Herman Baer on Ruth-Aaron pairs, an extract from a letter.

Moledet, August 1, 2005

I got Alper's list of 149 solutions and inspected it: to begin with the most remarkable: I found a “triplet”, that is the pairs 96 and 97 consisting of:

417162 = 2 x 3 x 251 x 277 -> 533 ------- a - 1
417163 = 17 x 53 x 463 -> 533 ------- a
 
417163 = 17 x 53 x 463 -> 533 ------- a
417164 = 22x 11 x 19 x 499 -> 533 ------- a + 1

Which may be combined to a new pairs (a2 -1 , a2)

Nr2050 : 174,024,968,568 = 23 x 11 x 19 x 251 x 277 x 499 -> 1066

174,024,968,569 = 172 x 532 x 4632 -> 1066 = 2 x 533

The estimated number of this solution is about 2050 according to a supposed continuation of Alpser's list.

Next Alper's list shows that I stopped my research less than 200 numbers before his pair Nr 47, while I estimated to be Nr40 ~ ; so I was induced to a very wrong impression concerning the density of solutions, instead of about 50, as I expected.

In the days, which passed while my first letter to you was on the way I found, using “external” combination, the solution Nrs48, 61, 67, 76, 117, 144 of Alper's list, the first 4 of which I supposed to follow Nr47 in direct succession. Only the first really did, the others caused me to believe - erroneously - in a very low density.

By the way Nr117 = Nr67 x Nr76 is a combination of the two preceding ones.

By combination of Alper's Nr46, Nr47 I arrived at a solution (Nr 692~)

48,996,584 = 23 x 7 x13 x 17 x 37 x 107 -> 187

48,996,585 = 32 x 5 x11 x 312 x 103 -> 187

Here is the place to mention a misprint, which occurred with the first big solution (Nr100~), which should be 2,923,920 (Nr 268~)

2,923,921

(by fault a digit “9” was inserted before the right most digit).

The then biggest solution 101,429,900 and 101,429,901 would get the estimated Nr812~.

Beside I discovered

Nr6 714 = 125 x 425 = 52 x 17 ->27

715 = 126 x 429 = 3 x 11 x 13 ->27

While d = 4, D = 75 = 3 x 52 ->13

Nr5 15 + 27 - 13 = 29,

Nr8 has two generating external combinations, for Nr10 I found one; Nr30 is an external combination with Nr6.

That is about all.

Yours sincerely Hermann Baer

P.S The provisional formula to estimate the current number of a solution I got inverting a geometric progression based on Alper's list;

Let q = 1.35, N the lesser of the pair of a solution, then

M = (((N - 70,000) / 370,000) x (q - 1)) + 1

~ Nr = (log M / log q + 1) x 50