vba - How can I find the first four consecutive numbers with four free denominators? -

the first 2 consecutive numbers 2 free denominators are:

14 = 2 × 7 15 = 3 × 5 

the first 3 consecutive numbers 3 free denominators are:

644 = 2 × 7 × 23 645 = 3 × 5 × 43 646 = 2 × 17 × 19 

how can find first 4 consecutive numbers 4 free denominators?

it isn't case 644 = 2 x 7 x 23. rather 644 = 2 x 2 x 7 x 23 -- looking numbers have distinct prime divisors which might repeated. in fact, 1309, 1310, 1311 first 3 numbers have 3 distinct non-repeated factors, , impossible 4 since run of 4 successive numbers have repeated factor of 4.

to solve problem of finding 4 successive numbers each of has 4 distinct (albeit possibly repeated) prime factors, can use modified version of sieve of eratosthenes, 1 keeps track of number of distinct divisors:

function sievek(n long, k long) long     'implements modified sieve of erasthones     'to return first of k successive numbers     'less or equal n, each of has     'k distinct prime factors.     'returns -1 if no such number exists      dim nums variant     dim long, p long     dim run long     dim limit long     dim primes variant      primes = array(2, 3, 5, 7, 11, 13) 'small primes     p = 1     = 0 k - 2         p = p * primes(i)     next     limit = int(1 + n / p)      redim nums(2 n) long     p = 2 'first prime     while p < limit         'mark subsequent multiples of p adding 1         = 2 * p n step p             nums(i) = nums(i) + 1         next         'find next p -- next 0         p = p + 1         while nums(p) <> 0             p = p + 1         loop     loop     'at stage, numbers marked.      'primes marked 0 , composites marked      'by number of distinct prime factors.     'check run of k ks     run = 0     = 2 n         if nums(i) = k             run = run + 1             if run = k 'we have winner!                 sievek = - k + 1                 exit function             end if         else             'reset run counter             run = 0         end if     next     sievek = -1 end function 

sievek(10^6,4) evaluates 134043 in less second. function doesn't yield factorization of or next 3 numbers, easy find. sievek(10^8,5) evaluates -1, there isn't run less 100 million of 5 consecutive numbers each of has 5 different prime factors.

remark: in earlier version of answer had logic flaw (which didn't prevent output being correct). namely -- had sieved square root of n, though if e.g. number m looks 2x3x5xp p prime m have 4 distinct prime factors though p might exceed square root of n. revised algorithm take care of possibility.