Quote:
Originally Posted by Happy5214
Conjecture 103104 is false (and doesn't even make any sense).

You are right.
Thank you very much for this comment.
It is an error in the statement.
I have corrected and the conjecture 104 becomes :
The prime number 13 appears in the decomposition of the terms of index 1, 2 of all sequences that begin with the integers 11^(12*k).
I have verified it up to k=50.
But we will encounter with this conjecture the same problem as with the others of the same type.
Indeed, (11^121)/10 = 2^3*3^2*7*13*19*37*61*1117 and it is the number 1117 that maintains the 13 at the next iteration.
And, for the sequence 11^(12*1117), the factor 1117^2 appears in the factorization of the term in index 1.
Quote:
Originally Posted by warachwe
The problem is that when the factor 13 are with even power, it does not preserve the 7 for the second iteration.
This only happen when k is multiple of 13, for example 2^(12*13)1 =3^2*5*7*13^2*53*79*...
This is why conjecture 34 ( 3^(18*37) ), 35 ( 3^(36*37) ), and 106 ( 11^(6*37) ) are false.
But this doesn't mean all similar conjecture are false, as there maybe others prime(s) that preserve p.
When we try to 'get rid of' those primes, there maybe yet another that will preserve p instead. Since the size of first iteration grow very quickly, it is hard to find other contradiction this way.
If some of those are true, I imagine the proof might be similar to the proof of conjecture (2).

Many thanks again for these observations !
 Conjecture (10) : Maintained, by luck ! Yes, the 13^2 factor appears if k is a multiple of 13, but by luck, there must be other primes that maintain the 7 factor at the second iteration.
 Conjecture (34) : invalidated.
 Conjecture (35) : invalidated.
 Conjecture (106) : invalidated.