Factoring - Number Factorizer

Prime numbers like all natural numbers arise by adding to the preceding number always one, / 0 + 1 = 1 + 1 = 2 + 1 = 3 + 1 = 4 + 1 = 5 + 1 = 6 + 1 = 7 + 1 = 8 + 1 = 9 / hence we can write: each successive number is about 1 greater than the previous. By adding the successive natural numbers to each we obtain triangular numbers (0 + 1 = 1, 1 + 2 = 3/1, 3 + 3 = 6, 6 + 4 = 10/2, 10 + 5 = 15, 15 + 6 = 21/3), in which every second is housed still more bigger quantity of successive odd numbers (3/1 = 3, 10/2 = 5, 21/3 = 7).

In the sum of preceding number a given magnitude there houses n- number of times and this property will use for testing the triangular numbers into factorization of prime factors. Every triangle number greater than one can be represented as a product of prime numbers. If in order of prime factors to which decomposes the number of triangular the last factor is equal to a given magnitude N it is a sign, that the magnitude is a prime number. [16² + 16]/2 = (256 + 16)/2 = 136 = (16+1) + (15+ 2) + (14 + 3) + (13 + 4) + (12 + 5) + (11 + 6) + (10 + 7) + (9 + 8) = 8*17 = 2*2*2*17 When the last factor is smaller than given magnitude of N it is a sign, that the magnitude is composed of smaller factors. [14² + 14]/2 = (196 + 14)/2 = 105 = (14 + 1) + (13 + 2) + (12 + 3)/3 + (11 + 4) + (10 + 5)/5 + (6 + 9)/3 + (8 + 7) = 7 * 15 = (3 * 5) * 7

Triangular numbers as the sum of preceding numbers to given odd number consist from n-the quantities of sum of terms added pair wise from opposite ends of list of preceding numbers a given odd number, which if not have a common divisor greater than 1, yield identical intermediate sums only of primes / 10 = (4 +1)/1 + (3 + 2)/1 = 5 + 5 = 2*5/, and if they have at least one common divisor greater than 1, that yield identical intermediate sums only of complex numbers /36 = (8 + 1)/1 + (7 + 2)/1 + (6 + 3)/3 + (5 + 4)/1 = 9 + 9 + 9 + 9 = 4*9 = (2*2)(3*3). Realizing that adding pair wise of terms from opposite ends of list preceding numbers given odd number, always yield identical intermediate sums, will allow us to create a testing algorithm whether a given triangular number as the sum of preceding numbers to given magnitude, consists of only prime numbers or complex. t = (n’” + n )/1 + (n” + n’)/1… = p + p.. = n(p), t = (n”’ + n)/1 + (n” + n’)/n + … = p(p) + p(p)… = [n][p(p)], 105 = (14+1)/1 + (13 + 2)/1 + (12 + 3)/3 + (11 + 4)/1 + (10 + 5)/5 + (9+ 6)/3 + (8+ 7)/1 = 7*15 = (3*5)*7 And so you can to imagine a distribution of primes in 3D.