A New Discovery on Prime Numbers: A Test Based on Giuga Numbers.

[Davide Rotondo]

Dear number theory enthusiasts, I have empirically tested a fascinating new property linking modified binomial coefficients with Giuga numbers (where G(n) = sum of j^n for j from 1 to n).
By defining a modified triangle via the ratios T(n,k) = G(n) / (G(k) * G(n-k)) and looking at the reduced numerators of the resulting fractions, a striking regularity emerges that acts as a primality test!
What does the test state?
For a given integer n, we compute the reduced numerators of all internal terms in row n. The test states that:
All prime numbers n (except 2) produce reduced numerators that are always divisible by n.

I have verified this property up to n = 672:
All prime numbers greater than 2 pass the test.
The number 2 fails (the natural parity exception).
The only composite number that passes is n = 9 (an exceptional pseudoprime for this specific formula!). All other composites fail.
The First 11 Rows
Here is how the first 11 rows look in both decimal and reduced fraction formats:
n = 1: [1]
n = 2: [1, 5/1, 1] (Note: 5 is not divisible by 2, showing the exception for 2)
n = 3 (Prime): [1, 36/5, 36/5, 1] (Numerator 36 is divisible by 3)
n = 4 (Composite): [1, 59/6, 85/6, 59/6, 1] (Numerators 59 and 85 are NOT divisible by 4)
n = 5 (Prime): [1, 25/2, 295/12, 295/12, 25/2, 1] (Numerators 25 and 295 are divisible by 5)
n = 6 to 8 (Composites): Fail the divisibility test.
n = 9 (SPECIAL Composite): The only composite passing unexpectedly!
n = 10 (Composite): Fails.
n = 11 (Prime): All internal numerators are divisible by 11.

I would love to hear feedback from the Seqfan community on this pattern and whether n=9 acts as a unique base-anomaly. Thoughts?

Davide

[Davide Rotondo]

For example, in the row for the number five, the values are obtained as follows: (1^5+2^5+3^5+4^5+5^5)/((1^1)*(1^4+2^4+3^4+4^4)), which equals 12.5 (or the fraction 25/2), and the numerator—25—is divisible by 5. Also, (1^5+2^5+3^5+4^5+5^5)/((1^2+2^2)*(1^3+2^3+3^3)) equals 24.58333333... (or the fraction 295/12), and 295 is divisible by 5. The pattern then repeats... I hope I have been clear and thorough. Please don't hesitate to ask. Have a great day.

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[Geoffrey Caveney]

Davide, if you have computed the values of Sum_{j=1..n} j^n up to n = 672, that is a contribution in and of itself, since the b-file of that sequence A031971 on OEIS presently only goes up to n = 100. I hope that you contribute additional terms to the b-file of A031971.

But the standard definition of a "Giuga number" is different than the one you are using. See https://en.wikipedia.org/wiki/Giuga_number and https://mathworld.wolfram.com/GiugaNumber.html . The sums A031971 are the basis of what is called Giuga's conjecture (https://mathworld.wolfram.com/GiugasConjecture.html) or the Agoh-Giuga conjecture (https://en.wikipedia.org/wiki/Agoh%E2%80%93Giuga_conjecture).

Since you mention that n = 2 fails your test and n = 9 passes it, you should at least mention further that n = 15, etc., fail the test, to show that odd composites > 9 fail. I confirmed that n = 15 indeed fails when k = 3.

Geoffrey

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[Davide Rotondo]

Sorry, Geoffrey, you're right; the thing is, all the background information needed to explain the test made my explanation unclear. I apologize again.

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