20200803, 17:30  #12 
"Jeppe"
Jan 2016
Denmark
2·7·11 Posts 
It is perhaps also interesting to note the lengths of the "preperiods", or offsets. That is the number of terms in the LL sequence preceding the first occurrence of the period. With Batalov's data as input, that is easy (PARI/GP):
Code:
findOffset(p,t)=s=Mod(4,2^p1);S=s;for(i=1,t,S=S^22);i=0;while(s!=S,s=s^22;S=S^22;i++);i Result: Code:
findOffset(11,60) 1 findOffset(23,32340) 1 findOffset(29,252) 1 findOffset(37,516924) 5 findOffset(41,822960) 1 findOffset(43,420) 3 findOffset(47,20338900) 4 findOffset(53,1309620) 2 Not sure if the above values have any interesting interpretation. We see that for the cases I was able to resolve, the constant 120000 in Batalov's C program was on the safe side. Of course, Viliam Furik's approach (searching for "14") works exactly in the cases where findOffset gives 1. /JeppeSN 
20200803, 18:13  #13  
"Jeppe"
Jan 2016
Denmark
232_{8} Posts 
Quote:
Code:
findPeriod(p,seed=4)=s=Mod(seed,2^p1);for(i=1,120000,s=s^22);S=s;i=0;until(s==S,S=S^22;i++);i Code:
findPeriod(11) 60 findPeriod(11,10) 10 findPeriod(11,2/3) 5 Explicitly, the LL sequence for 2^11  1, starting from seed 2/3 == 683, is: 683 > 1818 > 1264 > 1034 > 620 > 1609 > 1471 > 160 > 1034 > etc. Similarly, for 2^29  1, and seed 2/3, the period length is 154, which is 14 mod 28. /JeppeSN 

20200803, 18:15  #14 
"Serge"
Mar 2008
Phi(4,2^7658614+1)/2
21644_{8} Posts 
You have a solid footing to retrace Tony's DiGraphs adventure, now! It is a bit of a déjà vu, but a good warm up to visit his old threads.

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