RMS - sequence

[Andrew Weimholt]

I took a closer look at Ctibor's RMS Numbers sequence A140480, and
found quite a few new terms.

1, 7, 41, 239, 287, 1673, 3055, 6665, 9545, 9799, 9855, 21385,
26095, 34697, 46655, 66815, 68593, 68985, 125255, 155287, 182665,
242879, 273265, 380511, 391345, 404055, 421655, 627215, 730145,
814463, 823537, 876785

Many of them are products of terms occurring previously in the sequence.

In fact, for any numbers, A and B, both appearing in the sequence,
if GCD(A,B)=1, then A*B is also in the sequence.

To demonstrate this, I'll start with the simplest case, in which both
factors are prime...

- If p is a term in the sequence, then p^2 + 1 = 2x^2 for some integer, x.
- Similarly, if q is also a term, then q^2 + 1 = 2y^2 for some integer, y.

- The divisors of pq are 1, p, q, pq (assuming p != q).
- The sum of the squares of the divisors of pq is

 1 + p^2 + q^2 + (pq)^2

- This can be re-written as

 (p^2 + 1)(q^2 + 1)

- and re-writtem yet again as

 (2x^2)(2y^2)  =  4(xy)^2

- Since we have 4 divisors, the mean square is (xy)^2

- and the Root Mean Square (RMS) is xy, an integer.

- Therefore pq belongs in the sequence.

Now consider the product of three distinct primes in the sequence...

- Again, the sum of the squares of the divisors of pqr can be written as

 (p^2 + 1)(q^2 + 1)(r^2 + 1)

 = (2x^2)(2y^2)(2z^2)

 = 8 (xyz)^2

- And since we have 8 divisors, the RMS is simply xyz (again an integer).

The number of divisors for a square-free integer with k prime factors is 2^k,
so this works, no matter how many distinct prime factors we take from
the sequence.

But what about terms in the sequence, which have prime factors not in
the sequence?

We can use them as well to generate new terms by multiplying them with
other terms from the sequence, as long as the terms being multiplied
have no common prime factors.

For now, let's assume the terms are also square free...

- Let P = p_1 * p_2 * ... * p_k,  where p_1 thru p_k are prime.
- Let Q = q_1 * q_2 * ... * q_m,  where q_1 thru q_m are prime.

- then the sum of the squares of P's divisors is

 ((p_1)^2 + 1)((p_2)^2 + 1)...((p_k)^2 + 1)

- And if P happens to be in the sequence, then

 ((p_1)^2 + 1)((p_2)^2 + 1)...((p_k)^2 + 1) = 2^k * x^2, for some interger, x.

- Similary, if Q is in the sequence, then

 ((q_1)^2 + 1)((q_2)^2 + 1)...((q_m)^2 + 1) = 2^m * y^2, for some interger, y.

- If the product, PQ, is also square-free,
 then it will have 2^(m*k) divisors,
 and the sum of squares of its divisors will be

 2^(m*k) (xy)^2

- Again, the RMS of the product, PQ, is simply xy, an integer, so PQ
also belongs in the sequence.

Now what about terms in the sequence that are NOT square-free?
Can we multiply them with other terms in the sequence to get a product
also in the sequence?

Yes - as long as both terms have no common factors, it does not matter
whether or not they are square-free...

- Let P = (p_1)^(c_1) * (p_2)^(c_2) * ... * (p_k)^(c_k)  , with u divisors
- Let Q = (q_1)^(d_1) * (q_2)^(d_2) * ... * (q_m)^(d_m)  , with w divisors

- Then the sum of the squares of P's divisors is

 [(p_1)^(2 c_1) + (p_1)^(2(c_1-1)) + ... + (p_1)^2 + 1] *
 [(p_2)^(2 c_2) + (p_2)^(2(c_2-1)) + ... + (p_2)^2 + 1] *
 ...
 [(p_k)^(2 c_k) + (p_k)^(2(c_k-1)) + ... + (p_k)^2 + 1]

- and if P is in the sequence, this is equal to  u*x^2, for some integer x.

- and similarly for Q.

- Again, as long as GCD(P,Q)=1, the sum of the squares of the divisors
of PQ is just the product of the sums of squares of the divisors of P
and Q, so we get

 uw * (xy)^2

- And also because GCD(P,Q)=1, the number of divisors of PQ is just
product of the numbers of divisors of P and Q, so we get uw divisors.

- So once again we have

 RMS(PQ) = xy, an integer, and PQ belongs in the sequence.

Andrew