Let a primeven be a positive integer that is the product of an even number of primes. Let a primeodd be a positive integer that is the product of an odd number of primes. Then, 1 is a primeven because it is the product of 0 primes. 2 is a primeodd because it is the product of 1 prime. 3 is a primeodd because it is the product of 1 prime. 4 is a primeven because it is the product of 2 primes. Here are the first 10 positive integers.
Number: Factorization: Number of primes: Type:
1 0 primeven
2 2 1 primeodd
3 3 1 primeodd
4 2*2 2 primeven
5 5 1 primeodd
6 2*3 2 primeven
7 7 1 primeodd
8 2*2*2 3 primeodd
9 3*3 2 primeven
10 2*5 2 primeven
Suppose the primevens and primeodds had a race. First, the primevens would be ahead because 1 is a primeven. Then, there would be a tie because 2 is a primeodd. Then, the primeodds would be ahead because 3 is a primeodd. Then, there would be a tie because 4 is a primeven. Here are the winners from 1 to 10.
Number: Type: Primevens: Primeodds: Winner:
1 primeven 1 0 primevens
2 primeodd 1 1 tie
3 primeodd 1 2 primeodds
4 primeven 2 2 tie
5 primeodd 2 3 primeodds
6 primeven 3 3 tie
7 primeodd 3 4 primeodds
8 primeodd 3 5 primeodds
9 primeven 4 5 primeodds
10 primeven 5 5 tie
The primevens were ahead at the start, but have not been ahead since then. Do the primevens ever become the winner again?
(In reply to
re: some stats extended by Charlie)
You just need one more power of 10.
https://en.wikipedia.org/wiki/Liouville_function
I doubt Haselgrove's proof is below D6.
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Posted by Jer
on 2016-01-10 20:18:57 |
re(2): some stats extended
