returns the index of its argument in the array of primes. So:

primecardinal(2) is 1

primecardinal(3) is 2

primecardinal(5) is 3

...

Now, note that D(1)=primecardinal(P1)-primecardinal(Q1) starts as a

non-negative integer, and that at each successive step, the

value of D(n)=primecardinal(Pn)-primecardinal(Qn) either stays the

same or gets smaller. This is because the P values only move

by one at each step, while the Q values move by one or more

at each step.

So to show that it always ends, you just need to show two things:

that D(n) can only have an infinite run of constant values if it

is 0, and that D(n) can't cross over from being a positive integer

to a negative integer. The details are left up to the reader. :)

On 1/26/2012 12:05 PM, w_sindelar@... wrote:

>

>

> One, start with any odd positive prime P1 and put it in row 1 column P of a two column (P, Q) array.

>

>

> Two, if the rightmost digit dp of P1 equals 1, put Q1=11 in row 1 column Q. If dp equals 3, put Q1=3 in row 1 column Q. If dp equals 7, put Q1=7 in row 1 column Q. If dp equals 9, put Q1=19 in row 1 column Q.

>

>

> Three, check if Q1 equals P1. If so stop.

>

>

> Four, if Q1 does not equal P1, calculate the smallest prime P2 greater than P1 and put it in row 2 column P.

>

>

> Five, find the smallest prime Q2 greater than Q1, and such that rightmost digit dq of Q2 equals the rightmost digit dp of P2. Put Q2 in row 2 column Q.

>

>

> Six, go to step three with Q2 replacing Q1 and P2 replacing P1. Keep repeating and tabulating until Qn equals Pn.

>

>

> Remarkably, this equality (Qn=Pn) always occurred for each of the many random P's I tried.

>

>

> Below is a small example showing the result of each iteration. The format is (P1, Q1, 1), (P2, Q2, 2), (P3, Q3, 4)…(Pn, Qn, n).

>

>

> (61, 11, 1), (67, 17, 2), (71, 31, 3), (73, 43, 4), (79, 59, 5), (83, 73, 6), (89, 79, 7), (97, 97, 8). P120116-00

>

>

> Notice that by design, column P contains a set of 8 consecutive primes with a certain rightmost digit pattern, and column Q contains a set of 8 distinct smallest possible primes with the same rightmost digit pattern.

>

>

> Here is a large example showing only (P1, Q1, Count 1) and (Pn, Qn, Count n).

>

>

> First P= 1073741827. First Q= 7. Count n= 1. Last P= 1511334911. Last Q= 1511334911. Count n= 20862576.

>

>

> Will the exercise always end with Qn=Pn for any odd positive prime P?

>

>

> Thanks folks.

>

>

> Bill Sindelar

>

>

>

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