# Sum of squares of prime factors

• Guffel
In summary: The limit is definitely >0 because there are infinitely many n values where d_n<=0 (e.g. all perfect squares). To prove the exact limit, we can use the Prime Number Theorem to estimate the number of primes <n and then use this to estimate the number of solutions for d_n<=0. However, this method may not give an exact result, so further investigation is needed.In summary, the conversation is about the properties of the difference d_n between the sum of the squares of the prime factors of n and n itself. There are three values for n where d_n=0 (1, 16, and 27), and it is possible to prove that these are the only numbers with this
Guffel
I recently got interested in number theory and have been fiddling around with Scilab trying to find interesting things. I came across the following mildly interesting property, which I couldn't find much about on Google.

Form the difference $$d_n$$ between the sum of the squares of the prime factors of n and n itself. For example, 44 = (2)(2)(11) => d44 = 2^2+2^2+11^2 - 44 = 85.

I've made calculations up to n=1,000,000.

Questions

1. I only found three values for n where $$d_n=0$$ (1, 16 and 27). Are these the only numbers that have this property? Proof?

2. The ratio between the number of values having the property that $$d_n<=0$$ and n seems to grow asymptotically to approximately 0.265. Any way to find this limit (or to prove that it is > 0)?

3. The number series formed by all n having the property that $$d_n<=0$$ starts with 1, 16, 24, 27, 32, 36, 40 and is similar to A166319 for all the values listed on oeis.org. The definition of the series is related, but not similar, to the one in this post. Is it obvious that they are similar?

Any input is welcome!

I reckon 1. can be proved , as the gap between the sum of squares & product increases. I'll get back to this soon.

The average order of d_n seems to grow like the sum of primes, from which we can deduce the existence of the constant in 2.

Thanks for your input Eynstone! Looking forward to more feedback regarding 1 (and the rest too).

Some additional info about the calculations up to n = 1,000,000:

As stated in the OP d_n = 0 for n = 1, 16 and 27

|d_n| = 2 for n = 2, 45, 552 and 33,304

Then there's a bunch of numbers with |d_n| = 4, of which the largest is n = 982,802.

I don't know what to do with this (other than writing it here ).

Searching for related stuff on the internet indicates that fantasy is the limit when it comes to doing stuff with prime factors. I find it a bit peculiar that I haven't found anything regarding the observations in the OP. 1 and 2 should be provable without much trouble, no?

Just realized fwiw that all n with factors 2, p and p give d_n = 4, since d_n = 4+p^2+p^2 - 2p^2 = 4.

Sorry for my massive bumping, this will be the last one. Noone? Eynstone, did you get ever get back to it?

Guffel said:
3. The number series formed by all n having the property that $$d_n<=0$$ starts with 1, 16, 24, 27, 32, 36, 40 and is similar to A166319 for all the values listed on oeis.org. The definition of the series is related, but not similar, to the one in this post. Is it obvious that they are similar?

Yes, it's obvious, because in series A166319 it is allowed to incude 1's in the factorization. That means that you can make any sum of squares that is deficient (d_n<0) equal to the required number by adding an arbitrary number of 1's.
A166319 also allows non-prime factors, but this is just a shortcut. You can replace them by prime factors, so the sum will be even smaller, and then just add more 1's to restore equality.

Wim Nobel said:
Yes, it's obvious, because in series A166319 it is allowed to incude 1's in the factorization. That means that you can make any sum of squares that is deficient (d_n<0) equal to the required number by adding an arbitrary number of 1's.
A166319 also allows non-prime factors, but this is just a shortcut. You can replace them by prime factors, so the sum will be even smaller, and then just add more 1's to restore equality.
Thanks for your response, I realize now that the series are identical.

Ok, so number 3 is done. Any takers on 1 and 2?

1. If n=a.b.c...=a2+b2+c2 then a=b=c... (more explanation later when I have more time). Then if n has x prime factors, xa2=ax and ax-2-x=0. x<=4 since a>=2 (smallest prime number) and 25-2-5=3 and if a or x increases this answer increases. Then it only remains to check for solutions to all x values <=4, which are 3,3,3 and 2,2,2,2 (1 is not defined as a prime number).

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## 1. What is the definition of "Sum of squares of prime factors"?

The sum of squares of prime factors refers to the sum of the squares of all the prime numbers that can divide a given number without leaving a remainder.

## 2. Why is finding the sum of squares of prime factors important?

Finding the sum of squares of prime factors is important because it can help in understanding the factors of a number and can be used in solving various mathematical problems such as finding the prime factorization of a number or determining if a number is a perfect square or not.

## 3. How is the sum of squares of prime factors calculated?

The sum of squares of prime factors is calculated by first finding the prime factorization of the given number and then squaring each prime factor and adding them together. For example, the sum of squares of prime factors of 24 would be (2^2 + 3^2) = 13.

## 4. Can the sum of squares of prime factors be negative?

No, the sum of squares of prime factors cannot be negative as it involves squaring positive prime numbers.

## 5. What is the significance of the sum of squares of prime factors in number theory?

The sum of squares of prime factors has various applications in number theory, such as in proving theorems related to prime numbers and perfect squares. It is also used in algorithms for finding the greatest common divisor of two numbers.

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