Subject: Fat shredding techniques famous actors and actresses use.

From: Weight Loss Tricks&Tips <[email-masked]>
Date: Thu, 05 Feb 2015 09:07:39 -0800

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Djikstra's algorithm on the 'graph' of possible exponents  (and iteratively increasing them) and ing each 'node' (list of exponents) by its product value (N).

Here's w the above algorithm works:

Start with the smallest valid number and the smallest list of exponents (all zero) for our primes: N = 1 = 20 * 30 * ... * 190
Create a list of exponents by adding 1 to each exponent by index [(0+1, 0, 0, 

guarentee that I never double count, and that every single number I count has the correct factors. (I could also write a test for "didn't miss any" if you like, but I obviously cannot do that reasonably up to the full million).
You don't need a "did not miss any" test. (maybe I do)

Altugh I see that your code attempts to do it, can you confirm that numbers such as 46 or 678, 682 are not in the list?

our lists up to 4000th suld diverge already.
Here are the first 10,000 numbers. Cheers
I get the same list.  I reran my code and it turns out that we both have the correct answer at 1M.

#  H2 24807446830080; p: i.8
1000000

I forgot to include 17 in my original prime list, and was getting wrong answer due to that.
Pytn 2, takes around 35 seconds:

import sys
from operator import itemgetter
from collections import defaultdict
import heapq

primes = (2,3,5,7,11,13,17,19)
powers = tuple(0 for i in primes)

def compute(primes, powers):
    result = 1
    for i in xrange(len(primes)):
        result *= primes[i] ** powers[i]
    return result

def memoize_unique(fn):
   
haven't done the problem yet, but the one you got seems way too small
I did some reading on these types of number, and it looks like there's a pretty good formula for estimating w many there are.

The nth number that is not divisible by any prime greater than 20 suld be slightly greater than

exp(pow(5000000.0*n, 1.0/8)) / 3114
where five million ~ 8 * ln2*ln3...*ln19
and 3114 ~ sqrt(2*3...*19)

Pytn

#/usr/bin/pytn

results = [1]

def value_gen(n):
    index, value = 0, n
    while True:
        yield value
        if value == results[-1]:
            index += 1
            value = n * results[index]

gens = [value_gen(primes) for primes in (2,3,5,7,11,13,17,19)]

for i in xrange(999999):
    results.append(min(gen.next() for gen in gens))

print results[-1]

Algorithmically its just 
       Primes.Add(NextPrime(Primes.Max));

        static long NextPrime(long n)
        
            return Series(n + 1, i => i + 1).First(IsPrime);

really, Voyager is a dramatic step down in quality from the first three sws. To go from watching DS9 to VOY could give anyone vertigo. Of course, TV sws have their ups and downs - but you're stating a truism. Voyager had so many more downs integral to the sw and its characters as to make a qualitative difference.

You cannot watch a good episode of TOS, TNG (early seasons aside) or DS9 and accuse the writers of laziness or lack of vision, even if they misfire. Yet even a great episode of Voyager still bears witness to all sorts of stupidities in characterisation and so forth that is just at an unacceptable level. That is why Tim Lynch refused to  any more VOY past a certain point, in despair at w atrocious it was. VOY wasn't just a little worse. It sucked.
Yes, you can. It all comes down to w one 
return 0;

A fast and simpler J version

creates a list of the numbers that are below a set value (0 X). 1 X is prime factors 

X =: 1 : 'm &::@:['

  # a=. (11830 ; 2 3 5 7 11 13 19) (0 X ([: ~. ] #~ >:) 1 X (] , [: , */) ])(^:_) 1
  1000
  _10 . /:~ a
 11550 11552 11583 11616 11648 11664 11700 11704 11760 11830
   as a named function
 H =: (0 X ([: ~. ] #~ >:) 1 X (] , [: , */) ])(^:_)&1

 # H 24807446830080; p: i.8
  1000000   NB. # 24807446830080 is 1Mth numbr

If you forget 17 in the prime list then 1Mth number is: (10 seconds or so)

  999999 ( /:~)  H 3876855354725535 ; 2 3 5 7 11 13 19
  1945430938091520
  q: 1945430938091520  
  2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 
x;
    return 0;

unsigned long long SmoothNumber(int term)

    int powers[] =  2, 3, 5, 7, 11, 13, 17, 19 ;
    unsigned long long calcs[] =  2, 3, 5, 7, 11, 13, 17, 19 ;
    unsigned  indices[] =  0, 0, 0, 0, 0, 0, 0, 0 ;
    vector<unsigned long long> smoothNums (term, 0);

    smoothNums[0] = 1;

    for (int i = 1; i < term; i++)
    
        smoothNums[i] = SmallestInt(calcs, (sizeof(calcs) / sizeof(unsigned long long)));
        if (smoothNums[i] ==