[seqfan] Re: Another sequence.

[Ed Jeffery]

Oops, I am sorry to respond to my own post, but I forgot to add A(n) in
Step 1b to b from Step 4 without which the conjecture is obviously true. So
I corrected it and listed the procedure again, with a couple of other minor
corrections, as follows:


For N a positive integer, let f(N) = p_1^{k_1}*...*p_m^{k_m} be the unique
factorization of N, where p_1, ..., p_m exhaust the distinct prime divisors
of N and k_1, ..., k_m >= 1. Let g(f(N))  = p_1*k_1 + ... + p_m*k_m be the
sum of all prime divisors of N including multiplicities. Let A = A002808
[1] (composite numbers).

For n = 1,2,..., define a sequence {a(n)} using the following procedure.

START:

Step 1a. Let c_0 = g(f(A(n))). Go to Step 2.

Step 1b. (Loop) Put c_0 = g(f(A(n) + b)), if this is a repetition of the
procedure.

Step 2. If c_0 is a prime then we are done and we put a(n) = 1. Exit the
procedure and move on to the next n.

Step 3. Otherwise, let  c_j = g(f(c_{j-1})), for j > 0, and if for any j we
have c_j = c_{j-1}, with c_j a composite number, then put a(n) = 0, exit
the procedure and move on to the next n; otherwise continue with iterations
in this step until the first occurrence for which c_j is a prime.

Step 4. Let b = sum_{i = 0,...,j} c_j. If b is a prime, then put a(n) = 1,
exit the procedure and move on to the next n; otherwise loop to Step 1b.

END.

Question: If in each iteration of the procedure there is some j for which
c_j is a prime, then does the procedure eventually terminate (with b equal
to a prime)?