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In number theory, a polite number is a positive integer that can be written as the sum of two or more consecutive positive integers. A positive integer which is not polite is called impolite . [ 1 ] [ 2 ] The impolite numbers are exactly the powers of two , and the polite numbers are the natural numbers that are not powers of two.
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; [17] the largest number now known not to be a sum of ...
The sum of four cubes problem [1] asks whether every integer is the sum of four cubes of integers. It is conjectured the answer is affirmative, but this conjecture has been neither proven nor disproven. [2] Some of the cubes may be negative numbers, in contrast to Waring's problem on sums of cubes, where they are required to be positive.
In number theory, a Smith number is a composite number for which, in a given number base, the sum of its digits is equal to the sum of the digits in its prime factorization in the same base. In the case of numbers that are not square-free , the factorization is written without exponents, writing the repeated factor as many times as needed.
The two smallest consecutive powerful numbers are 8 and 9. Since Pell's equation x 2 − 8y 2 = 1 has infinitely many integral solutions, there are infinitely many pairs of consecutive powerful numbers (Golomb, 1970); more generally, one can find consecutive powerful numbers by solving a similar Pell equation x 2 − ny 2 = ±1 for any perfect ...
The number 19 is not a harshad number in base 10, because the sum of the digits 1 and 9 is 10, and 19 is not divisible by 10. In base 10, every natural number expressible in the form 9R n a n , where the number R n consists of n copies of the single digit 1, n > 0, and a n is a positive integer less than 10 n and multiple of n , is a harshad ...
Apart from 1 + 2 = 3 any subsequent Ulam number cannot be the sum of its two prior consecutive Ulam numbers. Proof: Assume that for n > 2, U n−1 + U n = U n+1 is the required sum in only one way; then so does U n−2 + U n produce a sum in only one way, and it falls between U n and U n+1.
The sum of the reciprocals of the powerful numbers is close to 1.9436 . [4] The reciprocals of the factorials sum to the transcendental number e (one of two constants called "Euler's number"). The sum of the reciprocals of the square numbers (the Basel problem) is the transcendental number π 2 / 6 , or ζ(2) where ζ is the Riemann zeta ...