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If p ≤ 0, then the nth-term test identifies the series as divergent. If 0 < p ≤ 1, then the nth-term test is inconclusive, but the series is divergent by the integral test for convergence. If 1 < p, then the nth-term test is inconclusive, but the series is convergent by the integral test for convergence.
Proof without words of the arithmetic progression formulas using a rotated copy of the blocks.. An arithmetic progression or arithmetic sequence is a sequence of numbers such that the difference from any succeeding term to its preceding term remains constant throughout the sequence.
The n th term describes the length of the n th run A000002: Euler's totient function φ(n) 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 ...
In mathematics, a recurrence relation is an equation according to which the th term of a sequence of numbers is equal to some combination of the previous terms. Often, only previous terms of the sequence appear in the equation, for a parameter that is independent of ; this number is called the order of the relation.
An integer sequence is computable if there exists an algorithm that, given n, calculates a n, for all n > 0. The set of computable integer sequences is countable.The set of all integer sequences is uncountable (with cardinality equal to that of the continuum), and so not all integer sequences are computable.
For example, consider the triple [20,21,29], which can be calculated from the Euclid equations with values m = 5 and n = 2. Also, arbitrarily put the coefficient of 4 in front of the x in the m term. Let m 1 = (4x + m), and let n 1 = (x + n). Hence, substituting the values of m and n:
The Fermat–Catalan conjecture concerns a certain Diophantine equation, equating the sum of two terms, each a positive integer raised to a positive integer power, to a third term that is also a positive integer raised to a positive integer power (with the base integers having no prime factor in common). The conjecture asks whether the equation ...
Hence the n-th solution is a n = H 2n +1 − 1 / 2 and c n = P 2n +1. The table above shows that, in one order or the other, a n and b n = a n + 1 are H n H n +1 and 2P n P n +1 while c n = H n +1 P n + P n +1 H n.