 7). There are no == 1,4 (mod 7). I wonder how this fits with

> http://primepuzzles.net/problems/prob_003.htm ?

>

> N, (mod 11), N/6 == (mod 11), (mod 7)

> 96, 8, 5, 5

> 402, 6, 1, 3

> 516, 10, 9, 5

> 786, 5, 10, 2

> 906, 4, 8, 3

> 1116, 5, 10, 3

> 1146, 2, 4, 5

> 1266, 1, 2, 6

> 1356, 3, 6, 5

> 3246, 1, 2, 5

> 4206, 4, 8, 6

>

> Add to these statement a look a MR1745569,

> http://primes.utm.edu/references/refs.cgi?author=Suzuki

> I have a scaned it if anyone wants a copy.

>

> Because of the above, 4*n+/-1 and 6*m+/-1 are the only ways to

> all primes factors with only two Dirichlet equations, and because

P2

> 4206 < 6!*6, I don't think there are any more possible. All other

> numbers > 6!*6 have some way of having numbers > 36 = 6^2 in the

> = p1*p2 product (prime factors are of 6*n +/- 0, 1, 2, 3, 4, or 5,

n

> < 6) + prime as to Chen's theorem (all even = P2 + p3).

GC

>

> I do see how this can be used as part of the proof of the t-prime-

> if anyone is up to it.

that

>

> Can anything more be said?

>

> John

>

>

> --- In primenumbers@yahoogroups.com, Harvey Dubner <harvey@d...>

> wrote:

> > I had a paper published in Journal of Recreational Mathematics,

> Vol.30(3),

> > 1999-2000, entitled

> > "Twin Prime Conjectures" which included the conjecture,

> >

> > Every even number greater than 4208 is the sum of two t-primes

> where a

> > t-prime is a prime which has a twin.

> >

> > This was verified up to 4.10^11 and included strong evidence

> the

includes

> > conjecture would hold for larger numbers. The paper also

> > interestng comparative Goldbach data.

asks.

> >

> > I will be happy to send a pdf copy of the paper to anyone who

> >

> > Harvey Dubner

> >

> > PS: No, the paper does not include a proof of the above

> conjecture or GC.

> > PPS: No, I am not working on such a proof.

> > - << Previous post in topic