Hi All Here is some interesting things to examine. We all know that if the digit sum of a number is 9 then the number is divisible by 9. So 0% of these numbers are prime But what if the digit sum is 8

If you’re reading this, I really hope you will enter my new high-IQ-like quiz contest. For two reasons: first, there is a significant cash prize (US$100) for the winner; and second, and more important

Hello, Last weeks I did an analysis on the intersection of n-random walks in a single point. There are quit some new (to me) observations I made. A short question is posted on mathematics stacks excha

Hi All I wonder if there exist a formula that can calculate the number of coprimes in a 2N interval by using the positions of the primes. One could use a empirical approach to this problem. Suppose th

Thank you, Tim. I see from the smallest three numbers that you posted that none of my own constructed factoring algorithms can compete with Pollard Rho for factoring the large difficult-to-factor posi

Hi all, I've been asked by a correspondent for some large numbers that can be used to test their factorization method. This may be useful to others in this group, so here are a few I've concocted: 723

This is a report on the constant patterning of use of a specific ciphertext strand and template that holds for the Z340 and other ciphers in the mix Included is the use of the steganography of Pi in t

Currently, I expect the difficulty, in general, of factoring an odd positive integer to be proportional to the cube of its logarithm. Kermit

Hello, Usually when determining pi a circle is drawn with a radius, then polygons are inscribed and out scribed. Lasts months I attempted to find a more intuitive way explaining the irrational and tra

Hi All Here is something, I find interesting. It looks like the following is true with some exceptions : Let N^2<N^2+r<(N+1)^2 then (N+k)^2+r+k is either prime ( possibly it is only coprime with the p

Hi All I wonder if it is possible to find the integer solutions to diophantine equations of this form : Z=N*(2x+1)-((N+1+2x)/2)*y Example : N=5 Z=5*(2x+1)-((6+2x)/2)*y=10x+5-3y-x*y I want to find all

How efficient would this factoring algorithm be for factoring large integers? z = 33 (z-1) = 32 t1 *(t1 * t2 * t3 + t2 + t3) = 32 t1 = 2 (t1*t2*t3 + t2 + t3) = 16 (2*t2*t3+t2+t3)=16 t2 and t3 are same

An illustration of the complicated relationship between addition and multiplication. 10*30 + 1*10 + 1*30 +1*1 = 341 10*29 +2*10 + 1*29 + 1*2 = 341 9*30 + 1*9 + 2*30 +2*1 = 341 10*28 + 3*10 + 1*28 +1*3

An interesting identity: ((x−a)(x−b)) / ((c−a)(c−b)) + ((x−b)(x−c)) / ((a−b)(a−c)) + ((x−c)(x−a)) / ((b−c)(b−a))=1

Dear colleagues, Attached is my manuscript containing a proposed partial proof of the binary Goldbach conjecture. Your constructive comments are most welcome. Best regards, Tatenda Kubalalika.

>>> z=(z-1)*53*59+1 >>> Factor2021(z) x = 277 y = 6941373105249865123 z = 1922760350154212639071 Number of iterations = 2 Probable Prime Test Algorithm Factor, using base, 3 >>> z=(z-1)*61*67+1 >>> Fa

Yesterday, I thought of this "square root algorithm", but the test shows that it is not as good as the older probable prime test factor algorithm. >>> z=2*3*5*7*11*13*17*19+1 >>> Factor2021(z) x = 347

I'm having difficulty expanding the following expression I want to write it as a 4th degree polynomial in y1 so that I can take the next step of assigning functional values to y1 that will make z2^2 a

take the perfect solution: a|b|c d|e|f g|h|i which yields (f.p.p) : 2e = c + g [0] The perfect solution above can be re-written: b|a|c h|g|i e|d|f which yields 2g = c+e [1] subtracting [0] & [1] toget

Hi All I hope someone will help me out here. Legendre conjecture is true if this is true (see attachment). This involves a summation series, which is using Ln(N). Kim