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Is there a bound for the number of products of powers of a set of prime numbers less than a given number n? For instance, if I am given the set of prime numbers {2,7,11,13} and I am only interested in products of powers of primes less than 32, then there are twelve products of powers of these primes numbers: $2=2^1$, $4=2^2$, $7=7^1$, $8=2^3$, $11=11^1$, $13=13^1$, $14=2^1\cdot 7^1$, $16=2^4$, $22=2^1\cdot 11^1$, $26=2^1\cdot 13^1$, $28=2^2\cdot 7^1$, and $32=2^5$.

I know people have asked about estimating the number of products of primes less than n and estimating the number of prime powers less than n. I am looking for a combination of these two questions.

John Asplund
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  • N is given, are the primes given as well?Or you mean ANY set of primes under n? – vanmeri Dec 05 '18 at 13:07
  • If the primes aren't given, then you would end up knowing quantity of primes under every natural! – vanmeri Dec 05 '18 at 13:09
  • I guess I was not clear enough. If I am given a set of prime numbers as I did above {2,7,11,13}, how many numbers are a product of powers of those prime numbers? In the example above, there are twelve. I will edit the problem to reflect my original intention. – John Asplund Dec 05 '18 at 17:31

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I thougth of this : you want to maximize the function Z = 2$^a$ + 7$^b$ + 11$^c$ + 13$^d$ , whenever z $\leq$ n. You may "linearize" this function by taking natural logarithm. By means of linear simplex method, you may find the máxima... Then you may try and analize which 4 - tuples (a,b,c,d) with integer coordinates are nearest the máxima found . Since exponentiating and taking logarithm are continous, taking "nearby" tuples should work.

I will try to find a more combinatorial answer, though. It is a very interesting question.

vanmeri
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