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A number where some but not all prime factors have multiplicity above 1 is neither square-free nor squareful. The Liouville function λ(n) is 1 if Ω(n) is even, and is -1 if Ω(n) is odd. The Möbius function μ(n) is 0 if n is not square-free. Otherwise μ(n) is 1 if Ω(n) is even, and is −1 if Ω(n) is odd.
1006 = semiprime, product of two distinct isolated primes (2 and 503); unusual number; square-free number; number of compositions (ordered partitions) of 22 into squares; sum of two distinct pentatope numbers (5 and 1001); number of undirected Hamiltonian paths in 4 by 5 square grid graph; [5] record gap between twin primes; [6] number that is ...
The square root of a positive integer is the product of the roots of its prime factors, because the square root of a product is the product of the square roots of the factors. Since p 2 k = p k , {\textstyle {\sqrt {p^{2k}}}=p^{k},} only roots of those primes having an odd power in the factorization are necessary.
144 (one hundred [and] forty-four) is the natural number following 143 and preceding 145. It is coincidentally both the square of twelve (a dozen dozens , or one gross .) and the twelfth Fibonacci number , and the only nontrivial number in the sequence that is square.
A powerful number is a positive integer m such that for every prime number p dividing m, p 2 also divides m.Equivalently, a powerful number is the product of a square and a cube, that is, a number m of the form m = a 2 b 3, where a and b are positive integers.
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).
Kunerth's algorithm is an algorithm for computing the modular square root of a given number. [ 1 ] [ 2 ] The algorithm does not require the factorization of the modulus, and uses modular operations that are often easy when the given number is prime.
A seemingly weaker yet equivalent statement to Bunyakovsky's conjecture is that for every integer polynomial () that satisfies (1)–(3), () is prime for at least one positive integer : but then, since the translated polynomial (+) still satisfies (1)–(3), in view of the weaker statement () is prime for at least one positive integer >, so ...