Posted 6 Mar 2014 currently i'm studying the sequence 1 2 3 5 7 8 11 12 13 17 19 20 23 27 28 29 30 31.... which is the odd product of prime numbers. i'm curious to know if there is a good counting function for the number of them below say 10,000,000. if so, can it be reversed to find approximately the nth odd product? 0 Share this post Link to post Share on other sites

0 Posted 6 Mar 2014 phil1882, 1) The sequence starts off with "2," not "1," because 1 is not a prime number. 2) There are commas between the numbers in the sequence and a comma before the ellipsis. 3) You are missing the number 18. 18 = 2*3*3 (The number of prime factors is odd.) Therefore, the sequence is: 2, 3, 5, 7, 8, 11, 12, 13, 17, 18, 19, 20, 23, 27, 28, 29, 30, 31, ... 0 Share this post Link to post Share on other sites

0 Posted 6 Mar 2014 When you're looking at large numbers, I would guess they'll usually have a large number of prime factors and would probably be equally likely to have an even as an odd number of factors. If you have reasonably efficient code to calculate factors, you could try counting them for say 50 consecutive large numbers to get a sense of whether or not they're equally likely to be odd as even.If so, the counting function's approximation f(n) ~= number of terms less than n with a odd number of prime factors ~= n/2. And the n-th odd product would be roughly 2n. 0 Share this post Link to post Share on other sites

Posted

currently i'm studying the sequence

1 2 3 5 7 8 11 12 13 17 19 20 23 27 28 29 30 31....

which is the odd product of prime numbers.

i'm curious to know if there is a good counting function for the number of them below say 10,000,000.

if so, can it be reversed to find approximately the nth odd product?

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