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Name First elements Short description OEIS Mersenne prime exponents : 2, 3, 5, 7, 13, 17, 19, 31, 61, 89, ... Primes p such that 2 p − 1 is prime.: A000043 ...
The number of points (n), chords (c) and regions (r G) for first 6 terms of Moser's circle problem. In geometry, the problem of dividing a circle into areas by means of an inscribed polygon with n sides in such a way as to maximise the number of areas created by the edges and diagonals, sometimes called Moser's circle problem (named after Leo Moser), has a solution by an inductive method.
A mathematical constant is a key number whose value is fixed by an unambiguous definition, often referred to by a symbol (e.g., an alphabet letter), or by mathematicians' names to facilitate using it across multiple mathematical problems. [1]
1 ⁄ 5: 0.2 Vulgar Fraction One Fifth 2155 8533 ⅖ 2 ⁄ 5: 0.4 Vulgar Fraction Two Fifths 2156 8534 ⅗ 3 ⁄ 5: 0.6 Vulgar Fraction Three Fifths 2157 8535 ⅘ 4 ⁄ 5: 0.8 Vulgar Fraction Four Fifths 2158 8536 ⅙ 1 ⁄ 6: 0.166... Vulgar Fraction One Sixth 2159 8537 ⅚ 5 ⁄ 6: 0.833... Vulgar Fraction Five Sixths 215A 8538 ⅛ 1 ⁄ 8: 0 ...
Thus F 6 consists of F 5 together with the fractions 1 / 6 and 5 / 6 . The middle term of a Farey sequence F n is always 1 / 2 , for n > 1. From this, we can relate the lengths of F n and F n−1 using Euler's totient function φ(n): | | = | | + ().
Unit fractions can also be expressed using negative exponents, as in 2 −1, which represents 1/2, and 2 −2, which represents 1/(2 2) or 1/4. A dyadic fraction is a common fraction in which the denominator is a power of two, e.g. 1 / 8 = 1 / 2 3 . In Unicode, precomposed fraction characters are in the Number Forms block.
Most modern lenses use a standard f-stop scale, which is an approximately geometric sequence of numbers that corresponds to the sequence of the powers of the square root of 2: f /1, f /1.4, f /2, f /2.8, f /4, f /5.6, f /8, f /11, f /16, f /22, f /32, f /45, f /64, f /90, f /128, etc. Each element in the sequence is one stop lower than the ...
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:
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