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How do you find the power of a perfect number?
An integer n > 1 is a perfect power if there are integers x and k > 1 with n = xk. Note that k ≤ log2 n; also, the minimal k is prime. A perfect-power detection algorithm is an algorithm that, given an integer n > 1, figures out whether n is a perfect power.
Is 8 a perfect power?
The first perfect powers without duplicates are: (sometimes 0 and 1), 4, 8, 9, 16, 25, 27, 32, 36, 49, 64, 81, 100, 121, 125, 128, 144, 169, 196, 216, 225, 243, 256, 289, 324, 343, 361, 400, 441, 484, 512, 529, 576, 625, 676, 729, 784, 841, 900, 961, 1000, 1024, (sequence A001597 in the OEIS)
What is the sum of all positive integers from 1 to infinity?
For those of you who are unfamiliar with this series, which has come to be known as the Ramanujan Summation after a famous Indian mathematician named Srinivasa Ramanujan, it states that if you add all the natural numbers, that is 1, 2, 3, 4, and so on, all the way to infinity, you will find that it is equal to -1/12.
Is there such a thing as a perfect power?
Perfect Powers. A perfect power is the more general form of squares and cubes. Specifically, it is any number that can be written as the product of some non-negative integer multiplied by itself at least twice. In other words, it is of the form nmn^mnm for some integers n≥0n\\ge 0n≥0 and m>1.m > 1.m>1.
How to create a sequence of perfect powers?
A sequence of perfect powers can be generated by iterating through the possible values for m and k. The first few ascending perfect powers in numerical order (showing duplicate powers) are (sequence A072103 in the OEIS):
Which is the only pair of consecutive perfect powers?
In 2002 Romanian mathematician Preda Mihăilescu proved that the only pair of consecutive perfect powers is 2 3 = 8 and 3 2 = 9, thus proving Catalan’s conjecture . Pillai’s conjecture states that for any given positive integer k there are only a finite number of pairs of perfect powers whose difference is k.
The sums of powers S m(z,n) = z m+(z+1) m+…+(z+n-1) m is related to the Bernoulli polynomials B m(z) by (∂ n-∂ z) S m(z,n) = B m(z) and (∂ 2λ-∂ Z) S 2k+1(z,n) = Ŝ’ k+1(Z) where Z = z(z-1), λ = S 1(z,n), Ŝ k+1(Z) ≡ S 2k+1(0,z).