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m is a divisor of n (also called m divides n, or n is divisible by m) if all prime factors of m have at least the same multiplicity in n. The divisors of n are all products of some or all prime factors of n (including the empty product 1 of no prime factors). The number of divisors can be computed by increasing all multiplicities by 1 and then ...
Also, every known pair shares at least one common prime factor. It is not known whether a pair of coprime amicable numbers exists, though if any does, the product of the two must be greater than 10 65. [9] [10] Also, a pair of co-prime amicable numbers cannot be generated by Thabit's formula (above), nor by any similar formula.
The same prime factor may occur more than once; this example has two copies of the prime factor When a prime occurs multiple times, exponentiation can be used to group together multiple copies of the same prime number: for example, in the second way of writing the product above, 5 2 {\displaystyle 5^{2}} denotes the square or second power of 5 ...
d() is the number of positive divisors of n, including 1 and n itself; σ() is the sum of the positive divisors of n, including 1 and n itselfs() is the sum of the proper divisors of n, including 1 but not n itself; that is, s(n) = σ(n) − n
If none of its prime factors are repeated, it is called squarefree. (All prime numbers and 1 are squarefree.) For example, 72 = 2 3 × 3 2, all the prime factors are repeated, so 72 is a powerful number. 42 = 2 × 3 × 7, none of the prime factors are repeated, so 42 is squarefree. Euler diagram of numbers under 100:
110 is a sphenic number and a pronic number. [1] Following the prime quadruplet (101, 103, 107, 109), at 110, the Mertens function reaches a low of −5. 110 is the sum of three consecutive squares, = + +. RSA-110 is one of the RSA numbers, large semiprimes that are part of the RSA Factoring Challenge.
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If the Euclidean algorithm requires N steps for a pair of natural numbers a > b > 0, the smallest values of a and b for which this is true are the Fibonacci numbers F N+2 and F N+1, respectively. [98] More precisely, if the Euclidean algorithm requires N steps for the pair a > b, then one has a ≥ F N+2 and b ≥ F N+1.