Thursday, October 1, 2026

Bertrand's Postulate and a conjecture for any three consecutive primes - key to formula for the next largest prime?

The postulate states that for any integer n >1, there is at least one prime p such that n < p < 2n. 

In a similar way, I conjecture that for any three consecutive primes p, q, and r with p < q < r:

(p + q)/r = 2 (approximately) 

For large values of p, q, and r, the expression approaches 2. 

Checking this conjecture for some sample values:

(991 + 997)/1009 = 1.97

(9967 + 9973)/10007 = 1.992

(99971 + 99989)/99991 = 1.999 

A crude approximation for the next largest prime r would then be that r is slightly greater than (p + q)/2. 

(99971 + 99989)/2 = 99980

(99991 - 99980)/99980 = 0.00011 or about 0.011%

Because of the Goldbach Conjecture, p + q will always be even and thus (p + q)/2 will always be an integer. 

If we wanted to find the next prime larger than 99991, a good initial guess would be 1.00011*99991 = 100002. That isn't prime but 100003 is. 

So if you have three consecutive primes, you can get a very good estimate of the next largest prime
by calculating the percent difference as shown above. 






Euler's prime formula, the Bunyakovsky Conjecture, and n^2 + n + 1

Euler's prime formula of n^2 + n + 41 produces prime numbers for all integers n from 1 to 40. 

If 41 is substituted for another larger prime, it produces a larger proportion of the first 1,000 primes.






n^2 + n + 1 produces prime numbers for all n that are congruent to n = 0 or 2 in modulo 6. 



If we understood primes perfectly, we'd understand pi perfectly. Every prime is a little slice of pi. 

It all fits together somehow.