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

Moledet, Aug 26, 2005

I found two more big solutions by combining Nrs(75,76) and Nrs(117,121) of Alpers list:

So there are now 6 solutions above one million, two stemming from the twins Nrs(13.14) and Nrs(29,30) and four based on Alper's list only, including the biggest originating from the “triplet” Nrs(96,97).

Concerning the provisional numbering I mentioned in the P.S. to my previous letter the parameter q = 1.35 should have been q = 1.48 but probably still better would be apartition into 5 intervals of 30 each, which leads to

M = {(N – 93,200)/233,400*(q-1)} +1, q = 1.22

~Nr = {(log M/log q) + 2 }*30

Accordingly we get the following table:

By the way, for the matter of exactness: in the list of my first letter I perceived another misprint: in the solution Nr8, second part, apparently should be

1331 = 11³...

Concerning the combination of two pairs of solutions to get a new and bigger solution I found after the original “internal” case and the “external” case a third possibility using sequences as well, arriving at the following resume:

1. original case: d=1, δ=0.

   a = P(a) -> S(a)

   a + d = P(a + 1) -> S(a), let b > a and another pair of solution

   b = P(b) -> S(b)

   b + 1 = P(b + 1) -> S(b)

   D = | a(b+1) – b(a+1)| = b-a;

   if D/a(b+1), D/b(a+1) (1) then

   c = a(b+1)/D, c+1 = b(a+1)/D is a new and bigger solution with S(c) = S(b) – S(D) = S(c+1) (1a)

   If condition (1) is not fulfilled, let be D₁ = [a(b+1), b(a+1), D] the g,c,d of the three and D = D'x D'', then

   c = a(b+1)/D' , c + D'' = b(a+1)/D'

   S( c ) = S(a) + S(b) – S(D') = S(c+ D'')

   Is an “extended” external solution.

2. external case: d=d(a) = d₁

   d(b) = d₂

   while δ=0 and d₁ ≠d₂ if notd_{1 =} d_{2 = 1}

   a = P(a) -> S(a)

   a+d₁ = P(a+ d₁₎ -> S(a)

   b = P(b) -> S(b)

   b+d₂ = P(b+ d₂₎ -> S(b)

   D = | a(b+ d₂) – b(a+ d₁) | = | a d₂ – b d₁| = d₁ d₂ | a / d₁ – b / d₂| =

   = | R₁ – R₂| where R is a comparative measure value for a solutions;

   if D/ a(b+ d₂), D/ b(a+d₁) (2)

   then c = a(b+ d₂)/D, c+1 = b(a+ d₁)/D is new and bigger solution with

   S(c ) = S(a) + S(b) – S(D) = S(c+1)(2a);

   (possibly c, c+1 must be interchanged)

   As before in ( 1 ) D = D'D'' can be split, if ( 2 ) is not fulfilled. Finally

3. last case: d_1,d_2; δ ≠0

   a = P(a) -> S(a)

   a+d₁ = P(a+ d₁) -> S(a) + δ

   b = P(b) ->S(b)

   b+d₂ = P(b+ d₂₎ -> S(b) + δ

D as before as well as condition (2)and c, c+1 , and we get

S(c ) = S(a) + (S(b+ δ)) – S(D) = S(c+1) = (S(a) + δ)) + S(b) – S(D) (3a)

for the new solution. Again D =D'D'' eventually can be split, and then

case (3) leads back to be a part of case (2).

Case (3) fits particularly well to generate the very low solutions:

2 = 2

3 = 3

4 = 2² -> 4

5 = 5

While d₁ = 1 = d_2, δ = 1, D = | 2x5 – 3x4 | = 2

c = 2x5/2 = 5

c+1 = 3x4/2 = 6

which is Nr1.

4 = 2² -> 4

5 = 5

9 = 3² -> 6

10 = 2x5-> 7

While d₁ = 1 = d_2, δ = 1, D = | 4x10– 5x9 | = 5

c = 4x10/5 = 8

c+1 = 5x9/5 = 9

which is Nr2.

9 = 3² -> 6

10 = 2x5-> 7

24 = 2³x3-> 9

25 = 5² -> 10

While d₁ = 1 = d_2, δ = 1, D = | 9x25– 10x24 | = 15

c = (9x25)/(3x5) = 15

c+1 = (10x24)/(3x5) = 16

which is Nr3.

35 = 5x7-> 12

36 = 2²x3² -> 10

65 = 5x13-> 18

66 = 2x3x11-> 16

While d₁ = 1 = d_2, δ = -2, D = | 35x66– 36x65 | = 30x | 7x11–6x13| = 30

c = 77

c+1 = 78

which is Nr4.

80 = 2⁴x5-> 13

81 = 3⁴ -> 12

224 = 2⁵x7-> 17

66 = 3²x5² -> 16

While d₁ = 1 = d_2, δ = -1, D = | 80x225–81x224 | = 2⁴x3² x| 5³–9x14| = 144

c = 125

c+1 = 126

which is Nr5.

For Nr6 = (714,715) = Nr5x {425 = 5²x17

429 = 3x11x13}

we already had a simple external combination with Nr5 in my previous letter.

In my first letterI mentioned Nrs(20,21) -> Nr36

Now I found another possibility

Nr36 = (Nr8 x Nr10) x Nr13 with d=2.