

A140480


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


45



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, 1069895, 1087009, 1166399, 1204281, 1256489
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OFFSET

1,2


COMMENTS

For any numbers, A and B, both appearing in the sequence, if gcd(A,B)=1, then A*B is also in the sequence.  Andrew Weimholt, Jul 01 2008
The primes in this sequence are the NSW primes (A088165). For the terms less than 2^31, the only powers greater than 1 appearing in the prime factorization of numbers are 3^3 and 13^2. It appears that all terms are +1 (mod 8). See A224988 for even numbers.  T. D. Noe, Jul 06 2008, Apr 25 2013
A basis for this sequence is given by A002315. This can be considered as the convergents of quasiregular continued fractions or a special 6ary numeration system (see A. S. Fraenkel) which gives the characterization of positions of some heap or Wythoff game. What is the SpragueGrundy function of this game?
Sequence generalized: sigma_rnumbers are numbers n for which sigma_r(n)/sigma_0(n) = c^r. Sigma_r(n) denotes sum of rth powers of divisors of n; c,r positive integers. This sequence are sigma_2numbers, A003601 are sigma_1numbers. In a weaker form we have sigma_r(n)/sigma_0(n) = c^t; t is an integer from <1,r>.  Ctibor O. Zizka, Jul 14 2008
The primes in this sequence are prime numerators with an odd index in A001333. The RMS values (A141812) of prime RMS numbers (this sequence) are prime Pell numbers (A000129) with an odd index.  Ctibor O. Zizka, Aug 13 2008
From Ctibor O. Zizka, Aug 30 2008: (Start)
The set of RMS numbers n could be split into subsets according to the number and form of divisors of n. By definition, RMS(n) = sqrt(sigma_2(n) / sigma_0(n)) should be an integer. Now consider some examples. For n prime number, n has 2 divisors [1,n] and we have to solve Pell's equation n^2 = 2*C^2  1; C positive integer. The solution is a prime n of the form u(i) = 6*u(i1)  u(i2), i >= 2, u(0)=1, u(1)=7, known as an NSW prime (A088165). For n = p_1*p_2, p_1 and p_2 primes, n has 4 divisors {1; p_1; p_2; p_1*p_2}. There are 2 possible cases. Firstly p^2 = (2*C)^2  1 which does not hold for any prime p; secondly p_1^2 = 2*C_1^2  1 and p_2^2 = 2*C_2^2  1; C_1 and C_2 positive integers.
The solution is that p_1 and p_2 are different NSW primes. If n = p^3, divisors of n are {1; p; p^2; p^3} and we have to solve the Diophantine equation (p^8  1)/(p  1) = (2*C)^2. This equation has no solution for any prime p. RMS numbers n with 4 divisors are only of the form n = p_1*p_2, with p_1 and p_2 NSW primes. The general case is n = p_1*...*p_t, n has 2^t divisors, and for t >= 3, NSW primes are not the only solution. If some of the prime divisors are equals p_i = p_j = ... = p_k, the general case n = p_1*...*p_t is "degenerate" because of the multiplicity of prime factors and therefore n has fewer than 2^t divisors. (End)
General recurrence is a(n) = (a(1)1)*a(n1)  a(n2), a(1) >= 4, lim_{n>infinity} a(n) = x*(k*x+1)^n, k = a(1)  3, x = (1 + sqrt((a(1)+1)/(a(1)3)))/2. Examples in OEIS: a(1)=4 gives A002878, whose prime terms give A121534. a(1)=5 gives A001834, whose prime terms give A086386. a(1)=6 gives A030221, whose prime terms {29, 139, 3191, ...} are not a sequence on the OEIS. a(1)=7 gives A002315, whose prime terms give A088165. a(1)=8 gives A033890; the OEIS does not have its prime terms as a sequence (do there exist any prime terms?). a(1)=9 gives A057080, whose prime terms {71, 34649, 16908641, ...} are not a sequence in the OEIS. a(1)=10 gives A057081, whose prime terms {389806471, 192097408520951, ...} are not a sequence in the OEIS.  Ctibor O. Zizka, Sep 02 2008
16 of the first 1660 terms are even (the smallest is 2217231104). The first 16 even terms are all divisible by 30976.  Donovan Johnson, Apr 16 2013
All the 83 even terms up to 10^13 (see A224988) are divisible by 30976.  Giovanni Resta, Oct 29 2019


LINKS

Giovanni Resta, Table of n, a(n) for n = 1..7430 (terms < 10^13, first 455 terms from T. D. Noe, terms 456..1660 from Donovan Johnson)
A. S. Fraenkel, Heap games, numeration systems and sequences, arXiv:math/9809074 [math.CO], 1998; Annals of Combinatorics, 2 (1998), 197210.
Aviezri S. Fraenkel, On the recurrence f(m+1)= b(m)*f(m)f(m1) and applications, Discrete Mathematics 224 (2000), pp. 273279.
H. W. Lenstra Jr., Solving the Pell Equation, Notices of the AMS, Vol.49, No.2, Feb. 2002, p.182192. [Ctibor O. Zizka, Aug 30 2008]
Eric Weisstein's World of Math, Root Mean Square


MAPLE

with(numtheory); List140480:=proc(q) local a, k, n;
for n from 1 to q do a:=divisors(n);
if type(sqrt(add(a[k]^2, k=1..nops(a))/tau(n)), integer) then print(n);
fi; od; end: List140480 (10^6); # Paolo P. Lava, Apr 11 2013


MATHEMATICA

rmsQ[n_] := IntegerQ[Sqrt[DivisorSigma[2, n]/DivisorSigma[0, n]]]; m = 160000; sel1 = Select[8*Range[0, m]+1, rmsQ]; sel7 = Select[8*Range[m]1, rmsQ]; Union[sel1, sel7] (* JeanFrançois Alcover, Aug 31 2011, after T. D. Noe's comment *)
Select[Range[1300000], IntegerQ[RootMeanSquare[Divisors[#]]]&] (* Harvey P. Dale, Mar 24 2016 *)


PROG

(Haskell)
a140480 n = a140480_list !! (n1)
a140480_list = filter
((== 1) . a010052 . (\x > a001157 x `div` a000005 x)) a020486_list
 Reinhard Zumkeller, Jan 15 2013


CROSSREFS

Cf. A002315, A001653, A001834, A001835, A001599, A000005, A000040, A003601, A010052, A001157, A020486, A158294, A224988.
Sequence in context: A026002 A173409 A057009 * A327055 A002315 A141813
Adjacent sequences: A140477 A140478 A140479 * A140481 A140482 A140483


KEYWORD

nonn,nice


AUTHOR

Ctibor O. Zizka, Jun 29 2008, Jul 11 2008


EXTENSIONS

More terms from T. D. Noe and Andrew Weimholt, Jul 01 2008


STATUS

approved



