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For instance, the first counterexample must be odd because f(2n) = n, smaller than 2n; and it must be 3 mod 4 because f 2 (4n + 1) = 3n + 1, smaller than 4n + 1. For each starting value a which is not a counterexample to the Collatz conjecture, there is a k for which such an inequality holds, so checking the Collatz conjecture for one starting ...
For example, 3 is a Mersenne prime as it is a prime number and is expressible as 2 2 − 1. [ 1 ] [ 2 ] The exponents p corresponding to Mersenne primes must themselves be prime, although the vast majority of primes p do not lead to Mersenne primes—for example, 2 11 − 1 = 2047 = 23 × 89 .
For instance, 6 has proper divisors 1, 2 and 3, and 1 + 2 + 3 = 6, so 6 is a perfect number. The next perfect number is 28, since 1 + 2 + 4 + 7 + 14 = 28. The first four perfect numbers are 6, 28, 496 and 8128. [2] The sum of proper divisors of a number is called its aliquot sum, so a perfect number is one that is equal to its aliquot sum.
When applied to meters, the terms perfect and imperfect are sometimes used as the equivalents of divisive and additive, respectively . [2] Additive and divisive meters. For example, 4 may be evenly divided by 2 or reached by adding 2 + 2. In contrast, 5 is only evenly divisible by 5 and 1 and may be reached by adding 2 or 3. Thus, 4 8 (or, more ...
It can be proven that: . For a given prime number p, if n is p-perfect and p does not divide n, then pn is (p + 1)-perfect.This implies that an integer n is a 3-perfect number divisible by 2 but not by 4, if and only if n/2 is an odd perfect number, of which none are known.
Goldbach's weak conjecture, every odd number greater than 5 can be expressed as the sum of three primes, is a consequence of Goldbach's conjecture. Ivan Vinogradov proved it for large enough n (Vinogradov's theorem) in 1937, [1] and Harald Helfgott extended this to a full proof of Goldbach's weak conjecture in 2013.
For example, 2 × 3 2 × 29 × 823 = 429606 is practical, because the inequality above holds for each of its prime factors: 3 ≤ σ(2) + 1 = 4, 29 ≤ σ(2 × 3 2) + 1 = 40, and 823 ≤ σ(2 × 3 2 × 29) + 1 = 1171. The condition stated above is necessary and sufficient for a number to be practical.
A non-trivial perfect group, however, is necessarily not solvable; and 4 divides its order (if finite), moreover, if 8 does not divide the order, then 3 does. [2] Every acyclic group is perfect, but the converse is not true: A 5 is perfect but not acyclic (in fact, not even superperfect), see (Berrick & Hillman 2003).