Claude to the Rescue

The number 27983, associated with my diurnal age today, has a property than confers it membership in OEIS A335752: the lexicographically earliest sequence of primes, each two of which sum to a triprime. The sequence begins: 3, 5, 47, 239, 3023, 27983, 53231, ... so let's confirm that these numbers satisfy using 3 as a starting point:

• \(3 + 5 = 8 =2^3\)
• \(3 + 47 = 50 = 2 \times 5^2\)
• \(3 + 239 = 242 = 2 \times 11^2 \)
• \(3 + 3023 = 3026 = 2 \times 17 \times 89 \)
• \(3+27983 = 27986 = 2 \times 7 \times 1999\)
  
  

The sums above are all triprimes and we will find that any number in the list paired with any other number will produce a triprime. The code for this was included in the OEIS comments but was in Mathematica. I got Gemini to convert it to Python code and then Grok but in both cases the primeomega function was misunderstood with both claiming that primeomega(12) =2 and not 3. 

I didn't fully understand the Python code and was looking to Gemini and then Grok to explain it to me. It was only Claude that succeeded to my satisfaction. Here is the code followed by Claude's explanation. I think Claude will be my go to resource for anything to do with coding.

from sympy import nextprime, primeomega

ps = [3]

for _ in range(10):

    p = nextprime(ps[-1]) 

    while not all(primeomega(x + p) == 3 for x in ps):

        p = nextprime(p)

    ps.append(p)

print(ps)

I'd be happy to explain this code line by line! This code generates a sequence of prime numbers with a special property.

Code Breakdown

from sympy import nextprime, primeomega

Imports two functions from SymPy:

• nextprime(\(n\)): Returns the next prime number greater than \(n\)
• primeomega(\(n\)): Returns the total number of prime factors of \(n\) (counting multiplicity). For example, primeomega(12) = 3 because 12 = 2² × 3 (two 2's and one 3)

ps = [3]

Initializes the sequence with the first prime number 3. This list will store our special sequence of primes.

for _ in range(10):

Loops 10 times to find 10 more primes (so we'll have 11 total including the initial 3). The underscore _ is used when we don't need the loop variable.

p = nextprime(ps[-1])

Gets the next prime after the last element in our list. 

• ps[-1] accesses the last element of the list
• Initially, this gives nextprime(3) = 5
• This is our candidate prime to add to the sequence

while not all(primeomega(x + p) == 3 for x in ps):

    p = nextprime(p)

This is the key part! It checks if the candidate prime \(p\) satisfies a special condition:

• For every number \(x\) already in our list ps
• The sum \(x + p\) must have exactly 3 prime factors (counting multiplicity)
• If this condition fails for any \(x\), we move to the next prime candidate
• This continues until we find a prime where all sums have exactly 3 prime factors

ps.append(p)

Adds the qualified prime to our sequence once it passes the test.

print(ps)

Prints the final sequence** of primes.

Example Walkthrough

Let's trace the first iteration:

1. Start with ps = [3]

2. Try p = 5: Check if 3 + 5 = 8 = 2³ has 3 prime factors ✓ (yes, three 2's)

3. Add 5 to list: ps = [3, 5]

4. Try p = 7: Check 3 + 7 = 10 = 2 × 5 (only 2 factors) ✗

5. Try p = 11: Check 3 + 11 = 14 = 2 × 7 (only 2 factors) ✗

6. Continue until finding a prime that works with both 3 and 5...

This creates a sequence where each new prime maintains the property that when added to any previous prime in the sequence, the result always has exactly 3 prime factors!

*************************

Now if we change the condition to primeomega = 2, we get the following sequence: 3, 7, 19, 139, 859, 8179, 173059, ... where it can be seen there is a big jump from 8179 to 173059. Testing this out using 3 again we see that:

• \(3+7=10=2 \times 5\)
• \(3+19=22=2 \times 11\)
• \(3+139=142=2 \times 71\)
• \(3+859=862=2 \times 431\)
• \(3+8179=8182=2 \times 4091\)
• \(3+173059=173062=2 \times 86531\)

The sequence is included in the OEIS as A113875: slowest growing sequence of primes having the prime-pairwise-average property: if i<j, (a(i)+a(j))/2 is prime. This makes sense in that 2 seems to be the common factor is all of the sums. 

The sequence with additional (very large) terms looks like this:

3, 7, 19, 139, 859, 8179, 173059, 1026199, 1827139, 15828679, 13187242759, 18732483199, 912492556939, 9130567625119

When we set primeomega = 4 we get the following sequence: 3, 13, 137, 1193, 4373, 26017, 33889, 495433, ... which is NOT listed in the OEIS. Again let's use 3 as our test number:

• \(3+13=16=2^4\)
• \(3+137=140=2^2 \times 5 \times 7\)
• \(3+1193=1196=2^2 \times 13 \times 23\)
• \(3+4373=4376=2^3 \times 547\)
• \(3+26017=26020=2^2 \times 5 \times 1301\)
• \(3+33889=33892=2^2 \times 37 \times 229\)
• \(3+495433=495436=2^2 \times 29 \times 4271\)

That's probably a good place to stop.