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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.
The number of representations of a natural number n as the sum of four squares of integers is denoted by r 4 (n). Jacobi's four-square theorem states that this is eight times the sum of the divisors of n if n is odd and 24 times the sum of the odd divisors of n if n is even (see divisor function ), i.e.
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 ...
Lemma: The sum of any non-empty set of distinct, non-consecutive Fibonacci numbers whose largest member is F j is strictly less than the next larger Fibonacci number F j + 1 . The lemma can be proven by induction on j. Now take two non-empty sets and of distinct non-consecutive Fibonacci numbers which have the same sum, =.
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. This contradicts the condition that U n+1 is the next smallest Ulam number. [5] For n > 2, any three consecutive Ulam numbers (U n−1, U n, U n+1) as integer ...
With the advent of computers, many more values of n have been checked; T. Oliveira e Silva ran a distributed computer search that has verified the conjecture for n ≤ 4 × 10 18 (and double-checked up to 4 × 10 17) as of 2013. One record from this search is that 3 325 581 707 333 960 528 is the smallest number that cannot be written as a sum ...