With precision in my inquiry, I am interested to learn whether there is any mathematical framework, method, or algorithm capable of efficiently identifying all prime numbers around a specific numerical value, such as 125,468,923, within a predetermined range of [-10,000, +10,000]. This approach would ideally bypass the exhaustive testing of every candidate within this interval, utilizing a more strategic mechanism for isolating and confirming primes. Insights into such methods are earnestly... Seeking Efficient Prime Identification Methods for Constrained Numerical Ranges ..read more

105840792690619648459681960002881 126840792690619648459681960002881 131840792690619648459681960002881 135840792690619648459681960002881 231840792690619648459681960002881 288840792690619648459681960002881 296840792690619648459681960002881 324840792690619648459681960002881 356840792690619648459681960002881 363840792690619648459681960002881 380840792690619648459681960002881 399840792690619648459681960002881 425840792690619648459681960002881 434840792690619648459681960002881 Execution time:... Innovating Primes: How the Prime Heart Theorem Could Change Everything ..read more

CHAT GTP1+2+4+8+16+32+5063+10126+20252+40504+81008=162496.1+2+4+8+16+32+5063+10126+20252+40504+81008=162496.Since the sum does indeed equal 162496, based on this calculation, it supports the claim that 162496 is a perfect number ..read more

Let's call a number N 'sub-perfect' if the sum of its divisors less than N is N-1. (There may be a different term for such numbers). It is easy to see that powers of 2 are sub-perfect. Does anyone know a proof or counter-example to the conjecture that all sub-perfect numbers are powers of 2 ..read more

I am sorry for my English. This is a repost from one forum member from another mathematical forum. I'll try to translate it into English (as best I can). [this is a text from another person on the forum]: Since Petrov is being quoted here, he also has a much more interesting article devoted to prime numbers: “Petrov I.B. METAREMULTION (general superficial numerical study of an interesting prime number)” Author’s article, self-publishing, 2023, 5 pp. (attached to... Is Petrov’s “metaremultion” finite ..read more

My idea of proving this theorem is presented in the attacheded document. May be it is not worth your time but I was facinated by the simplicity of the proposition and devoted a good amount of time and thinking ..read more

Hi think I may have found a new formula for the Riemann hypothesis. This is how you find the real side of the limit of zeta(n+bi): https://www.desmos.com/calculator/rcyapfsmuz Here is how you can find the imaginary side of the limit: https://www.desmos.com/calculator/uaociu20r4 The y=a(60) on the top is your final answer, you can increase its accuracy by setting the v to a larger number or by plugging in a higher number into that y=a(60). You can check the final answer... New formula for the Riemann Hypothesis ..read more

Addition, subtraction, multiplication and exponentiation modulo n are no problem. Division modulo n (often) IS a problem since fractions are not allowed, only integers. Normally you would calculate the answer to a modular division by calculating the multiplicative inverse of the denominator and then multiply the numerator by that inverse. (everything mod n of course) This is perfectly okay and works fine. My question is: is there an algorithm that gives me the answer to a modular division... Modular Division Algorithm ..read more

Hey guys so I was wondering what you thought of this number Meum. I met it while talking to myself - ironically in a jail. I was trying to compute the shapes of fractal equations in my head and got a shiver down my spine. "Why does this number show up everywhere, who are you? "I am meum, 1.19758, I exponentiate to infinity perfectly. Years later I found the top level equations. Apparently the value doesn't vary ..read more

Diagonaliztion as a process involves constructing a number that cannot possibly exist in an infinite list of numbers of a set such as the reals, then because that list was assumed to have a bijection with the naturals it concludes that a bijection is impossible. This conclusion however is flawed in that it is never tests if diagonalization will also create a new natural number not in the list of natural numbers that we can then use to continue the bijection. Say we have a list of all... The flaw in Cantor's Diagonalization Argument ..read more