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For any integer n, n ≡ 1 (mod 2) if and only if 3n + 1 ≡ 4 (mod 6). Equivalently, n − 1 / 3 ≡ 1 (mod 2) if and only if n ≡ 4 (mod 6). Conjecturally, this inverse relation forms a tree except for the 1–2–4 loop (the inverse of the 4–2–1 loop of the unaltered function f defined in the Statement of the problem section of ...
This series was used as a representation of two of Zeno's paradoxes. [2] For example, in the paradox of Achilles and the Tortoise, the warrior Achilles was to race against a tortoise. The track is 100 meters long. Achilles could run at 10 m/s, while the tortoise only 5. The tortoise, with a 10-meter advantage, Zeno argued, would win.
where f (2k−1) is the (2k − 1)th derivative of f and B 2k is the (2k)th Bernoulli number: B 2 = 1 / 6 , B 4 = − + 1 / 30 , and so on. Setting f ( x ) = x , the first derivative of f is 1, and every other term vanishes, so [ 15 ]
1, 1, 2, 2, 4, 2, 6, 4, 6, 4, ... φ(n) is the number of positive integers not greater than n that are coprime with n. A000010: Lucas numbers L(n) 2, 1, 3, 4, 7, 11, 18, 29, 47, 76, ... L(n) = L(n − 1) + L(n − 2) for n ≥ 2, with L(0) = 2 and L(1) = 1. A000032: Prime numbers p n: 2, 3, 5, 7, 11, 13, 17, 19, 23, 29, ... The prime numbers p ...
where c 1 = 1 / a 1 , c 2 = a 1 / a 2 , c 3 = a 2 / a 1 a 3 , and in general c n+1 = 1 / a n+1 c n . Second, if none of the partial denominators b i are zero we can use a similar procedure to choose another sequence { d i } to make each partial denominator a 1:
Today, a more standard phrasing of Archimedes' proposition is that the partial sums of the series 1 + 1 / 4 + 1 / 16 + ⋯ are: + + + + = +. This form can be proved by multiplying both sides by 1 − 1 / 4 and observing that all but the first and the last of the terms on the left-hand side of the equation cancel in pairs.
6 1 2 1 1 −1 4 5 9. and would be written in modern notation as 6 1 / 4 , 1 1 / 5 , and 2 − 1 / 9 (i.e., 1 8 / 9 ). The horizontal fraction bar is first attested in the work of Al-Hassār (fl. 1200), [35] a Muslim mathematician from Fez, Morocco, who specialized in Islamic inheritance jurisprudence.
The idea becomes clearer by considering the general series 1 − 2x + 3x 2 − 4x 3 + 5x 4 − 6x 5 + &c. that arises while expanding the expression 1 ⁄ (1+x) 2, which this series is indeed equal to after we set x = 1.