[seqfan] Another sequence.

[Ed Jeffery]

SeqFans:

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).

Let n in {1,2,...}. Define a sequence {a(n)} using the following procedure.
For each n,

START:

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

Step 1b. (Loop.) c_0 = g(f(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 evaluation of the next n.

Step 3. Otherwise, let  c_{j+1} = g(f(c_{j})), for j > 0, and if for any j
we have c_{j+1} = 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+1} is a prime.

Step 4. Let b = sum_{i = 0,...,j} c_j. If b is a prime, then put a(n) = 1,
otherwise loop to Step 1b.

END.

CONJECTURE: If c_{j+1} =!= c_j for any j, then the procedure always
terminates.

I hope the conjecture is true, but I really would not mind seeing a
counterexample. If there are counterexamples, then are they finite in
number and do they otherwise have properties in common?

Other more interesting sequences could be the sequence of those n for which
a(n) = 1, or the sequence of primes encountered: I wonder how exhaustive of
the primes that sequence might be. Finally, another sequence could be the
number of steps required to reach termination of the procedure, with 0
denoting an infinite number of steps as in the first part of Step 3.

LEJ

REFERENCES:

[1] Sequence A002808 in OEIS, https://oeis.org/A002808