Puzzle 1142 p(n+1) = floor[p(n)#*k - p(n)*floor(p(n-1)#k-1)]; P(1)=2, k=2.920050977316134...

Puzzle 1142 p(n+1) = floor[p(n)#*k - p(n)*floor(p(n-1)#k-1)]; P(1)=2, k=2.920050977316134...
Davide Rotondo sent the following puzzle:

In A249270 is reported and studied the function:

p(n+1) = floor[p(n)#*(2.920050977316134...) - p(n)*floor(p(n-1)#*(2.920050977316134...)-1)]; p(1)=2

From the cited sequence A249270 we may learn that this function was published in:

Dylan Fridman, Juli Garbulsky, Bruno Glecer, James Grime and Massi Tron Florentin, A Prime-Representing Constant, The American Mathematical Monthly, Vol. 126, No. 1 (2019), pp. 70-73, ResearchGate link.

In the cited 2020 Numberphile video by James Grime and Brady Haran, we may learn much more details.

Three of these details are:

a) This function provides all of the primes in order from 2 to infinite without any gap...as far as we count with an infinite constant 2.920050977316134...
b) But this function is not "predictive" in the sense that the calculation of this infinite constant depends of our previous knowledge of the infinite series of the prime numbers.
c) Nowadays there are known 10000 terms of this constant. In A249270, see Alois P. Heinz, Table of n, a(n) for n = 1..10000

Rotondo provided the following examples:

5 = floor[(3#)*(2.920050977316134...)-3*floor(2#*(2.920050977316134...)-1)]

7 = floor[(5#)*(2.920050977316134...)-5*floor(3#*(2.920050977316134...)-1)]

11 = floor[(7#)*(2.920050977316134...)-7*floor(5#*(2.920050977316134...)-1)]

13 = floor[(11#)*(2.920050977316134...)-11*floor(7#*(2.920050977316134...)-1)]

and its largest example he could compute:

Let’s compute p(31)=127
Let’s k=2.920050977316134…

p(31)=floor[p(30)#*k-p(30)*floor(p(29)#*k-1)]=
P(31)=floor[113#*k-113*floor(109#*k-1)]=
P(5)=floor[31610054640417607788145206291543662493274686990*k-113*
floor(279734996817854936178276161872067809674997230*k-1)]=
P(31)=floor[
92302970945767854832754821514910241686192128535.031856989000560574659842288968180186492388842728633
...
-113*floor(
816840450847503140112874526680621607842408217.124175725566376642253626922911222833508782202148041
-1
)]=
P(31)=floor[
92302970945767854832754821514910241686192128535.031856989000560574659842288968180186492388842728633
......
92302970945767854832754821514910241686192128408
=
P(31)=floor[92302970945767854832754821514910241686192128535.031856989000560574659842288968180186492388842728633-
92302970945767854832754821514910241686192128408]=
P(31)=floor[127.031...]
P(31)=127

Q Can you explain why this function works as has been told above?