OFFSET
1,1
COMMENTS
The intention was to generate a sequence of uninteresting numbers. - John R Phelan, Dec 01 2014
LINKS
Wikipedia, Complement (set theory)
Wikipedia, Interesting number paradox
FORMULA
A \ B represents set "subtraction", all the elements in A that are not in B.
In other words, start with the Natural numbers (A000027).
Remove the prime numbers (A000040).
Remove the 11-smooth numbers, numbers whose prime divisors are all <= 11 (A051038).
Remove the base-10 palindromes (A002113).
Remove the perfect powers, m^k where m > 0 and k >= 2 (A001597).
And what's left is this sequence.
a(n) ~ n; in particular, a(n) = n + n/log n + o(n/log n). - Charles R Greathouse IV, Nov 27 2013
EXAMPLE
16 is not in the sequence since it's a perfect power, 2^4.
19 is not in the sequence since it's prime.
18 is not in the sequence since it's 2*3*3, so it's 11-smooth.
22 is not in the sequence since it's a base 10 palindrome.
26 is in the sequence since it's 2*13, so it's not prime, not 11-smooth, not a base-10 palindrome, and not a perfect power.
PROG
(Java) public class Nnn {public static void main(String[] args) {String str = ""; for (int i = 0; i < 1000000 && str.length() < 250; i++) {if (isPrime(i) || isSmooth(11, i) || isPerfectPower(i) || isPalindrome(i)) {} else {str += i + ", "; }} System.out.println(str); } static boolean isPalindrome(int i) {return ((i+"").equals(new StringBuilder(i+"").reverse().toString())); } static boolean isSmooth(int s, int n) {if (n<2) return true; for (int i=2; i<=s; i++) {while (n%i==0) n=n/i; } return n==1; } static boolean isPerfectPower(int n) {for (int i=2; i<=Math.sqrt(n); i++) {int j=i*i; while (j<n) j*=i; if (j==n) return true; } return false; } static boolean isPrime(int n) {if (n<2) return false; for (int i=2; i<=Math.sqrt(n); i++) {if (n%i==0) return false; } return true; }}
CROSSREFS
KEYWORD
base,easy,nonn
AUTHOR
John R Phelan, Nov 27 2013
STATUS
approved