Moledet, Aug 26, 2005
I found two more big solutions by combining Nrs(75,76) and Nrs(117,121) of Alpers list:
Nrs(75,76) | 17,657,288 = 2^{3}x11^{2}x17x29x37-> 111 |
17,657,289 = 3^{3}x13^{3}x19x47-> 111 | |
Nrs(117,121) | 6,723,584 = 2^{11}x7^{2}x67 -> 103 |
6,723,585 = 3^{2}X5x11x17^{2}X47 ->103; |
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:
Appr. | Diffr.Δ | |
---|---|---|
Nrs(13,14) | = 2,923,200 (+1) ~ Nr256 | |
Nrs(117,121) | = 6,723,584 (+1) ~ Nr359 | 103 |
Nrs(75,76) | = 17,657,288 (+1) ~ Nr492 | 133 |
Nrs(46,47) | = 48,996,584 (+1) ~ Nr641 | 149 |
Nrs(29,30) | = 101,429,900 (+1) ~ Nr749 | 108 |
Nrs(96,97) | = 174,024,968,568 (+1) ~ Nr1123 | 1123 |
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 = 11^{3}...
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:
general case
a = P(a) -> S(a)
a + d = P(a + d) -> S(a) + δ
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 D_{1 }= [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.
external case: d=d(a) = d_{1}
d(b) = d_{2}
while δ=0 and d_{1 }≠d_{2 }if notd_{1 = }d_{2 = 1}
a = P(a) -> S(a)
a+d_{1} = P(a+ d_{1) }-> S(a)
b = P(b) -> S(b)
b+d_{2} = P(b+ d_{2) }-> S(b)
D = | a(b+ d_{2}) – b(a+ d_{1}) | = | a d_{2 }– b d_{1}| = d_{1 }d_{2 }| a / d_{1 }– b / d_{2}| =
= | R_{1} – R_{2}| where R is a comparative measure value for a solutions;
if D/ a(b+ d_{2}), D/ b(a+d_{1}) (2)
then c = a(b+ d_{2})/D, c+1 = b(a+ d_{1})/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
a = P(a) -> S(a)
a+d_{1} = P(a+ d_{1}) -> S(a) + δ
b = P(b) ->S(b)
b+d_{2} = P(b+ d_{2) }-> 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 = 2^{2 }-> 4
5 = 5
While d_{1} = 1 = d_{2, }δ = 1, D = | 2x5 – 3x4 | = 2
c = 2x5/2 = 5
c+1 = 3x4/2 = 6
which is Nr1.
4 = 2^{2 }-> 4
5 = 5
9 = 3^{2 }-> 6
10 = 2x5-> 7
While d_{1} = 1 = d_{2, }δ = 1, D = | 4x10– 5x9 | = 5
c = 4x10/5 = 8
c+1 = 5x9/5 = 9
which is Nr2.
9 = 3^{2 }-> 6
10 = 2x5-> 7
24 = 2^{3}x3-> 9
25 = 5^{2 }-> 10
While d_{1} = 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 = 2^{2}x3^{2 }-> 10
65 = 5x13-> 18
66 = 2x3x11-> 16
While d_{1} = 1 = d_{2, }δ = -2, D = | 35x66– 36x65 | = 30x | 7x11–6x13| = 30
c = 77
c+1 = 78
which is Nr4.
80 = 2^{4}x5-> 13
81 = 3^{4 }-> 12
224 = 2^{5}x7-> 17
66 = 3^{2}x5^{2 }-> 16
While d_{1} = 1 = d_{2, }δ = -1, D = | 80x225–81x224 | = 2^{4}x3^{2} x| 5^{3}–9x14| = 144
c = 125
c+1 = 126
which is Nr5.
For Nr6 = (714,715) = Nr5x {425 = 5^{2}x17
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.