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The number of connected simple cubic graphs on 4, 6, 8, 10, ... vertices is 1, 2, 5, 19, ... (sequence A002851 in the OEIS). A classification according to edge connectivity is made as follows: the 1-connected and 2-connected graphs are defined as usual. This leaves the other graphs in the 3-connected class because each 3-regular graph can be ...
According to Brooks' theorem every connected cubic graph other than the complete graph K 4 has a vertex coloring with at most three colors. Therefore, every connected cubic graph other than K 4 has an independent set of at least n/3 vertices, where n is the number of vertices in the graph: for instance, the largest color class in a 3-coloring has at least this many vertices.
The tables below list all of the divisors of the numbers 1 to 1000. A divisor of an integer n is an integer m , for which n / m is again an integer (which is necessarily also a divisor of n ). For example, 3 is a divisor of 21, since 21/7 = 3 (and therefore 7 is also a divisor of 21).
Fourth powers are also formed by multiplying a number by its cube. Furthermore, they are squares of squares. Some people refer to n 4 as n tesseracted, hypercubed, zenzizenzic, biquadrate or supercubed instead of “to the power of 4”. The sequence of fourth powers of integers, known as biquadrates or tesseractic numbers, is:
A list of articles about numbers (not about numerals). Topics include powers of ten, notable integers, prime and cardinal numbers, and the myriad system.
This is a list of articles about prime numbers. A prime number (or prime) is a natural number greater than 1 that has no positive divisors other than 1 and itself. By Euclid's theorem, there are an infinite number of prime numbers. Subsets of the prime numbers may be generated with various formulas for primes.
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Because of the factorization (2n + 1)(n 2 + n + 1), it is impossible for a centered cube number to be a prime number. [3] The only centered cube numbers which are also the square numbers are 1 and 9, [4] [5] which can be shown by solving x 2 = y 3 + 3y, the only integer solutions being (x,y) from {(0,0), (1,2), (3,6), (12,42)}, By substituting a=(x-1)/2 and b=y/2, we obtain x^2=2y^3+3y^2+3y+1.