hamming-numbers

Display first N natural numbers, the divisors of which are only 2, 3 and 7. I wrote something like that. I am a beginner in Lisp. Thank you! defvar x 1 (defun numbers(n) if(mod x 2 ) (loop for x from 1 to n do(print x) ) ) print(numbers()) Because I just had some time, you could have a look at this. Might not be the perfect solution but should be a good starting point for a beginner. Check out the books in the info tab to get into the syntax etc. (defun divisible-by (n m) "Returns T if N is evenly divisible by M." (zerop (mod n m))) (defun numbers (n) "Print all number upto N which are

Regular numbers are numbers that evenly divide powers of 60. As an example, 60 2 = 3600 = 48 × 75, so both 48 and 75 are divisors of a power of 60. Thus, they are also regular numbers. This is an extension of rounding up to the next power of two . I have an integer value N which may contain large prime factors and I want to round it up to a number composed of only small prime factors (2, 3 and 5) Examples: f(18) == 18 == 2 1 * 3 2 f(19) == 20 == 2 2 * 5 1 f(257) == 270 == 2 1 * 3 3 * 5 1 What would be an efficient way to find the smallest number satisfying this requirement? The values involved

A friend of mine is interviewing for a job. One of the interview questions got me thinking, just wanted some feedback. There are 2 non-negative integers: i and j. Given the following equation, find an (optimal) solution to iterate over i and j in such a way that the output is sorted. 2^i * 5^j So the first few rounds would look like this: 2^0 * 5^0 = 1 2^1 * 5^0 = 2 2^2 * 5^0 = 4 2^0 * 5^1 = 5 2^3 * 5^0 = 8 2^1 * 5^1 = 10 2^4 * 5^0 = 16 2^2 * 5^1 = 20 2^0 * 5^2 = 25 Try as I might, I can't see a pattern. Your thoughts? user515430 Dijkstra derives an eloquent solution in "A Discipline of

Is this the 1 billionth ugly/hamming number? 62565096724471903888424537973014890491686968126921250076541212862080934425144389 76692222667734743108165348546009548371249535465997230641841310549077830079108427 08520497989078343041081429889246063472775181069303596625038985214292236784430583 66046734494015674435358781857279355148950650629382822451696203426871312216858487 7816068576714140173718 Does anyone have code to share that can verify this? Thanks! Will Ness this SO answer shows code capable of calculating it. the test entry on ideone.com takes 1.1 0.05 sec for 10 9 (2016-08-18: main speedup

I'm working on a problem that asks to generate a sequence using prime numbers 2, 3, and 5, and then displaying then nth number in the sequence. So, if I ask the program to display the 1000th number, it should display it. I can't be using arrays or anything like that, just basic decisions and loops. I started working on it and hit a wall... here's what I got: #include <iostream> using namespace std; int main() { unsigned int n=23; for(int i=2; i<n; i++){ if(i%2==0){ cout<<i<<", "; }else if(i%3==0){ cout<<i<<", "; }else if(i%5==0){ cout<<i<<", "; } } return 0; } Unfortunately, that code doesn't