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These 17 camels leave one camel left over, which the judge takes back as his own. [2] This is possible as the sum of the fractions is less than one: 1 / 2 + 1 / 3 + 1 / 9 = 17 / 18 . Some sources point out an additional feature of this solution: each son is satisfied, because he receives more camels than his ...
If one considers only the odd numbers in the sequence generated by the Collatz process, then each odd number is on average 3 / 4 of the previous one. [16] (More precisely, the geometric mean of the ratios of outcomes is 3 / 4 .) This yields a heuristic argument that every Hailstone sequence should decrease in the long run ...
In mathematics, the infinite series 1 / 2 + 1 / 4 + 1 / 8 + 1 / 16 + ··· is an elementary example of a geometric series that converges absolutely. The sum of the series is 1.
The latter means that, for Farey sequences of even order n, the number of fractions with numerators equal to n / 2 is the same as the number of fractions with denominators equal to n / 2 , that is (/) = (/).
"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.
The two sequences {Τ 2n−1} and {Τ 2n} might themselves define two convergent continued fractions that have two different values, x odd and x even. In this case the continued fraction defined by the sequence { Τ n } diverges by oscillation between two distinct limit points.
The sequence can be used to prove that there are infinitely many prime numbers, as any prime can divide at most one number in the sequence. More strongly, no prime factor of a number in the sequence can be congruent to 5 modulo 6, and the sequence can be used to prove that there are infinitely many primes congruent to 7 modulo 12. [20]
Problems 1–6 compute divisions of a certain number of loaves of bread by 10 men and record the outcome in unit fractions. Problems 7–20 show how to multiply the expressions 1 + 1/2 + 1/4 = 7/4, and 1 + 2/3 + 1/3 = 2 by different fractions. Problems 21–23 are problems in completion, which in modern notation are simply subtraction problems.
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