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Anonymous Sun 29 Nov 10:27:36 2020 No.12398635 View Reply Original Report

Quoted By:

Can anyone dumb this down for me please?
https://math.stackexchange.com/questions/58426/number-of-factors-less-than-a-number

This part I don't get:
"For every factor a of n2 such that a<n there is a factor b=n2/a that is greater than n... Thus, if it is "simple to find the total number of factors as a combinatorial sum", then just subtract one from this and divide by two."

Also would I need to do anything to this if I was looking for factors of 2n^2 < n instead? (as I am doing Project Euler #735)

Any pointers would be appreciated!

Anonymous

Anonymous Sun 29 Nov 2020 11:31:18 No.12398758 Report

Quoted By: >>12398809 >>12398826

In any pair of factors of a number, one will be less than the square root and the other greater. So the number of factors below the square root is the same as the number above it.
>2n^2 < n
The factors of 2n^2 are all the same as for n^2 with 2 and n^2 as well. It won't have factors between n and 2n.

Anonymous

Anonymous Sun 29 Nov 2020 12:04:29 No.12398809 Report

Quoted By:

>>12398758
He has almoost 10 answers for him in the comments

Anonymous

Anonymous Sun 29 Nov 2020 12:14:08 No.12398826 Report

Quoted By: >>12398940

>>12398758
Oh I see!
The function to find the number of factors <= n for 2n^2 is:
f(n) = round((2n^2 - 1) / 2) right?

but when doing this for 1 to 15, 72, 288, and 450 are off by one...

12 factors for 72 (n=6):
(12-1)/2 = 6

18 factors for 288 and 450 (n=12,15):
(18-1)/2 = 9

Anonymous

Anonymous Sun 29 Nov 2020 13:10:58 No.12398940 Report

Quoted By: >>12398957

>>12398826
>function to find the number of factors <= n for 2n^2 is:
>f(n) = round((2n^2 - 1) / 2) right?
No, that's just approximately n^2.
There isn't a function for number of factors. Think if n was a huge prime, only 2 and n would be factors of 2n^2.
Half the factors will be below the square root (n) and half above so if you knew the total number T of factors then the number below n would be (T-1)/2. You need to find T essentially by testing each (prime) number below n as factors.

Anonymous

Anonymous Sun 29 Nov 2020 13:18:44 No.12398957 Report

Quoted By:

>>12398940
To find the number of factors by finding the number of prime factors.

For 50, prime factors are 2^1 * 5^2. So the number of factors would be (1 + 1)(2 + 1).

help pls

help pls Sun 29 Nov 2020 16:49:50 No.12399593 Report

Quoted By: >>12400708

So I have to take the number of divisors for 72 (2*6^2) and then subtract one and then divide by two and then add one to include n. But this gives me:

(12 - 1)/2 + 1 = 6.5

but the 6th number in factors of 72 is 8...

Anonymous

Anonymous Sun 29 Nov 2020 21:37:22 No.12400708 Report

Quoted By:

>>12399593
You would only add or subtract 1 if it's a perfect square and depending on whether you're counting the square root or not.
>72
Factors of 72 are
1, 2, 3, 4, 6, 8, 9, 12, 18, 24, 36, 72.
sqrt(72)=8.5
12 in total and the number of factors below 8.5 is the same as the number above, 6 on each side. That's all the a<n, b>n stuff means. You don't need +/-1 because 72's not square.

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