Average of Three Primes Squared

The number associated with my diurnal age today is 28129 and it is a member of OEIS A075893:

     A075893: average of three successive primes squared with \(n > 3\) 
                

    \( \dfrac{\text{prime} (n)^2+ \text{prime}(n+1)^2+\text{prime}(n+2)^2)}{3}\)

The initial terms are:

65, 113, 193, 273, 393, 577, 777, 1057, 1337, 1633, 1913, 2289, 2833, 3337, 3897, 4417, 4953, 5537, 6153, 7017, 8073, 9177, 10073, 10753, 11313, 12033, 13593, 15353, 17353, 18417, 20097, 21441, 23217, 24673, 26369, 28129, 29953, 31577, 33761, 35497, 37513, 38553

The OEIS comments state that:

Unlike the average of three successive primes, the average of three successive primes (greater than 3) squared is always integral.

I asked Gemini why this was so:

The reason the average of the squares of three successive primes (greater than 3) is always an integer is based on a fascinating pattern about how primes behave when divided by 3.

In fact, this rule doesn't just apply to successive primes—it applies to any three primes greater than 3. Here is the step-by-step mathematical breakdown of why this happens.

1. All Primes Greater Than 3 Avoid Multiples of 3

By definition, a prime number is only divisible by 1 and itself. This means that any prime number strictly greater than 3 cannot be a multiple of 3.

If we divide any number by 3, we can only ever get one of three possible remainders: 0, 1, or 2.

• Because primes greater than 3 are not divisible by 3, they will never have a remainder of 0.
• Therefore, any prime number p > 3 must leave a remainder of either 1 or 2 when divided by 3.

In algebra, this means every prime p > 3 can be written in one of two ways, where k is an integer:

• p = 3k + 1 (leaves a remainder of 1)
• p = 3k + 2 (leaves a remainder of 2, which can also be written as p = 3k - 1)

2. Squaring the Primes

Now, let's look at what happens when we square these prime numbers.

Case A: The prime is 3k + 1
p² = (3k + 1)²
p² = 9k² + 6k + 1

We can factor out a 3 from the first two terms:
p² = 3(3k² + 2k) + 1

Because 3(3k² + 2k) is a multiple of 3, the entire expression leaves a remainder of 1 when divided by 3.

Case B: The prime is 3k - 1
p² = (3k - 1)²
p² = 9k² - 6k + 1

Again, factor out a 3 from the first two terms:
p² = 3(3k² - 2k) + 1

Just like the first case, this leaves a remainder of 1 when divided by 3.

The Key Rule: No matter which type of prime you have, as long as it is greater than 3, its square will always be exactly 1 more than a multiple of 3.

3. Averaging Three Squared Primes

Let p₁, p₂, and p₃ be any three primes greater than 3. Based on our rule above, we can express their squares as:

• p₁² = 3a + 1
• p₂² = 3b + 1
• p₃² = 3c + 1

Now, add them together to find the sum:
Sum = (3a + 1) + (3b + 1) + (3c + 1)
Sum = 3a + 3b + 3c + 3

We can factor out a 3 from the entire sum:
Sum = 3(a + b + c + 1)

Since the sum of the three squared primes is a perfect multiple of 3, dividing that sum by 3 to find the average will always result in a clean integer:
Average = 3(a + b + c + 1) / 3 = a + b + c + 1

Because a, b, and c are integers, the average itself is guaranteed to be an integer.