smooth-numbers

So the problem is: Write a program to find the nth super ugly number. Super ugly numbers are positive numbers whose all prime factors are in the given prime list primes of size k. For example, [1, 2, 4, 7, 8, 13, 14, 16, 19, 26, 28, 32] is the sequence of the first 12 super ugly numbers given primes = [2, 7, 13, 19] of size 4. So my algorithm basically finds all possible factors using the pattern they follow, pushes them to an array, sorts that array and then returns the nth value in the array. It accurately calculates all of them, however, is too slow with high nth values. My question is what

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