# RMS - sequence

Ctibor.ZIZKA Ctibor.ZIZKA at seznam.cz
Sat Jun 28 20:40:34 CEST 2008

```Dear all,

could the following sequence be of interest for OEIS ?  :

RMS numbers :  numbers n such that root mean square of divisors of n is an
integer .

The sequence starts with 1, 7, 41, .

Some examples :

1)

Numbers n  with 2 divisors. We know n must be prime, and we have to count
RMS for divisors of p  : (1, p).

RMS(n = 1*p)  = sqrt ( (1+p^2)/2 ) is a positive integer.

In this case the Root mean square is an integer  for primes 1,7,41,239,.
which are  primes in A002315 (prime NSW numbers).

The value of RMS(n=1*p) is then  1,5,29,169,. which is A(001653).

RMS(1,a1), a1 positive integer.

Then a1 for which RMS(1,a1) is an integer is from A002315 and the RMS(1,A1)
is A(001653).

2)

Numbers n with 3 divisors. We know n must be of the form p*p which gives
divisors (1, p, p*p)  :

RMS(n=1*p*p) = sqrt ( (1+p^2+p^4)/2 )  is a positive integer.

I do not know any p for which this equation holds.

Looking generaly on RMS (a1, a2, a3) , a1,a2,a3 positive integers  :

RMS(1,1,a3) is a positive integer for a3 from  (1,5,19,71,265,.) which is
A001834

and the integer value of RMS(1,1,a3) is then 1,3,11,41,153,. which is
A001835

RMS(1,a2,a3) has infinetely many integer solutions and interesting behavior
for sequences of a3 values with different a2 values.

3)

Numbers with 4 divisors . There are 2 possibilities

n = p*p*p with divisors (1, p, p*p, p*p*p)

n = 1*p1*p2 with divisors (1, p1, p2, p1*p2)

Again, I do not know if there is any solution for those n.

General case  RMS(a1, a2, a3, a4), a1, a2, a3, a4 positive integers has
solutions.

etc.

Articles I could recommend (but only for RMS(1, a1)  :

Aviezri S.Fraenkel - On the recurrence f(m+1)=b(m)*f(m)-f(m-1) and
applications

Newman,Shanks,Williams - Simple groups of square order and an interesting
sequence of primes.

Best Regards,

Ctibor O. ZIZKA

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