RMS - sequence

[Andrew Weimholt]

Minor Correction...
On the last line, where I said..

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

I meant to say...

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

Andrew

On 7/1/08, Andrew Weimholt <andrew at weimholt.com> wrote:
> 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
>