HOME plateWON | World!OfNumbers Sets of three composites in bi-directional'Sum Of Prime Factors' pro-/ retrogression

Introduction

This page is the continuation of the topic started in WONplate 78.
Remarkable progress was made by Roberto Botrugno from Italy.
and enjoy the larger solutions he discovered using his algorithm.

The Tables

Despite the extraordinary efforts of Roberto Botrugno
( please take a look at his remarkable solutions of up to length 62 !! )
the FOURTH triplet still hasn't been determined.

The following triplet is the smallest one whereby the terms have exactly two prime factors
immediately followed by his record triplet with terms of length 102 !
In the words of Roberto Botrugno :

" I have verified every possible combination between sopf_840 and sopf_99588340320
(my smallest solution) and there aren't any other P_sopf solutions.
I don't know if there is a F_sopf."

Definition of P-sopf and F_sopf.
F_sopf is a number that creates more than 2 factors numbers in every term (sopf_716).
P_sopf is a number that creates only 2 prime factors for every term (sopf_24, sopf_840, ...).
At the moment to find P_sopf is easier for me than to find F_sopf
so I'll work on the F_sopf ehehe :-)

Read on for a more detailed explanation of his P_sopf algorithm

 Roberto Botrugno made a connection with Puzzle 97 of PP&P [ January 13, 2002 ] Note that 298687992/716 = 417162 298688708/716 = 417163 298689424/716 = 417164 if 1) sopf(a) = sopf(a+1) = sopf(a+2) ; 2) the equation sopf(a)+b+c = b*c has integer solutions 3) sopf(a*b*c) = sopf(a*b*c+1) then a*b*c is a F_sopf. Example 533+(535+3) = 535*2 but sopf(417162*535*2) = 647 533+(268+3) = 268*3 but sopf(417162*268*3) = 607 533+(90+7) = 90*7 but sopf(417162*90*7) = 553 533+(179+4) = 179*4 and sopf(417162*179*4) = sopf(417163*179*4) Maybe four consecutive numbers with the same sopf create a quadrisopf of the form F_sopf. (ps. quadrisopf of the form P_sopf is impossible because there doesn't exist 4 squares in arithmetic progression. )

Here is Roberto Botrugno's method for discovering these huge P_sopf solutions

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First example consider sopf_24 (95, 119, 143).

Note that :
95 + 24 + 24 = 143
95 + 24 + 24 + 1 = 144
95 + 49 = 12^2
95 + 7^2 = 12^2

119 + 24 = 143
119 + 24 + 1 = 144
119 + 25 = 12^2
119 + 5^2 = 12^2

143 + 1 = 144
143 + 1 = 12^2
143 + 1^2 = 12^2

Rewrite (95, 119, 143) as follows
(12^2-7^2, 12^2-5^2, 12^2-1^2)
Note that 1^2, 5^2 and 7^2 are squares in arithmetic progression (24).

a^2 - b^2 = (a+b)(a-b)
(12+7)(12-7) = 95,
(12+5)(12-5) = 119,
(12+1)(12-1) = 143.

sopf_((12+7)+(12-7)) = sopf_((12+5)+(12-5)) = sopf_((12+1)+(12-1)) = 24
if and only if
12+-7, 12+-5 and 12+-1 are all primes.

Counterexample consider 60^2-17^2, 60^2-13^2 and 60^2-7^2 =>
77 * 43,  47 * 73 , 67 * 53.
giving
77 * 43 = 3311
47 * 73 = 3431
67 * 53 = 3551
and
3551 - 3431 = 120
3431 - 3311 = 120
but, sopf_(77 * 43) = 61 because 77 is composite.
sopf_(47 * 73) = 120 and sopf_(67 * 53) = 120 because 47 and 67
are all prime.
17^2, 13^2 and 7^2 are square numbers in arithmetic progression (120).

Second example consider sopf_840 (174191, 175031, 175871)
These triplet terms can be expressed as a difference of 2 squares :
174191 = 420^2 - 47^2
175031 = 420^2 - 37^2
175871 = 420^2 - 23^2

420+-47=373,467
420+-37=383,457
420+-23=397,443

sopf_(467 * 373) = sopf_(457 * 383) = sopf_(443 * 397) = 840
because all six resulting numbers are prime.
47^2, 37^2 and 23^2 are in arithmetic progression (840).

In general the problem is to find 3 square numbers in
arithmetic progression so that
a^2-b^2=k , b^2-c^2=k , (e.g. 47^2-37^2=840, 37^2-23^2=840)
and (k/2)+-a, (k/2)+-b, (k/2)+-c are prime.

Consider 3 integers a,b,c.
If a^2-b^2=k and b^2-c^2=k
then a^2, b^2 and c^2 are 3 square numbers in arithmetic
progression with difference k.
If k/2+-a, k/2+-b and k/2+-c are all primes
then (k/2)^2-a^2,(k/2)^2-b^2 and(k/2)^2-c^2 are 3 numbers
in arithmetic progression
with difference k having the same sopf = k.

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Enjoy the following P_sopf solutions kindly sent to me by Roberto !
It is not an exhaustive list. Roberto has in the mean time found many more triplets.

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Contributions

Roberto Botrugno (email) from Italy.

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