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Proof by exhaustion can be used to prove that if an integer is a perfect cube, then it must be either a multiple of 9, 1 more than a multiple of 9, or 1 less than a multiple of 9. [3] Proof: Each perfect cube is the cube of some integer n, where n is either a multiple of 3, 1 more than a multiple of 3, or 1 less than a multiple of 3. So these ...
The maximal number of face turns needed to solve any instance of the Rubik's Cube is 20, [2] and the maximal number of quarter turns is 26. [3] These numbers are also the diameters of the corresponding Cayley graphs of the Rubik's Cube group. In STM (slice turn metric) the minimal number of turns is unknown, lower bound being 18 and upper bound ...
In number theory, a perfect number is a positive integer that is equal to the sum of its positive proper divisors, that is, divisors excluding the number itself. [1] 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.
A scrambled Rubik's Cube. An algorithm to determine the minimum number of moves to solve Rubik's Cube was published in 1997 by Richard Korf. [10] While it had been known since 1995 that 20 was a lower bound on the number of moves for the solution in the worst case, Tom Rokicki proved in 2010 that no configuration requires more than 20 moves. [11]
The only tetrahedral number that is also a square pyramidal number is 1 (Beukers, 1988), and the only tetrahedral number that is also a perfect cube is 1. The infinite sum of tetrahedral numbers' reciprocals is 3 / 2 , which can be derived using telescoping series:
Sociable Dudeney numbers and amicable Dudeney numbers are the powers of their respective roots. The number of iterations i {\displaystyle i} needed for F p , b i ( n ) {\displaystyle F_{p,b}^{i}(n)} to reach a fixed point is the Dudeney function's persistence of n {\displaystyle n} , and undefined if it never reaches a fixed point.
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The aliquot sequence starting with a positive integer k can be defined formally in terms of the sum-of-divisors function σ 1 or the aliquot sum function s in the following way: [1] = = = > = = = If the s n-1 = 0 condition is added, then the terms after 0 are all 0, and all aliquot sequences would be infinite, and we can conjecture that all aliquot sequences are convergent, the limit of these ...