Isaiah 53:2 — “He had no beauty or majesty to attract us to him, nothing in his appearance that we should desire him.”
Derpetology
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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.
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