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Numbers n such that [sum_i=1..r (p(i)^2)]/r = c^2. p(i) prime divisors of n, c integer.
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%I #21 Mar 03 2015 10:45:02

%S 2,3,4,5,7,8,9,11,13,16,17,19,23,25,27,29,31,32,37,41,43,47,49,53,59,

%T 61,64,67,71,73,79,81,83,89,97,101,103,107,109,113,119,121,125,127,

%U 128,131,137,139,149,151,157,161,163,167,169,173,179,181,191,193,197,199,211

%N Numbers n such that [sum_i=1..r (p(i)^2)]/r = c^2. p(i) prime divisors of n, c integer.

%C A005063(n)/A001221(n) = c^2.

%C Also numbers n such that the root mean square (quadratic mean) of the prime divisors of n is an integer.

%C These numbers are power of primes (p^k with k>=1) (A000961) or in A255580. - _Daniel Lignon_, Feb 26 2015

%H Daniel Lignon, <a href="/A144711/b144711.txt">Table of n, a(n) for n = 1..1000</a>

%p A005063 := proc(n) add(p^2,p=numtheory[factorset](n)) ; end: A001221 := proc(n) nops(numtheory[factorset](n)) ; end: isA144711 := proc(n) local sofpr ; sofpr := A001221(n) ; if sofpr <> 0 then issqr(A005063(n)/sofpr) ; else false ; fi; end: for n from 1 to 500 do if isA144711(n) then printf("%d,",n) ; fi; od: # _R. J. Mathar_, Sep 20 2008

%t Select[Range[2,1000],IntegerQ[RootMeanSquare[Select[Divisors[#],PrimeQ]]]&] (* _Daniel Lignon_, Feb 26 2015 *)

%Y Cf. A005063, A001221, A140480.

%K easy,nonn

%O 1,1

%A _Ctibor O. Zizka_, Sep 19 2008

%E More terms from _R. J. Mathar_, Sep 20 2008