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The multiplicity of a prime factor p of n is the largest exponent m for which p m divides n. The tables show the multiplicity for each prime factor. If no exponent is written then the multiplicity is 1 (since p = p 1). The multiplicity of a prime which does not divide n may be called 0 or may be considered undefined.
[1] [2] 126 is a sum of two cubes, and since 125 + 1 is σ 3 (5), 126 is the fifth value of the sum of cubed divisors function. [3] [4] 126 is the fifth -perfect Granville number, and the third such not to be a perfect number. Also, it is known to be the smallest Granville number with three distinct prime factors, and perhaps the only such ...
Continuing this process until every factor is prime is called prime factorization; the result is always unique up to the order of the factors by the prime factorization theorem. To factorize a small integer n using mental or pen-and-paper arithmetic, the simplest method is trial division : checking if the number is divisible by prime numbers 2 ...
Suppose, to the contrary, there is an integer that has two distinct prime factorizations. Let n be the least such integer and write n = p 1 p 2... p j = q 1 q 2... q k, where each p i and q i is prime. We see that p 1 divides q 1 q 2... q k, so p 1 divides some q i by Euclid's lemma. Without loss of generality, say p 1 divides q 1.
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 ...
The factorizations are often not unique in the sense that the unit could be absorbed into any other factor with exponent equal to one. The entry 4+2i = −i(1+i) 2 (2+i), for example, could also be written as 4+2i= (1+i) 2 (1−2i). The entries in the table resolve this ambiguity by the following convention: the factors are primes in the right ...
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A definite bound on the prime factors is possible. Suppose P i is the i 'th prime, so that P 1 = 2, P 2 = 3, P 3 = 5, etc. Then the last prime number worth testing as a possible factor of n is P i where P 2 i + 1 > n; equality here would mean that P i + 1 is a factor. Thus, testing with 2, 3, and 5 suffices up to n = 48 not just 25 because the ...