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In number theory, an abundant number or excessive number is a positive integer for which the sum of its proper divisors is greater than the number. The integer 12 is the first abundant number. The integer 12 is the first abundant number.
where f (2k−1) is the (2k − 1)th derivative of f and B 2k is the (2k)th Bernoulli number: B 2 = 1 / 6 , B 4 = − + 1 / 30 , and so on. Setting f ( x ) = x , the first derivative of f is 1, and every other term vanishes, so [ 15 ]
Then the triangle is in Euclidean space if the sum of the reciprocals of p, q, and r equals 1, spherical space if that sum is greater than 1, and hyperbolic space if the sum is less than 1. A harmonic divisor number is a positive integer whose divisors have a harmonic mean that is an integer. The first five of these are 1, 6, 28, 140, and 270 ...
d() is the number of positive divisors of n, including 1 and n itself; σ() is the sum of the positive divisors of n, including 1 and n itselfs() is the sum of the proper divisors of n, including 1 but not n itself; that is, s(n) = σ(n) − n
A number-line visualization of the algebraic addition 2 + 4 = 6. A "jump" that has a distance of 2 followed by another that is long as 4, is the same as a translation by 6. A number-line visualization of the unary addition 2 + 4 = 6. A translation by 4 is equivalent to four translations by 1.
G(3) is at least 4 (since cubes are congruent to 0, 1 or −1 mod 9); for numbers less than 1.3 × 10 9, 1 290 740 is the last to require 6 cubes, and the number of numbers between N and 2N requiring 5 cubes drops off with increasing N at sufficient speed to have people believe that G(3) = 4; [22] the largest number now known not to be a sum of ...
The aliquot sum function can be used to characterize several notable classes of numbers: 1 is the only number whose aliquot sum is 0. A number is prime if and only if its aliquot sum is 1. [1] The aliquot sums of perfect, deficient, and abundant numbers are equal to, less than, and greater than the number itself respectively. [1]
While the partial sums of the reciprocals of the primes eventually exceed any integer value, they never equal an integer. One proof [6] is by induction: The first partial sum is 1 / 2 , which has the form odd / even . If the n th partial sum (for n ≥ 1) has the form odd / even , then the (n + 1) st sum is