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Sum of Natural Numbers (second proof and extra footage) includes demonstration of Euler's method. What do we get if we sum all the natural numbers? response to comments about video by Tony Padilla; Related article from New York Times; Why –1/12 is a gold nugget follow-up Numberphile video with Edward Frenkel
For example, the sum of the first n natural numbers can be denoted as ∑ i = 1 n i {\displaystyle \sum _{i=1}^{n}i} For long summations, and summations of variable length (defined with ellipses or Σ notation), it is a common problem to find closed-form expressions for the result.
The n-th harmonic number, which is the sum of the reciprocals of the first n positive integers, is never an integer except for the case n = 1. Moreover, József Kürschák proved in 1918 that the sum of the reciprocals of consecutive natural numbers (whether starting from 1 or not) is never an integer.
In number theory, Ramanujan's sum, usually denoted c q (n), is a function of two positive integer variables q and n defined by the formula c q ( n ) = ∑ 1 ≤ a ≤ q ( a , q ) = 1 e 2 π i a q n , {\displaystyle c_{q}(n)=\sum _{1\leq a\leq q \atop (a,q)=1}e^{2\pi i{\tfrac {a}{q}}n},}
The algorithm performs summation with two accumulators: sum holds the sum, and c accumulates the parts not assimilated into sum, to nudge the low-order part of sum the next time around. Thus the summation proceeds with "guard digits" in c , which is better than not having any, but is not as good as performing the calculations with double the ...
A perfect totient number is an integer that is equal to the sum of its iterated totients. That is, we apply the totient function to a number n, apply it again to the resulting totient, and so on, until the number 1 is reached, and add together the resulting sequence of numbers; if the sum equals n, then n is a perfect totient number.
It is used to prove Kronecker's lemma, which in turn, is used to prove a version of the strong law of large numbers under variance constraints. It may be used to prove Nicomachus's theorem that the sum of the first n {\displaystyle n} cubes equals the square of the sum of the first n {\displaystyle n} positive integers.
The number of steps to calculate the GCD of two natural numbers, a and b, may be denoted by T(a, b). [96] If g is the GCD of a and b, then a = mg and b = ng for two coprime numbers m and n. Then T(a, b) = T(m, n) as may be seen by dividing all the steps in the Euclidean algorithm by g. [97]