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 :-)

Note that

if

Example

Maybe four consecutive numbers with the same sopf create a quadrisopf
of the form F_sopf.

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

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.






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.
If you want an up_to_date list please contact Roberto himself.