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For instance, the sequence 5, 7, 9, 11, 13, 15, . . . is an arithmetic progression with a common difference of 2. If the initial term of an arithmetic progression is a 1 {\displaystyle a_{1}} and the common difference of successive members is d {\displaystyle d} , then the n {\displaystyle n} -th term of the sequence ( a n {\displaystyle a_{n ...
The sequence of primes numbers contains arithmetic progressions of any length. This result was proven by Ben Green and Terence Tao in 2004 and is now known as the Green–Tao theorem. [3] See also Dirichlet's theorem on arithmetic progressions. As of 2020, the longest known arithmetic progression of primes has length 27: [4]
Equivalently, a sequence is a harmonic progression when each term is the harmonic mean of the neighboring terms. As a third equivalent characterization, it is an infinite sequence of the form 1 a , 1 a + d , 1 a + 2 d , 1 a + 3 d , ⋯ , {\displaystyle {\frac {1}{a}},\ {\frac {1}{a+d}},\ {\frac {1}{a+2d}},\ {\frac {1}{a+3d}},\cdots ,}
Goldbach’s Conjecture. One of the greatest unsolved mysteries in math is also very easy to write. Goldbach’s Conjecture is, “Every even number (greater than two) is the sum of two primes ...
Sequences dn + a with odd d are often ignored because half the numbers are even and the other half is the same numbers as a sequence with 2d, if we start with n = 0. For example, 6n + 1 produces the same primes as 3n + 1, while 6n + 5 produces the same as 3n + 2 except for the only even prime 2. The following table lists several arithmetic ...
Recamán's sequence: 0, 1, 3, 6, 2, 7, 13, 20, 12, 21, 11, 22, 10, 23, 9, 24, 8, 25, 43, 62, ... "subtract if possible, otherwise add": a(0) = 0; for n > 0, a(n) = a(n − 1) − n if that number is positive and not already in the sequence, otherwise a(n) = a(n − 1) + n, whether or not that number is already in the sequence. A005132: Look-and ...
Van der Waerden's theorem is a theorem in the branch of mathematics called Ramsey theory.Van der Waerden's theorem states that for any given positive integers r and k, there is some number N such that if the integers {1, 2, ..., N} are colored, each with one of r different colors, then there are at least k integers in arithmetic progression whose elements are of the same color.
For the congruum problem, the parameterized solution reduces this testing problem to checking a finite set of parameter values, as the two parameters and must both divide the given number. In contrast, for the congruent number problem, a finite testing procedure is known only conjecturally, via Tunnell's theorem , under the assumption that the ...
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