Sum of squares of prime factors

[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!

 

[Eynstone]

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

 

[Eynstone]

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.

 

[Guffel]

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 ).

 

[Guffel]

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.

 

[Guffel]

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

 

[Wim Nobel]

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.

 

[Guffel]

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?

 

[cubzar]

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=a^x and a^x-2-x=0. x<=4 since a>=2 (smallest prime number) and 2^5-2-5=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).