Penultimate Lucas-Lehmer step - Page 5

ATH · 2018-12-19, 21:26

4 and 10 is the beginning of 2 infinite series of universal starting values:
A(1)=4, A(2)=52, A(N)=14*A(N-1)-A(N-2)
B(1)=10, B(2)=970, B(N)=98*B(N-1)-B(N-2)

and in the article linked in post #31 they find other series.

In the infinite series the penultimate step is the same or reverse the initial value 4 and 10 alternating every 2nd step in the series, see bottom half of: penultimatellstep.txt

Once I calculated the antepenultimate (2nd to last) and anteantepenultimate (3rd to last) steps up to M35, and they are not the same or reverse of 4, 10 in the series, but there might be some other connection I did not find.

Batalov · 2018-12-19, 23:32

It is easier to mnemonize them as
A(0)=A(1)=4, A(N)=a*A(N-1)-A(N-2)
B(0)=B(1)=10, B(N)=b*B(N-1)-B(N-2)
Also notice that a=14 = A(1)^2-2 and b=98 = B(1)^2-2 (in other words, they are the second iteration value in LL process).

And of I remember correctly the 2nd series (the B(N) series) is simply the reflection of the A(N) series off the top of the range (in modular sense). Either A series of B series is enough to generate all seeds.

2018-12-19, 21:26	#45
ATH Einyen Dec 2003 Denmark 3⁵·13 Posts	4 and 10 is the beginning of 2 infinite series of universal starting values: A(1)=4, A(2)=52, A(N)=14A(N-1)-A(N-2) B(1)=10, B(2)=970, B(N)=98B(N-1)-B(N-2) and in the article linked in post #31 they find other series. In the infinite series the penultimate step is the same or reverse the initial value 4 and 10 alternating every 2nd step in the series, see bottom half of: penultimatellstep.txt Once I calculated the antepenultimate (2nd to last) and anteantepenultimate (3rd to last) steps up to M35, and they are not the same or reverse of 4, 10 in the series, but there might be some other connection I did not find. Last fiddled with by ATH on 2018-12-19 at 22:02

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2018-12-19, 23:32	#46
Batalov "Serge" Mar 2008 Phi(4,2^7658614+1)/2 3⁶·13 Posts	It is easier to mnemonize them as A(0)=A(1)=4, A(N)=aA(N-1)-A(N-2) B(0)=B(1)=10, B(N)=bB(N-1)-B(N-2) Also notice that a=14 = A(1)^2-2 and b=98 = B(1)^2-2 (in other words, they are the second iteration value in LL process). And of I remember correctly the 2nd series (the B(N) series) is simply the reflection of the A(N) series off the top of the range (in modular sense). Either A series of B series is enough to generate all seeds.