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Calculus. In mathematics, the harmonic series is the infinite series formed by summing all positive unit fractions : The first terms of the series sum to approximately , where is the natural logarithm and is the Euler–Mascheroni constant. Because the logarithm has arbitrarily large values, the harmonic series does not have a finite limit: it ...
The Pell numbers have P n = 2P n−1 + P n−2. If the coefficient of the preceding value is assigned a variable value x, the result is the sequence of Fibonacci polynomials. Not adding the immediately preceding numbers. The Padovan sequence and Perrin numbers have P(n) = P(n − 2) + P(n − 3).
Apéry's theorem. In mathematics, Apéry's theorem is a result in number theory that states the Apéry's constant ζ (3) is irrational. That is, the number. cannot be written as a fraction where p and q are integers. The theorem is named after Roger Apéry . The special values of the Riemann zeta function at even integers ( ) can be shown in ...
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In mathematics, the n -th harmonic number is the sum of the reciprocals of the first n natural numbers : Starting from n = 1, the sequence of harmonic numbers begins: Harmonic numbers are related to the harmonic mean in that the n -th harmonic number is also n times the reciprocal of the harmonic mean of the first n positive integers.
The Basel problem is a problem in mathematical analysis with relevance to number theory, concerning an infinite sum of inverse squares. It was first posed by Pietro Mengoli in 1650 and solved by Leonhard Euler in 1734, [ 1] and read on 5 December 1735 in The Saint Petersburg Academy of Sciences. [ 2] Since the problem had withstood the attacks ...
Ramanujan's sum. In number theory, Ramanujan's sum, usually denoted cq ( n ), is a function of two positive integer variables q and n defined by the formula. where ( a, q) = 1 means that a only takes on values coprime to q . Srinivasa Ramanujan mentioned the sums in a 1918 paper. [ 1]
2n < p, because every factor must divide (2n)!; p = 2n, because 2n is not prime; n < p < 2n, because we assumed there is no such prime number; 2n / 3 < p ≤ n: by Lemma 3. Therefore, every prime factor p satisfies p ≤ 2n / 3. When >, the number () has at most one factor of p. By Lemma 2, for any prime p we have p R(p,n) ≤ 2n, and () since ...
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