Each x is prime with probability 1/ln(x), each x has M/x multiples less than M, as a fraction of M that is just 1/x. Together that makes 1/(x ln(x)) with the indefinite integral ln(ln(x)). If we plug in 2^32 and 2^64 [1], we get ln(2). So about 69.3 % of all 64 bit integers should have a prime factor larger than 2^32 and therefore not be the product of two 32 bit integers. That leaves about 13 % unaccounted. Three prime factors all larger than 2^32/2, five prime factors all larger than 2^32/3, and so on cannot be packed into two 32 bit integers. Not sure to how much this will add up.
[1] The bounds are important because they guarantee that there is at most one prime factor from that range and this ensures that we are not double counting anything. If the upper bound was larger than the square of the lower bound, then we would have to worry about double counting numbers with more than one large prime factor.
This is just looking at it from a different perspective. Both, one 64 bit integer and the product of two 32 bit integers represent a number up to 2^64 with 64 bits. But while all 64 bit integers are unique, there are, as you say, several representation for some numbers as the product of two 32 bit integers and therefore it is impossible to represent all 64 bit integers. Commutativity alone costs you about 50 percent of all numbers in the range as x * y and y * x represent the same 64 bit number but with two different representation as a product of two 32 bit numbers, at least if x and y are different. But this tells you nothing about the numbers that you can not represent, only that they must exist. I was looking at it from this other perspective, which numbers are not representable as a product and why.
But also all of their multiples. I suspect that those account for the vast majority.
Each x is prime with probability 1/ln(x), each x has M/x multiples less than M, as a fraction of M that is just 1/x. Together that makes 1/(x ln(x)) with the indefinite integral ln(ln(x)). If we plug in 2^32 and 2^64 [1], we get ln(2). So about 69.3 % of all 64 bit integers should have a prime factor larger than 2^32 and therefore not be the product of two 32 bit integers. That leaves about 13 % unaccounted. Three prime factors all larger than 2^32/2, five prime factors all larger than 2^32/3, and so on cannot be packed into two 32 bit integers. Not sure to how much this will add up.
[1] The bounds are important because they guarantee that there is at most one prime factor from that range and this ensures that we are not double counting anything. If the upper bound was larger than the square of the lower bound, then we would have to worry about double counting numbers with more than one large prime factor.
Nice work. I guess "vast majority" is overstating the case, but majority anyway.
The simple explanation is just that lot of ways that numbers can multiply to produce the same product
Like an odd number x times an even number y, x* y produces the same product as x* 2 and y/2
Same for a multiple of 3 c and and a non multiple of 3 d, c * d = c/3 * d*3
This is just looking at it from a different perspective. Both, one 64 bit integer and the product of two 32 bit integers represent a number up to 2^64 with 64 bits. But while all 64 bit integers are unique, there are, as you say, several representation for some numbers as the product of two 32 bit integers and therefore it is impossible to represent all 64 bit integers. Commutativity alone costs you about 50 percent of all numbers in the range as x * y and y * x represent the same 64 bit number but with two different representation as a product of two 32 bit numbers, at least if x and y are different. But this tells you nothing about the numbers that you can not represent, only that they must exist. I was looking at it from this other perspective, which numbers are not representable as a product and why.