## 25567Re: Divisibility of the Sum of Two Prime Terms of Arithmetical Progres sions

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• Jun 25, 2014
Thank you Tom. I now realize to my embarrassment that my hasty reply to David's message 25561 is ridiculously confusing. I apologize to David and the group, and hope the following revised version of my message 25560 better explains my intent.

As used here, “divisible” means division with no remainder, and “twin” means either the smaller or larger prime of a twin prime pair.

One, If an arithmetical progression of the form AP=A+B*n, n=0, 1, 2, 3...n, contains prime terms, and B is divisible by 12, then the sum S of any 2 prime terms is never divisible by any twin prime middle number.

Two, If an even positive integer N is not divisible by 4 and not divisible by 6, then it is not divisible by any twin prime middle number.

For both of the above statements, S and N very likely will meet at least one of these 3 conditions:

1) They equal the sum of 2 twins, both being either small or large twins.

2) They divide a twin prime middle number.

3) They divide a twin prime middle number, a twin prime middle number of times.

Here is a large example for statement one:

Pick A=15! +1=1307674368001 and B=15*12=180.

For the AP=A+B*n,
The 15th prime term= 1307674400041.
The 30th prime term= 1307674418221.
The sum S of prime terms= 2615348818262.

For this example, the 3 conditions are met.

S is the sum of large twin= 3529 and large twin= 2615348814733.

S divides the twin prime middle number 133382789731362, 51 times.

S divides the twin prime middle number= 124752138631097400, 47700 times. 47700 is a twn prime middle number.

I feel there is a simple explanation of why the exercises work, but I can’t think of one. Can anyone educate me? Thanks folks.

Bill Sindelar

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