A Revision

In my previous post, I made a modification to the following algorithm:

Suppose we take any positive integer \(n \gt 1\) 

• if prime, double it and add 1: \(n \rightarrow 2n+1\)
• if composite, determine its number of divisors \(d\)
• if \( n \pmod d \equiv 0\) then \(n \rightarrow \dfrac{n}{d} \)
• if \( n \pmod d \not\equiv 0 \) then \(n \rightarrow n \times d\)

Keep repeating this process until a loop is reached or call a stop after a fixed number of iterations. 

The modification I made prevented the numbers generated from becoming too large too quickly. Instead of doubling a prime and adding 1, I decided to do this only if the number was a \(4k+1\) prime. If it was a \(4k+3\), I subtracted 1 and divided by 2. The new algorithm looks like this:

• if a \(4k+1\) prime, double it and add 1: \(n \rightarrow 2n+1\)
• if a \(4k+3\) prime, subtract 1 and divide by 2: \(n \rightarrow (n-1)/2\)
• if composite, determine its number of divisors \(d\)
• if \( n \pmod d \equiv 0\) then \(n \rightarrow \dfrac{n}{d} \)
• if \( n \pmod d \not\equiv 0 \) then \(n \rightarrow n \times d\) 

Here is an example using 28069 (permalink):

--- Loop detected at value 149708 ---

Divisors to Sequence:

28069, 56139, 224556, 18713, 37427, 149708, 1796496, 71859840, 561405, 8982480, 112281, 898248, 28743936, 2069563392, 10778976, 149708

------------------------------

Sequence Length: 16

Highest Value:   2069563392

This method of dealing with primes is also better suited to the sequences I mentioned earlier in my two posts: 

• Number's Factors to Sequence Algorithm 1
• Number's Factors to Sequence Algorithm 2

Here are the two new algorithms. 

Suppose we take any positive integer \(n \gt 1\) and apply the following rules to it:

• if a \(4k+1\) prime, double it and add 1: \(n \rightarrow 2n+1\)
• if a \(4k+3\) prime, subtract 1 and divide by 2: \(n \rightarrow (n-1)/2\)
• if composite, determine its number of factors \(f\) counted \( \textbf{with multiplicity}\)
• if \( n \pmod f \equiv 0\) then \(n \rightarrow \dfrac{n}{f} \)
• if \( n \pmod f \not\equiv 0 \) then \(n \rightarrow n \times f\)

Keep repeating this process until a loop is reached or call a stop after a fixed number of iterations. 

Here is an example using 28069 (permalink):

--- Loop detected at value 37427 ---

Number of factors to sequence with multiplicity:

28069, 56139, 112278, 37426, 18713, 37427, 74854, 224562, 898248, 149708, 37427

--------------------

Sequence Length: 11 

Highest Value:   898248 

Suppose we take any positive integer \(n \gt 1\) and apply the following rules to it:

• if a \(4k+1\) prime, double it and add 1: \(n \rightarrow 2n+1\)
• if a \(4k+3\) prime, subtract 1 and divide by 2: \(n \rightarrow (n-1)/2\)
• if composite, determine its number of factors \(f\) counted \( \textbf{without multiplicity}\)
• if \( n \pmod f \equiv 0\) then \(n \rightarrow \dfrac{n}{f} \)
• if \( n \pmod f \not\equiv 0 \) then \(n \rightarrow n \times f\)

Keep repeating this process until a loop is reached or call a stop after a fixed number of iterations. 

Here is an example using 28069 (permalink):

--- Loop detected at value 224562 ---

28069, 56139, 112278, 37426, 18713, 37427, 74854, 224562, 898248, 224562

--------------------

Sequence Length: 10

Highest Value:   898248